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Find the area of the crescent-shaped region (called a $ lune $) bounded by arcs of circles with radii $ r $ and $ R $. (See the figure.)...

Question

Find the area of the crescent-shaped region (called a $ lune $) bounded by arcs of circles with radii $ r $ and $ R $. (See the figure.)

Find the area of the crescent-shaped region (called a $ lune $) bounded by arcs of circles with radii $ r $ and $ R $. (See the figure.)



Answers

Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii $r$ and $R$. (See the figure.) (FIGURE CAN'T COPY)

Let's find the area of the crescent shaped region, which we call a loon bounded by the arcs of the circles with radius. Little are and big are and here's the corresponding figure for convenience. We could go ahead and just positioned the larger circle with Radius Capital R so that its center at the origin. The nice thing about doing that is that you have the equation X squared plus y squared equals R Square. This represents the large circle of radius are on black And then from there, actually, we can go ahead because in this problem where basically well, we'Ll find the area between these two circles that will just be an area between two curves. So we'LL actually want a function for this upper half of the black circle. So just take the square root here, positive square root because around the upper half circle now for the small circle, it's not quite centered at the origin The X value zero, but we've gone up by some value here. Let's just do notice, five b So notice that b distance between the sensors of the circles now here, looking at the picture, we could actually use the fact that if we look at this right part of the blue circle, the right most part that this is also on the bicycle let me go ahead and just draw right triangle here. If this is our little R Well, since this points on the black circle this new segment right here, the high pop news has links are the radius of the big circle. And then I could use that diagram to find me B Square equals, Let me actually take a step back B Square plus little Arthur Claire equals big r squared by pit, agree a CE go ahead and sell for B So that's the distance between the two centers. So now let's write an expression for the area that we want. So here we liked the area between the two circles That will be everything that underneath the other blue circle, but also at the same time above the blacks. Terrible. So really hear what we'LL need is where we have equation for the upper half as a black circle, which I'm doing right now. This was we got this earlier by solving for why and the Black Circle. So this is the lower function, the function that will be subtracting. But we need the upper function to subtract from. So in this case, the upper function is just the upper half of this blue circle. So I get that by solving the equation for the small circle which I should have written over here. Let me come back to the small circle we found B. So the equation for the small circle is X squared. Plus why minus b square equals little are square because we shifted up by a little bit in the wind direction and then we can go ahead and simplify this using our value. Be so First, let's just go ahead and write this integral. The integral will be from negative art are Then we have the upper circle, so the other and blue, the blue upper circle. So let's solve this small circle for why, and doing so, we should have B plus the square root of R squared minus x word. So this is the upper curve, and we know it B is equal to before I. Now let's just leave. It is B, but we're again way saw for be a few minutes ago. So This is the upper curve. And then I'm subtracting the curve in black and we saw that for why up here? So this is just this radical capital R squared minus little r squared no d x. So Edgar minus lower for the area between two curves. Let's make this a little easier by Instead of finding the theory of the entire loon, let's just find the area of the right half right hand side and then just multiplied by two. So instead of going from negative, our two are we'LL just go from zero are multiplied by two. Should be Sierra are there should be a two hour here. Silly, silly typo there and same expression over here. Same radicals. Let's go to the next page. I'm running out of room here. Oh, let's go ahead and distribute that integral to each of the three expressions So we end up with two. There are B dx two zero R little R squared minus X squared the X and then finally for the minus sign oh, capital R Square minus X square d x Now these last two hundred girls are very similar, so we could probably just go ahead and solve both of them at the same time, actually, But before we do that, let's do the simpler and a girl over here. So this and a girl of is just bx X is going from Sierra are, and so we just have to be our And then, if we want, we can go ahead and plug in the value of B. On the previous page, we saw that BEA was equal to the square root capital R squared minus R squared all inside their ankle. So that evaluates the first integral and the end points Now for the second and rule to avoid doing unnecessary repetition. Let's just really look at the integral a a squared minus x squared inside the radical. If we can evaluate this integral, then we can just plugin a equals little are and then plug in a equals big R and that also evaluate the other two hundred girls because they're so similar. So now let's ignore the ours for a moment and put in a day our tricks up for this in a rolling green, X equals a sign data. So D X is a co scientific data, and we know that this So this after the trigger. Natural trick substitution. We have a squared minus a square sine squared data. And then the ex, which was a co sign Taito. So I can pull a square outside the radical. I take the square room. I just get a Have another A over here. So give me a square and then I have one minus sign square on the inside and we know that one minus sign squared is co sign squared someone. We take the square room, which is have co sign left over another co sign over here. So we get co science where when you have ah, integral of a co signer assigned that has the even power. It will be best to use the half angle formula. So I pull out that one half, I get a squared over two and then I have, in a roll of data of one which is just data in the end, the integral of co sign tooth dater, which assigned to theta over too. And I guess here because I didn't put limits of integration for right now, let's just proceed. But way could ignore the sea when we actually evaluate these integral is at the end because these air definite intervals now coming back to this latest expression where we left off, we know this is equal to to scientific co signed data by the double angle formula so we could go ahead and cancel those tunes. We get a squared over two State A plus sign data, coastline data, plus he. And now finally, the last thing to do here. We're basically done, but we'd like to get things back in terms of X, so there will actually go ahead and use this trick. So draw the triangle, get everything back in terms of X, and then once we plug in little R or big are we could just go plug in our original limits. Zero's a little art. So let me go on to the next page, form our truth Substitution. We no sign of data is X ray, so let's draw that triangle. There's our theta so sine opposite over adjacent X over, eh? And we could use potash grand there up to find out opposite side nature that excuse me, that Jason Side a squared plus X squared is a square. So software age. So this is all due to the Pythagorean. Darryl. Now we have enough to say what? Co sign it. So we left off that one half a squared data plus Scientific co signed it on the previous speech. So now just go ahead and use the triangle. Here we have one half a square Seita. You could have get by just solving this equation up here for data. So take Sinan verse on the side. So here we have data equals sign Inverse X Over. Eh? Again, that's coming from this equation up here. Signed peoples ex operate. And then for the next term, we have signed data where we know that that's X operator. Let me go ahead and actually just distribute the one have a square. So that's the first term. And then we have for the next term, we have sign, which is X over, eh? And then we have co sign, which is h overlay Oh, each other and in our policy, and then just simplify. Here. All that's necessary is that cancel those a square and then to simplify this one more time. Yeah. Okay, so now is this is where, at the point, we're to evaluate those two in a girls that we were interested in. We want to go ahead. So this was from the previous page. Now we have to go ahead and take a to be a little R and eight to be big bar. And we could also go back to our usual limits of integration, which was zero little r So plugging this all in and going to the next page the first integral that we had was which was to be our And we also written that as two are radical, no r squared, minus little are square. So that was the first in a girl we already evaluated enough for the last two of the girls. These were the same in a girl. Just one had a little r one had big r so plugging in a equals r into our formula. We have our squared arc sine X over our plus ex radical and then a squared minus X squared becomes little R squared minus X squared. And this is our original and point zero r. So this is replacing the previous integral that represented this term here. This was coming from the two integral zero are radical R squared minus X squared. You might have noticed that there's the one half that we had in the previous formula. Those disappeared because we multiplied by this two year. So here we just plugged in a equals little R into our formula. And then now we go ahead and about and subtract the other in the girl. So then we have the minus. Then we have now its capital R squared. So a equals big are on this integral in this term sign in verse X over capital are this time plus X Basically the same term is last time in the second term that we just replaced. Little are with big are But this little are Here's the limit of integration is still our little R Now we're at the point where we could just go ahead and plug in R and zero into the second to terms. So this term out here we leave alone. This already been evaluated from zero dar. So we have to our route our square minus little are square. And then let's go ahead and plug in little r appear for X. So we have our squared off in the front and then we have sign in verse of our over our, which is just one. And actually we could even replace this. We can evaluate this. This is just pie over too Sign of fire or two is one, and pie or two is in the range of the signing Chris function. By definition, somebody race that, and then the next term will be plug in. Little are to the radical for X. We get squirt of Zero, which is zero. So there's nothing else to add. And then when I subtract, we plug in our equals. X equals zero. So then we have sign inverse of zero, which is era. And then we have a zero over here in front of the radical, so we just get zero again. So there's actually nothing else to write here. That's the second term, and then now we could do the same thing. But for the last term, so let's plug in the last one minus. So we have that capital R squared sign in verse, plugging that little are for X and then we also plug in. Little are for Exxon decide, and we'LL have a little R square room. Big R squared minus little r squared. And then when we plug in zero, we get signed inverse of zero, which is zero. And then we get a zero in front of the radical. So we get the Zero Kane. So this is our expression. And let's go ahead and simplify what we can. We see that here we have a two are radical. And then over here, we have a negative radical, so those could be combined. So we just have one left. So one R capital R squared minus R squared in the radical Plus we evaluated side inverse of one. That's a pie over to our square. And then we have minus capital are squares sign in, verse are over, little are over bigger. And this is our final answer. This is the area of the loon.

So if we want to find the area of this like, shaded blue region on top, here I'm going to do first is turn this, um, or not. Turn it. But place it on to this XY plane here, like this. So this is X. And then this Here's why. So this red equation we can represent by X squared plus y squared is equal to capital R squared. And then this blue one will first, let's just go ahead and call this distance right here. Uh, h we can figure out how to solve for that in a bit. Um, but the blue line is there. Going to be war center is going to be at zero age, so this would be X squared. Plus y minus. H squared is equal to little r squared. Okay, Okay. Now, in order for us to solve for what H is going to be, we can actually create a triangle right here. I'll draw that triangle of top here like this. So this is a church. This is our I didn't notice going from here to here. Well, this is just going to be a radius of Oregon, so that black line is just capital are. So to get h h is going to be equal to, um are actually. Well, let's first, just so we can use Pythagoras. And so this is a right triangle. It would be eight squared plus r squared is equal to capital r squared and we just attract over square root. So h is going to be capital r squared my little r squared square root. Okay, um, so we have all of this information now, I might just leave it as a judge because I don't want to actually have to write that out, but just kind of keep in mind that we found this value for each already. Okay, so now, um, we are interested in finding just that area there. So notice it's like if we cut out this and then we just did the area of the top minus the area of the bottom. So what we would want to do is first notice that if we erase this bottom portion of this, this is just the top half of our circle. So this top part, if we go ahead and solve for the positive route here, it would be that and then this red part of top is just the positive root of why as well. So let's go ahead and solve for Y in both cases. So first here in red, um, this would just give us why is equal to r squared minus X squared square rooted down here, this would give us why is it too little r squared, minus X squared, square rooted. But then remember, we're adding each of us would be plus age. Okay, so we have that now and our bounds of integration. Well, this is going to be negative, little r and this is going to be positive. Little are because again, that's just the radius we're working with. So this area is going to be the integral from negligible our to our of our top function, which is this blue one. So it would be square root of R squared, minus x squared, plus h and then we're going to do minus the integral of the function we have over here or not. The integral just r squared, minus x word that and then we'll integrate with respect to X. Okay, now, one thing to make note of or actually let me check out fast if y'all have learned this yet. Okay. Yeah. So I'm pretty sure, um, I'll have learned this, but notice that this here is an even function. So it's even. And so remember, what even means is this implies that, um so that implies that we would have f of negative X equal to f Quebec's. And if this is the case, we're integrating something from, like, negative beta A of F of x. Yes, this is just equal to two times the integral from zero to a of f of x dx. And I just want to use this here, uh, to make it a little bit easier so you can go in and check to make sure that this is even. But I'm just gonna go ahead and apply directly, so that's just gonna be zero or 902 times from zero to our of. So you have the square root of R squared minus x squared, plus h then minus the square root of capital R squared minus x word DX. OK, now, at this point, what we will need to end up doing is actually distributing this across everything. So let's go ahead and do that. So, um I mean, technically, I guess we already know how to integrate all of this because this is just an area of a circle of radius r r squared. And then that would just be H. But since we're learning about these tricks substitutions, um, we can just use that to kind of double check in the end. So this would be two times the integral from zero are of the square root of our square by six squared DX plus two times the entrepreneur of zero to our of H dx and then minus the interval, two times the interval of zero. Little are to the square root of capital R squared, minus x word. Or actually, I guess I should say this first one is at least that area. This will actually have to do, like a trick substitution to get. Uh, so now let's see, um, for this first one, actually break this up into, like, different intervals, So this would be 12 and then three. So for the first one, the substitution we're going to want to make is the following. So any time we have, like, something squared minus x squared. We're going to want to say X is equal to That's something so would be our And then we're going to use sign Fada. Okay, so then taking the derivative of this, this gives us DX is able to are co sign data of d theta for our differential as we're taking the derivative retrospective data. Uh, and now we have to go and plug everything in. So over here, this would be two times. So if I come over here and plug in zero into this, that's just going to be zero. And then if I go ahead or actually maybe I should show what I'm doing. So when X is equal to zero, then this would be zero is equal to our signed data. So we divide each side by our. So we get signed data and then sign is going to be equal to zero at data is equal to zero. So that's why are lower bound is zero. And now when X is able to are, we get our is a guitar sign of data and then we divide each side by our. So we get one is equal to science data, and then we no sign of data is going to be equal to one at high half so that our upper bound is now pie half. Okay, now we can go ahead and make our substitution. So this would be the square root of R squared minus. And then our substitution was our sign of data. And then d x, um, becomes our co sign data d say to. So now let's go ahead and clean all this up. Um, so we still have I'm gonna pull that are out. So the two are integral from zero to pi, half of So we have our squared minus r sine squared, which, if we use Pythagoras, remember cosine squared. Why was rounded up co sign square data plus signs. Where data is he going to one. Um so if we go ahead and multiply everything by our squared r squared r squared and then we subtract that over and then take the square root on each side, we would just get our co signed data is equal to R squared minus R squared sine squared, Data square rooted. So this is just going to be our Hussein theater, and then we have times cosine theta do you say to? So now we can go ahead and pull that other are out to the two R squared, integral from zero to pi half of co sine squared theta d theta. Yeah, and then co sine squared. We would need to go ahead and use the power reducing identity to integrate this. And so doing that should give us, um b one plus co sign of two theta all over to dictator. Uh, we can go in and pull this two out. So actually, those just cancel out and then we would just have the integral from zero to high half of one plus co sign two theta d theta. Now we go ahead and integrate these. So one is just going to be data co sign of tooth data. Well, first, that would be sign of tooth data. And then, if you did a use of we would need to actually just divide that by two. Then we evaluate this from zero to pi half. So this is going to give us r squared. So data is pie half. If I plug pie up into there, that would be signed up. I, which is zero plus zero and then minus r squared when we plug in zero, which is just zero. So we would end up with where this is just going to be pi r squared over two. So this first part, I just kind of write this below. So this is two pi r squared over two Now for the second part. Well, h is just a constant. So it would just be plus two times h two times H and R minus zero, which would just be to h r. And then for this third one. Let's go ahead and move that down. So for this third one, so two times integral from zero two little are of the square root of r squared the other way around, I mean, Oh, yeah, that's right. Capital R squared minus little X word DX. And so now we'll need to go ahead and make our substitution pretty much the same as before. But instead of little are now, will be capital are of side data, our capital, our time scientist. Um, So again, our differential is going to be capital are cosign theta d theta. And if we were to go ahead and plug our bounds in so it would be so. An ax is zero. This is going to give us zero is equal to capital or of science data, which again just as data zero. And now when x is equal to little are well, there's going to be r is equal to capital are of signs data. So we'd have to divide by our So the little r over capital r is equal to sine data And then we would just have to take our design on each side. So this is going to say theta is equal to park sign of little are over capital are and we know this is going to be defined since little are is strictly less than capital are, um because otherwise, if little R was larger than our sun would be undefined. But in this case, it doesn't matter. So let's go ahead and use all that to plug in over here. So this would be two times the integral of zero two. Actually, I'll just write that sine inverse sine inverse of little are over capital or and then this would become the square root of capital R squared, minus capital r sine of theta squared and then DX becomes capital are co signed data data and then similar to what we did before. This just becomes capital are of co side data so we can go ahead and pull that r squared out. It'll be too r squared, integral from zero decide and verse of little are over capital are and then that becomes co sign squared data. And, well, we already figured out what this is supposed to be, uh, here, so we can just go ahead and take this and kind of move it down. So I'm going to be a little bit lazy and also just so I don't actually accidentally plug in something wrong, I'm gonna go ahead and just replace this if capital R squared. Actually, we write that in blue to kind of show that that was the only difference, Really? So we worked really hard, and then we got to here, um, and then our bounds this instead should be signed. Inverse of little are over capital are all right. So now we can just go ahead and plug these in, and then we would be done. So when we plug in sine inverse first that should give us We have r squared and they do which is be signed Inverse of little are over capital are and then plus sign of two side and verse of little are over capital are all over to and then zero. If we plug this in well, that would just be zero again so we could leave it like this if we wanted, but I don't know if that doesn't look too pretty to me. So what I'm gonna do is use a couple of trig identities to make it look a little bit prettier. So remember, any time we have, like, sign of two theta, we can rewrite this ads to sign data Times Co sign data. So in this case, our data is sine inverse. So this so data is equal to sine inverse of little are over capital are and doing this would give us to so that would just be little are over capital R and then co sign of that. Well, let's actually draw a triangle really fast to figure out what that should be. So we have sine of theta is equal to the little r over capital are. And if we draw this. So capital R is our hypothesis. Little our is our height. And so then that means our heart pot news over here is just going to be r squared minus for a little r squared all the way around capital r squared. I don't know why it went that far, but yeah, capital R little r and then capital R squared minus little r squared square rooting so called science supposed to be adjacent. So square root of r squared minus the lost work over our pot news, which is our And so that should be what co sign is, uh, so we can go ahead and collect our ours. So this is going to be too little Are square root of r squared minus little r squared all over Capitol R squared. All right, so let me just get rid of that and then plug into little are square root of r squared my little R squared all over Capitol Square, and then also knows those twos they're cancel out. So, um, if we actually go ahead and distribute that r squared notice that would cancel out with that one and then we would just have r squared there. So let me clean that up because I just wrote a ton of stuff. So would be capital r squared times sine inverse of little are over capital are and then plus little are especially. What would I have left there? So just be little are square root of r squared minus r squared. Yeah, and so that would be our second interval. Now let's actually come back up here. I'm going to just pick all of this up and move it down to the very bottom. Okay, so now we just need to take this and then plug it in right here. So then this is going to be remember, it was supposed to be minus That would be minus R squared sign in verse minus sine inverse of our over capital R and then minus little are square root of R squared, minus little r squared. And now this should be our area. Actually, one thing I forgot, um, this h was capital R squared. Minus little are square, So yeah, let me go ahead and replace that really fast. So two. So this was supposed to be square root of r squared by this little r squared. And actually, what did I have here again? I think it was a little r. Let me check Real fast can. Yeah. So it would be little Are there? Uh oh. And actually notice We have two copies of this and one copy here. So those just simplified down to one copy, Actually. Glad I remember to plug that in because this makes it even a little bit more simple. So this cancels out with that. So let me write this out one last time. So it'll be Hi little r squared over two plus and then I'll move that are out front. So little are square root of capital r squared, minus little r squared and then minus capital r squared times sine inverse of little are over counter. And now this is going to be our area, actually blue again. So that is our area.

So if we want to find the area of this like, shaded blue region on top, here I'm going to do first is turn this, um, or not. Turn it. But place it on to this XY plane here, like this. So this is X. And then this Here's why. So this red equation we can represent by X squared plus y squared is equal to capital R squared. And then this blue one will first, let's just go ahead and call this distance right here. Uh, h we can figure out how to solve for that in a bit. Um, but the blue line is there. Going to be war center is going to be at zero age, so this would be X squared. Plus y minus. H squared is equal to little r squared. Okay, Okay. Now, in order for us to solve for what H is going to be, we can actually create a triangle right here. I'll draw that triangle of top here like this. So this is a church. This is our I didn't notice going from here to here. Well, this is just going to be a radius of Oregon, so that black line is just capital are. So to get h h is going to be equal to, um are actually. Well, let's first, just so we can use Pythagoras. And so this is a right triangle. It would be eight squared plus r squared is equal to capital r squared and we just attract over square root. So h is going to be capital r squared my little r squared square root. Okay, um, so we have all of this information now, I might just leave it as a judge because I don't want to actually have to write that out, but just kind of keep in mind that we found this value for each already. Okay, so now, um, we are interested in finding just that area there. So notice it's like if we cut out this and then we just did the area of the top minus the area of the bottom. So what we would want to do is first notice that if we erase this bottom portion of this, this is just the top half of our circle. So this top part, if we go ahead and solve for the positive route here, it would be that and then this red part of top is just the positive root of why as well. So let's go ahead and solve for Y in both cases. So first here in red, um, this would just give us why is equal to r squared minus X squared square rooted down here, this would give us why is it too little r squared, minus X squared, square rooted. But then remember, we're adding each of us would be plus age. Okay, so we have that now and our bounds of integration. Well, this is going to be negative, little r and this is going to be positive. Little are because again, that's just the radius we're working with. So this area is going to be the integral from negligible our to our of our top function, which is this blue one. So it would be square root of R squared, minus x squared, plus h and then we're going to do minus the integral of the function we have over here or not. The integral just r squared, minus x word that and then we'll integrate with respect to X. Okay, now, one thing to make note of or actually let me check out fast if y'all have learned this yet. Okay. Yeah. So I'm pretty sure, um, I'll have learned this, but notice that this here is an even function. So it's even. And so remember, what even means is this implies that, um so that implies that we would have f of negative X equal to f Quebec's. And if this is the case, we're integrating something from, like, negative beta A of F of x. Yes, this is just equal to two times the integral from zero to a of f of x dx. And I just want to use this here, uh, to make it a little bit easier so you can go in and check to make sure that this is even. But I'm just gonna go ahead and apply directly, so that's just gonna be zero or 902 times from zero to our of. So you have the square root of R squared minus x squared, plus h then minus the square root of capital R squared minus x word DX. OK, now, at this point, what we will need to end up doing is actually distributing this across everything. So let's go ahead and do that. So, um I mean, technically, I guess we already know how to integrate all of this because this is just an area of a circle of radius r r squared. And then that would just be H. But since we're learning about these tricks substitutions, um, we can just use that to kind of double check in the end. So this would be two times the integral from zero are of the square root of our square by six squared DX plus two times the entrepreneur of zero to our of H dx and then minus the interval, two times the interval of zero. Little are to the square root of capital R squared, minus x word. Or actually, I guess I should say this first one is at least that area. This will actually have to do, like a trick substitution to get. Uh, so now let's see, um, for this first one, actually break this up into, like, different intervals, So this would be 12 and then three. So for the first one, the substitution we're going to want to make is the following. So any time we have, like, something squared minus x squared. We're going to want to say X is equal to That's something so would be our And then we're going to use sign Fada. Okay, so then taking the derivative of this, this gives us DX is able to are co sign data of d theta for our differential as we're taking the derivative retrospective data. Uh, and now we have to go and plug everything in. So over here, this would be two times. So if I come over here and plug in zero into this, that's just going to be zero. And then if I go ahead or actually maybe I should show what I'm doing. So when X is equal to zero, then this would be zero is equal to our signed data. So we divide each side by our. So we get signed data and then sign is going to be equal to zero at data is equal to zero. So that's why are lower bound is zero. And now when X is able to are, we get our is a guitar sign of data and then we divide each side by our. So we get one is equal to science data, and then we no sign of data is going to be equal to one at high half so that our upper bound is now pie half. Okay, now we can go ahead and make our substitution. So this would be the square root of R squared minus. And then our substitution was our sign of data. And then d x, um, becomes our co sign data d say to. So now let's go ahead and clean all this up. Um, so we still have I'm gonna pull that are out. So the two are integral from zero to pi, half of So we have our squared minus r sine squared, which, if we use Pythagoras, remember cosine squared. Why was rounded up co sign square data plus signs. Where data is he going to one. Um so if we go ahead and multiply everything by our squared r squared r squared and then we subtract that over and then take the square root on each side, we would just get our co signed data is equal to R squared minus R squared sine squared, Data square rooted. So this is just going to be our Hussein theater, and then we have times cosine theta do you say to? So now we can go ahead and pull that other are out to the two R squared, integral from zero to pi half of co sine squared theta d theta. Yeah, and then co sine squared. We would need to go ahead and use the power reducing identity to integrate this. And so doing that should give us, um b one plus co sign of two theta all over to dictator. Uh, we can go in and pull this two out. So actually, those just cancel out and then we would just have the integral from zero to high half of one plus co sign two theta d theta. Now we go ahead and integrate these. So one is just going to be data co sign of tooth data. Well, first, that would be sign of tooth data. And then, if you did a use of we would need to actually just divide that by two. Then we evaluate this from zero to pi half. So this is going to give us r squared. So data is pie half. If I plug pie up into there, that would be signed up. I, which is zero plus zero and then minus r squared when we plug in zero, which is just zero. So we would end up with where this is just going to be pi r squared over two. So this first part, I just kind of write this below. So this is two pi r squared over two Now for the second part. Well, h is just a constant. So it would just be plus two times h two times H and R minus zero, which would just be to h r. And then for this third one. Let's go ahead and move that down. So for this third one, so two times integral from zero two little are of the square root of r squared the other way around, I mean, Oh, yeah, that's right. Capital R squared minus little X word DX. And so now we'll need to go ahead and make our substitution pretty much the same as before. But instead of little are now, will be capital are of side data, our capital, our time scientist. Um, So again, our differential is going to be capital are cosign theta d theta. And if we were to go ahead and plug our bounds in so it would be so. An ax is zero. This is going to give us zero is equal to capital or of science data, which again just as data zero. And now when x is equal to little are well, there's going to be r is equal to capital are of signs data. So we'd have to divide by our So the little r over capital r is equal to sine data And then we would just have to take our design on each side. So this is going to say theta is equal to park sign of little are over capital are and we know this is going to be defined since little are is strictly less than capital are, um because otherwise, if little R was larger than our sun would be undefined. But in this case, it doesn't matter. So let's go ahead and use all that to plug in over here. So this would be two times the integral of zero two. Actually, I'll just write that sine inverse sine inverse of little are over capital or and then this would become the square root of capital R squared, minus capital r sine of theta squared and then DX becomes capital are co signed data data and then similar to what we did before. This just becomes capital are of co side data so we can go ahead and pull that r squared out. It'll be too r squared, integral from zero decide and verse of little are over capital are and then that becomes co sign squared data. And, well, we already figured out what this is supposed to be, uh, here, so we can just go ahead and take this and kind of move it down. So I'm going to be a little bit lazy and also just so I don't actually accidentally plug in something wrong, I'm gonna go ahead and just replace this if capital R squared. Actually, we write that in blue to kind of show that that was the only difference, Really? So we worked really hard, and then we got to here, um, and then our bounds this instead should be signed. Inverse of little are over capital are all right. So now we can just go ahead and plug these in, and then we would be done. So when we plug in sine inverse first that should give us We have r squared and they do which is be signed Inverse of little are over capital are and then plus sign of two side and verse of little are over capital are all over to and then zero. If we plug this in well, that would just be zero again so we could leave it like this if we wanted, but I don't know if that doesn't look too pretty to me. So what I'm gonna do is use a couple of trig identities to make it look a little bit prettier. So remember, any time we have, like, sign of two theta, we can rewrite this ads to sign data Times Co sign data. So in this case, our data is sine inverse. So this so data is equal to sine inverse of little are over capital are and doing this would give us to so that would just be little are over capital R and then co sign of that. Well, let's actually draw a triangle really fast to figure out what that should be. So we have sine of theta is equal to the little r over capital are. And if we draw this. So capital R is our hypothesis. Little our is our height. And so then that means our heart pot news over here is just going to be r squared minus for a little r squared all the way around capital r squared. I don't know why it went that far, but yeah, capital R little r and then capital R squared minus little r squared square rooting so called science supposed to be adjacent. So square root of r squared minus the lost work over our pot news, which is our And so that should be what co sign is, uh, so we can go ahead and collect our ours. So this is going to be too little Are square root of r squared minus little r squared all over Capitol R squared. All right, so let me just get rid of that and then plug into little are square root of r squared my little R squared all over Capitol Square, and then also knows those twos they're cancel out. So, um, if we actually go ahead and distribute that r squared notice that would cancel out with that one and then we would just have r squared there. So let me clean that up because I just wrote a ton of stuff. So would be capital r squared times sine inverse of little are over capital are and then plus little are especially. What would I have left there? So just be little are square root of r squared minus r squared. Yeah, and so that would be our second interval. Now let's actually come back up here. I'm going to just pick all of this up and move it down to the very bottom. Okay, so now we just need to take this and then plug it in right here. So then this is going to be remember, it was supposed to be minus That would be minus R squared sign in verse minus sine inverse of our over capital R and then minus little are square root of R squared, minus little r squared. And now this should be our area. Actually, one thing I forgot, um, this h was capital R squared. Minus little are square, So yeah, let me go ahead and replace that really fast. So two. So this was supposed to be square root of r squared by this little r squared. And actually, what did I have here again? I think it was a little r. Let me check Real fast can. Yeah. So it would be little Are there? Uh oh. And actually notice We have two copies of this and one copy here. So those just simplified down to one copy, Actually. Glad I remember to plug that in because this makes it even a little bit more simple. So this cancels out with that. So let me write this out one last time. So it'll be Hi little r squared over two plus and then I'll move that are out front. So little are square root of capital r squared, minus little r squared and then minus capital r squared times sine inverse of little are over counter. And now this is going to be our area, actually blue again. So that is our area.

So if we want to find the area of this like, shaded blue region on top, here I'm going to do first is turn this, um, or not. Turn it. But place it on to this XY plane here, like this. So this is X. And then this Here's why. So this red equation we can represent by X squared plus y squared is equal to capital R squared. And then this blue one will first, let's just go ahead and call this distance right here. Uh, h we can figure out how to solve for that in a bit. Um, but the blue line is there. Going to be war center is going to be at zero age, so this would be X squared. Plus y minus. H squared is equal to little r squared. Okay, Okay. Now, in order for us to solve for what H is going to be, we can actually create a triangle right here. I'll draw that triangle of top here like this. So this is a church. This is our I didn't notice going from here to here. Well, this is just going to be a radius of Oregon, so that black line is just capital are. So to get h h is going to be equal to, um are actually Well, let's first, just so we can use Pythagoras. And so this is a right triangle. It would be eight squared. Plus, our square is equal to capital r squared and we just attract over square root. So h is going to be capital r squared my little r squared square root. Okay, um, so we have all of this information now, I might just leave it as a judge because I don't want to actually have to write that out, but just kind of keep in mind that we found this value for each already. Okay, so now, um, we are interested in finding just that area there. So notice it's like if we cut out this and then we just did the area of the top minus the area of the bottom. So what we would want to do is first notice that if we erase this bottom portion of this, this is just the top half of our circle. So this top part, if we go ahead and solve for the positive route here, it would be that and then this red part of top is just the positive root of why as well. So let's go ahead and solve for Y in both cases. So first here in red, um, this would just give us why is equal to r squared minus X squared square rooted down here, this would give us why is it too little r squared, minus X squared, square rooted. But then remember, we're adding each of us would be plus age. Okay, so we have that now and our bounds of integration. Well, this is going to be negative, little r and this is going to be positive. Little are because again, that's just the radius we're working with. So this area is going to be the integral from negligible our to our of our top function, which is this blue one. So it would be square root of R squared, minus x squared, plus h and then we're going to do minus the integral of the function we have over here or not. The integral just r squared, minus x word that and then we'll integrate with respect to X. Okay, now, one thing to make note of or actually let me check out fast if y'all have learned this yet. Okay. Yeah. So I'm pretty sure, um, I'll have learned this, but notice that this here is an even function. So it's even. And so remember, what even means is this implies that, um so that implies that we would have f of negative X equal to f Quebec's. And if this is the case, we're integrating something from, like, negative beta A of F of x. Yes, this is just equal to two times the integral from zero to a of f of x dx. And I just want to use this here, uh, to make it a little bit easier so you can go in and check to make sure that this is even. But I'm just gonna go ahead and apply directly, so that's just gonna be zero or 902 times from zero to our of. So you have the square root of R squared minus x squared, plus h then minus the square root of capital R squared minus x word DX. OK, now, at this point, what we will need to end up doing is actually distributing this across everything. So let's go ahead and do that. So, um I mean, technically, I guess we already know how to integrate all of this because this is just an area of a circle of radius r r squared. And then that would just be H. But since we're learning about these tricks substitutions, um, we can just use that to kind of double check in the end. So this would be two times the integral from zero are of the square root of our square by six squared DX plus two times the entrepreneur of zero to our of H dx and then minus the interval, two times the interval of zero. Little are to the square root of capital R squared, minus x word. Or actually, I guess I should say this first one is at least that area. This will actually have to do, like a trick substitution to get. Uh, so now let's see, um, for this first one, actually break this up into, like, different intervals, So this would be 12 and then three. So for the first one, the substitution we're going to want to make is the following. So any time we have, like, something squared minus x squared. We're going to want to say X is equal to That's something so would be our And then we're going to use sign Fada. Okay, so then taking the derivative of this, this gives us DX is able to are co sign data of d theta for our differential as we're taking the derivative retrospective data. Uh, and now we have to go and plug everything in. So over here, this would be two times. So if I come over here and plug in zero into this, that's just going to be zero. And then if I go ahead or actually maybe I should show what I'm doing. So when X is equal to zero, then this would be zero is equal to our signed data. So we divide each side by our. So we get signed data and then sign is going to be equal to zero at data is equal to zero. So that's why are lower bound is zero. And now when X is able to are, we get our is a guitar sign of data and then we divide each side by our. So we get one is equal to science data, and then we no sign of data is going to be equal to one at high half so that our upper bound is now pie half. Okay, now we can go ahead and make our substitution. So this would be the square root of R squared minus. And then our substitution was our sign of data. And then d x, um, becomes our co sign data d say to. So now let's go ahead and clean all this up. Um, so we still have I'm gonna pull that are out. So the two are integral from zero to pi, half of So we have our squared minus r sine squared, which, if we use Pythagoras, remember cosine squared. Why was rounded up co sign square data plus signs. Where data is he going to one. Um so if we go ahead and multiply everything by our squared r squared r squared and then we subtract that over and then take the square root on each side, we would just get our co signed data is equal to R squared minus R squared sine squared, Data square rooted. So this is just going to be our Hussein theater, and then we have times cosine theta do you say to? So now we can go ahead and pull that other are out to the two R squared, integral from zero to pi half of co sine squared theta d theta. Yeah, and then co sine squared. We would need to go ahead and use the power reducing identity to integrate this. And so doing that should give us, um b one plus co sign of two theta all over to dictator. Uh, we can go in and pull this two out. So actually, those just cancel out and then we would just have the integral from zero to high half of one plus co sign two theta d theta. Now we go ahead and integrate these. So one is just going to be data co sign of tooth data. Well, first, that would be sign of tooth data. And then, if you did a use of we would need to actually just divide that by two. Then we evaluate this from zero to pi half. So this is going to give us r squared. So data is pie half. If I plug pie up into there, that would be signed up. I, which is zero plus zero and then minus r squared when we plug in zero, which is just zero. So we would end up with where this is just going to be pi r squared over two. So this first part, I just kind of write this below. So this is two pi r squared over two Now for the second part. Well, h is just a constant. So it would just be plus two times h two times H and R minus zero, which would just be to h r. And then for this third one. Let's go ahead and move that down. So for this third one, so two times integral from zero two little are of the square root of r squared the other way around, I mean, Oh, yeah, that's right. Capital R squared minus little X word DX. And so now we'll need to go ahead and make our substitution pretty much the same as before. But instead of little are now, will be capital are of side data, our capital, our time scientist. Um, So again, our differential is going to be capital are cosign theta d theta. And if we were to go ahead and plug our bounds in so it would be so. An ax is zero. This is going to give us zero is equal to capital or of science data, which again just as data zero. And now when x is equal to little are well, there's going to be r is equal to capital are of signs data. So we'd have to divide by our So the little r over capital r is equal to sine data And then we would just have to take our design on each side. So this is going to say theta is equal to park sign of little are over capital are and we know this is going to be defined since little are is strictly less than capital are, um because otherwise, if little R was larger than our sun would be undefined. But in this case, it doesn't matter. So let's go ahead and use all that to plug in over here. So this would be two times the integral of zero two. Actually, I'll just write that sine inverse sine inverse of little are over capital or and then this would become the square root of capital R squared, minus capital r sine of theta squared and then DX becomes capital are co signed data data and then similar to what we did before. This just becomes capital are of co side data so we can go ahead and pull that r squared out. It'll be too r squared, integral from zero decide and verse of little are over capital are and then that becomes co sign squared data. And, well, we already figured out what this is supposed to be, uh, here, so we can just go ahead and take this and kind of move it down. So I'm going to be a little bit lazy and also just so I don't actually accidentally plug in something wrong, I'm gonna go ahead and just replace this if capital R squared. Actually, we write that in blue to kind of show that that was the only difference, Really? So we worked really hard, and then we got to here, um, and then our bounds this instead should be signed. Inverse of little are over capital are all right. So now we can just go ahead and plug these in, and then we would be done. So when we plug in sine inverse first that should give us We have r squared and they do which is be signed Inverse of little are over capital are and then plus sign of two side and verse of little are over capital are all over to and then zero. If we plug this in well, that would just be zero again so we could leave it like this if we wanted, but I don't know if that doesn't look too pretty to me. So what I'm gonna do is use a couple of trig identities to make it look a little bit prettier. So remember, any time we have, like, sign of two theta, we can rewrite this ads to sign data Times Co sign data. So in this case, our data is sine inverse. So this so data is equal to sine inverse of little are over capital are and doing this would give us to so that would just be little are over capital R and then co sign of that. Well, let's actually draw a triangle really fast to figure out what that should be. So we have sine of theta is equal to the little r over capital are. And if we draw this. So capital R is our hypothesis. Little our is our height. And so then that means our heart pot news over here is just going to be r squared minus for a little r squared all the way around capital r squared. I don't know why it went that far, but yeah, capital R little r and then capital R squared minus little r squared square rooting so called science supposed to be adjacent. So square root of r squared minus the lost work over our pot news, which is our And so that should be what co sign is, uh, so we can go ahead and collect our ours. So this is going to be too little Are square root of r squared minus little r squared all over Capitol R squared. All right, so let me just get rid of that and then plug into little are square root of r squared my little R squared all over Capitol Square, and then also knows those twos they're cancel out. So, um, if we actually go ahead and distribute that r squared notice that would cancel out with that one and then we would just have r squared there. So let me clean that up because I just wrote a ton of stuff. So would be capital r squared times sine inverse of little are over capital are and then plus little are especially. What would I have left there? So we just be little are square root of r squared minus r squared? Yeah, and so that would be our second interval. Now let's actually come back up here. I'm going to just pick all of this up and move it down to the very bottom. Okay, so now we just need to take this and then plug it in right here. So then this is going to be remember, it was supposed to be minus That would be minus R squared sign in verse minus sine inverse of our over capital R and then minus little are square root of R squared, minus little r squared. And now this should be our area. Actually, one thing I forgot, um, this h was capital R squared. Minus little are square, So yeah, let me go ahead and replace that really fast. So two. So this was supposed to be square root of r squared by this little r squared. And actually, what did I have here again? I think it was a little r. Let me check Real fast can. Yeah. So it would be little Are there? Uh oh. And actually notice We have two copies of this and one copy here. So those just simplified down to one copy, Actually. Glad I remember to plug that in because this makes it even a little bit more simple. So this cancels out with that. So let me write this out one last time. So maybe hi little r squared over two plus and then I'll move that are out front. So little are square root of capital r squared, minus little r squared and then minus capital r squared times sine inverse of little are over counter. And now this is going to be our area, actually blue again. So that is our area.


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