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The figure shows the rcgion of integration for the integrl 6GK frY,-kdyrkr(Assume y(r) = vrand zW)=4-y) This integral . equal [0; (fill in the boxes )f(x,y,z)dydzdx...

Question

The figure shows the rcgion of integration for the integrl 6GK frY,-kdyrkr(Assume y(r) = vrand zW)=4-y) This integral . equal [0; (fill in the boxes )f(x,y,z)dydzdxEvaluate; KI dydr - Vx"+y"Note: J sec0d0 = In(Vz+1)

The figure shows the rcgion of integration for the integrl 6GK frY,-kdyrkr (Assume y(r) = vrand zW)=4-y) This integral . equal [0; (fill in the boxes ) f(x,y,z)dydzdx Evaluate; KI dydr - Vx"+y" Note: J sec0d0 = In(Vz+1)



Answers

The figure shows the region of integration for the integral
$$ \int_0^1 \int_0^{1 - x^2} \int_0^{1 - x} f(x, y, z)\ dy dz dx $$
Rewrite this integral as an equivalent iterated integral in the five other orders.

Were given the region of integration foreign integral, and we were asked to rewrite the center gorillas, an equivalent iterated integral in five other orders. So the region of integration is given in exercise. 34 of this section integral is first with respect to X, then with respect, dizzy and with respect to why of the function f x y z. So, first, let's write down the projections under the coordinate planes. So projecting onto the X Y plane, you see that this is rejecting solely into the first quadrant. And you say the projection is in fact a triangle with Vertex is of the origin 10 and 01 The equation for the hypotenuse of this triangle is X plus Y equals one or y equals one minus X. Now consider projection under the Z axis again, we see the slides in the first quadrant, and the projection is a area under a parabola. So the curve is C equals one minus X squared for projecting. And this has the intercept at C equals one has an X intercept at X equals one and looks something like this. Of course we know from the picture that X must be greater than zero Z must be greater than zero. This can also be written, by the way, as X equals square root of one minus C. What's the positive square? Because we're in the first quadrant. Finally, let's consider the projection into the Y Z plane sounds a little bit more complicated, so I saved it for last. So first, let's consider how this curve of intersection projects under the Y Z plane. Well, we see that surface Z equals one minus X squared intersex plane. Why it was one minus X in the curve with projection onto the Y Z plane. Find this by setting X equal to zero, actually zero of them by substituting. So we have that Z is equal to one minus X squared. Which X could be written as one minus? Why? So this is one minus one minus. Why squared or, in other words, Z is equal to to why minus y squared, which is a parabola. So we split up the projection onto the Y Z plane into two regions, so Z equals to I minus y squared. This has the intercept of zero the origin and then has let's see we're only going upto white walls, one according to our figure. So plugging in why equals one we get The Z is equal to one as well to a point at 11 and the Prabal looks like this spacing down. Of course it's prevalent. Also be written by solving for why, as why equals one minus Z square rooted. And then So it's one minus square root of one minus Izzy. Yeah, like I said before, you have a wise to be less than or equal to one, and Z has to be greater than or equal to zero. We also have our figure that Z has to be less than or equal to one and the wise to be greater than or equal to zero. So we get this square shape. So really, you project into two distinct regions, one which will call our one in this one, which will call are too, or I guess, the other way around. This is our too, and this will be our one, and we can describe the region's R one and R two as follows you have that are one is thes set of pairs. Why see such that we can describe this as why being between zero and one and then Z has to be between zero and the curve two y minus y squared and you see that yes, Y Z is in our one then it follows that from the graph of E X lies between zero and between the plane X equals one minus y solving for X and likewise we have that are to come. You described as the set pairs y Z such that now Z lies between zero and one. And why lies between the vertical line Y equals zero and curve y equals one minus the square root of one minus C and for Y Z in the sub region are too. We see that X lies between the plane X equals zero and between the plane, which can be solved from Z equals one minus X squared. It's actually not a plane, it's a surface. Solving Z equals one minus X squared. We get that X is equal to square root of one minus Z, therefore follows that are given integral, which is the integral from 0 to 1 integral from zero to one minus X squared, integral from 0 to 1 minus X of F of x y z de y de cdx. This can also be written as well. First, let's restrict dizzy production on the Z axis we see that Z lies between zero and one and then looking at a graph of the Y Z plane, we see that Z has to lie between zero and one minus the square root of one minus g. This is region are one and finally looking at the solid and noting what we said earlier for y Z in the region are one excess to lie between. I'm sorry this is in the region are to I'm sorry. So this is integrating from Z equals 01 and then from the line y equals here to the line. Why equals one minus the square root of one minus. Why she recognize is our too And we know that for uh huh points in our to value of X has to lie between zero and one minus y playing one minus Why? So we this integral of f x y z the x de wide is he and then consider the contribution from the other sub region. So we have again integrating Z from zero toe. One actually known. Yeah, integrating Z from 01 then integrating Why, from looking at her projection into the Y Z plane instead. From the curve y equals one minus the square of one minus C to the line y equals one and then integrating ex again from well, Now because y Z lies in are one it follows X has to lie between the planes. X equals zero and X equals one minus. Why may mistake this first integral Here we have that. Why is he lies in our two and their four x should lie between zero and the surface. X equals the square root of one minus c. My mistake. And here because why's he lies in our one x Lies between the plain X equals zero and the plain X equals one minus Y f of x y z dx dy y z z. So now let's consider another complicated form. Consider integrating first with respect toe. Why so from why equals zero toe? Why equals one? This is what y ranges from by protecting on the Y axis. Then, looking at her y Z plane, well, we have First we can integrate over the region are one. So if y ranges from 0 to 1, we then have zero range from the curve equals zero to the curves equals two y minus y squared. And then this means that Y Z lies in our one, which implies that X lies between the planes. X equals zero and X equals one minus y f of x y z d x d c d y. And then we have the other integral from why equals 01 and then now we're integrating over the other sub region are too. So from Z equals two y minus y squared to Z equals one. And then this means that y z lies in our two, which implies that X lies between the plain X equals zero and the surface X equals the square root of one minus Z of our function. F x y z the x dizzy d y. So so far we have to integrated. Inter iterated Integral tried to be expressed his sums. The rest of them, however, are simple enough to express a single intervals. This is that's integrates from X equal from X first projecting onto the X axis we see X ranges from 0 to 1 and then integrating with respect to Z. Looking at her X Z plane is your head zero range from zero to the curve, one minus X squared and finally looking at a region we see that why ranges from the plane. Why equals zero to the plane? Why equals one minus X of our function? F x y z de y dizzy DX, and then consider integrating with respect to Z first projecting onto the Z axis, we see that see ranges from 01 then integrating with respect to X. What we look at our XY plane. You see the X ranges from zero to the square root of one minus C and finally you said, Why ranges from plain Y equals zero to the plane? Why equals one minus X Earth of X y z de y dx dy DZ. Another way to write it is by integrating first with respect. Why so projecting Under the Y axis, you see the white ranges from 0 to 1, then looking at her. That's why plane you see the X ranges from X equals zero to the line X equals Y minus one. I'm sorry X equals one minus why? And finally looking at her solid, we see that Z ranges from the plane. Z equals zero surface C equals one minus X squared of f of x y z dizzy dx, dy y And finally consider integrating with respect to X First again, we see that X ranges after projecting on the X axis from 01 and then integrating with respect to why look at her X y plane said that Why ranges from what I equals zero to the line y equals one minus X and finally looking at our region, we see that Z ranges from the plane Z equals zero to the plane or surface. Z equals one minus x squared our function f x y z dizzy de y dx. And so we have six different ways of writing our integral as an integrated iterated triple integral, one of which was already given to us

For this problem. We have been given a triple integral. It is an iterated integral. We've put it already. Put it in order. Dizzy D Y d x, And this is related to, uh, the picture that's shown in your book. So this probably problem will probably make a little more sense if you do have your book open to problem 33 so you can see the shape that we're talking about now this is in a certain order. First would find the integral with respect to see Then we would integrate for why and then X. But that's not the only way we could have done this. We could have done a triple integral. I'm going to come back and put in these limits in a minute. I could have done dizzy d x d y. That's the second one. I didn't have to have fizzy First, I could do d y d z d x I could have done d Y d x DZ or I could have put the X first. I could have had d x d Y d c or d x d c d y. Any one of those six I could do in the order. What I really need to make sure I do, though, is get the limits of integration. Correct. So and in each of these cases, I am taking the integral of the function. I just didn't copy that piece down, especially because we don't know what the function actually is as far as being able to write it down there. Um, so that's what we have. So how do I right thes limits of integration? How do we find them? Well, the nice thing about the problem is they've told us how these variables relate to each other. If you look at the picture, you can see that why equals the square root of X? Or I could say X equals y squared. So that's one way that X and Y relate to each other. I also know that Z equals one minus y or I could say that why equals one minus Z. So that's another way that I could say that these will relate to each other. Okay. If I wanted to go from X to Z well, X is why squared? Why is one minus z? So I could say that X is one minus z squared so I can get from one variable to the others. So let's come back to our problem here. First one. Okay, we've been given that one, so we don't have to go back and look at that. So what if I did? DZ dx dy y? Well, let's do look a z first. How does Z Let's kind of look here? How does Z relate to the other variables? Well, Z equals one minus. Why? So I can say that Z is going to go from zero toe one minus Why, that's a later variable down the road so I can write it like that. Next, I have X. Now I have to put X in terms of why X is going to go from zero. So why squared? We know that X equals y squared. It's one of the things we put over here and ran it. Put a little green arrow next to us. You can see it there. Okay. And then I know then that why is going to go from 0 to 1. Okay, now, what if I did? Why first? So that's our next to going to do d y first. Well in either case. If you look at that picture, why is going to go from the square root of X two? Um one minus. Um, I'm sorry. Say that backward square root of X to one minus z. That's That's the the the limits there. So in either case, whether I'm going d y d c d x or d Y d x d c I'm going to go from the square root of X to one minus C square root of x toe, one minus z. Okay. What if I do Z next? Well, how does e relate to actual Z is going to go from zero toe X, which is one minus the square root of X. If Z is one minus, why why is the square root of X Z is going to be? And I should have written that down originally, Z is one minus the square root of X just based on the equations that they've given us and then X goes from 0 to 1. Okay, our fourth case, we know we already have those first limits of integration, but now we wanna put X in terms of Z, while X is going to go from zero and I know that X is one minus Z squared. That's how those relate to each other. And then Z would go from zero toe one. That's a square. Okay, now what about D X being our first integration that we dio well X is why squared? So I can say the X goes from zero toe y squared in both cases at now. What if I do? Why next? That's that. That's our number. Number five here dx dy y dizzy If why is next? I need to put why in terms of Z So why is going to go from zero toe one minus z and then Z goes from 0 to 1? If I do dx dy DZ d y, I need to put Z now in terms of why So Z is going to go from zero toe one minus y and then why finishes again by going from zero toe one so we can change the world that we do this integration. And as long as we're careful about our limits, sometimes some limits are easier to find than others, sometimes integrations easier if you do in one order versus another. So there might be a reason to pick one of these as preferential over another, but they will all end up giving you the same answer at the final step.

Hello. So here we have the integral going from zero to A. Of F. Of X. Dx is equal to negative. The integral we reverse the endpoint. So negatively integral from 80 of F. Of X. Dx. Which is equal to negative. The some of the integral going from A to B. And then the integral going from B 20 So given our areas of area A, B, C and D. This is then going to be equal to what we have the minus sign here. So negative the area of A minus the area of be. So area of A will give it as one point 408 and then minus the area of be in the area of B is 2.4 75. So we take this difference here and then hit the negative of that. Um which is going to be equal to 1.067.

Were given the region of integration foreign integral. And we were asked to write this integral as an equivalent iterated integral in five other orders. So we're already given one order as an equivalent Integral. This is the integral from 01 from the squared of X 21 and from 0 to 1 minus. Why of the function f of X y z Dizzy d Y d x So the region of integration is shown under exercise 33 of this section First project the solid region me onto the coordinate planes project onto the X Y plane Notice that this is only going to be in the first quadrant we obtain well, projection gives us a curve. Why equals the square root of X which has appointed the origin as well as at 11 It looks like this This is of course can also be written as X equals y squared and projecting. We also see that Z equals one minus. Why projects to the line y equals one men's equals zero. And so we get this region and the X Y plane. Likewise in the y z plane again, we're in the first quadrant projecting the solid onto the Y Z plane gives us a triangle which has vergis is at the origin a swell as at 11 Sorry, 1001 And the equation for the hypotenuse is Z equals one minus y and finally projecting onto the X Z plane again, we project into the first quadrant. We see that we get well, the parable a y equals the square root of X parabolic cylinder He's going to project well, we substitute this into the equation for the plane. Z equals one minus y. So we get Sequels one minus the square root of X, which can also be written as X equals one minus Z squared. And this has a Z intercept at one. And the next intercept of one is well, and it looks like this. We're sorry. That's the wrong way. And we also see that X has to be greater than zero and Z is great in the zero. So we get this shape. Therefore, it follows that our iterated, integral from X equals zero toe one from y equals square root of X to y equals one, and from Z equals zero C equals one minus y of our function f of X y z Is he dead? Body X. This can also be written as projecting onto the Y axis. This is the integral from y equals zero toe. Why equals one and observing the X Y plane. We see the X then lies between zero and why squared on the right and then looking at her solid, we see that Z lies between c equals zero in the plains equals one minus y. This is dizzy dx dy y. Then let's consider integrating first with respect to Z. So the projection on Dizzy tells us that Z lies between zero and Z equals one. Then looking at her y z plane projection, see that Why has to lie between zero and one minus Z and looking at our solid e, we see that X has to lie between the plain X equals zero and cylinder X equals y squared, says DX D. Why does he? And now let's consider integrating again with respect toe. Why first? So we have that protecting under the Y axis wide ranges from 01 and now consider R Y Z plane. We see that Z must range from zero to the lines equals one minus y, and finally that X again ranges from the plain. X equals zero to the parabolic cylinder. X equals y squared F of X y z. This is DX dy DZ d y, and now it's considering degrading First with respect to X. Projecting onto the X axis, we see that X lies between zero and one and then looking at our X Z plane. See that Z must lie between the line Z equals zero and Z equals one minus the square root of X and then looking at her solid. We see that why must lie between the, I guess in verse of the parabolic cylinder, which is square root X and between the plane, which is why equals one minus Z f of x y z Do you Why dizzy dx and finally consider integrating with respect to Z first again on the Z axis, we see that Z ranges from 0 to 1 and now consider rxc plane again. Get it X must range from the line X equals zero to the curve. X equals one minus Z squared and looking at are sold again. We see that wine a strange from the cylinder square root of X to the plain white was one minus C f of x y z d Y d x DZ And so we have six different ways of expressing this iterated integral, including the one that we're given.


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