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The region bounded by the graphs ofy = x2 + 1, 0 <x < land the X-axisis revolved about the line y = -1, Find the volume of the solid of revolution...

Question

The region bounded by the graphs ofy = x2 + 1, 0 <x < land the X-axisis revolved about the line y = -1, Find the volume of the solid of revolution

The region bounded by the graphs ofy = x2 + 1, 0 <x < land the X-axis is revolved about the line y = -1, Find the volume of the solid of revolution



Answers

Find the volume of the solid of revolution that is formed by revolving the region bounded by the graphs of $y=\frac{1}{\sqrt{x}+1}, x=0, x=4,$ and $y=0$ about the $y$ -axis.

So we are given The curved by is equal to equally power X and boundaries are why is equal to zero. X is equal to zero. X is equal to tooth. So we will people to the integral from 0 to 2 high times y skway, DX or easy could do in the grill from 0 to 2 and fortified in front of the grill and then eat the power X squared DX so that we is now equal to high times the integral from 0 to 2. Eat power, uh, two x dx and that would be equal to B is equal to five times e to the power to x over to limits of integration from 0 to 2. So we now have me is equal to pi kinds he to the power for over two minus e to the power zero over to 18 hours. Zero is one. So we have these equal to high over two and factor the one or two and Mackey's We would he to de power for minus one, and that is approximately equal to 84.1 nine to

So we have. Why is equal to one over the square root of X squared plus one and why is equal to one over squaring off to So here is our region where we want to find the volume and when rotated about the excesses. So if we take an elemental rectangle here than the lower case are here with people too square one over squid ink of to and the upper case are here would people to Why one over squaring off x squared plus one. So we know that people human this case is equal to the integral from negative 1215 times R squared minus R squared dx So easy co two We put the pie in front of clinical signs So musical two pi times in a dream from negative 1 to 1 and our square would be one over x squared plus one and a nor kissed Our spirit would be negative 1/2 and any cricket with respect to X. So these equal to high times the integral of one of our experts one is changing and worse off x the gulf negative whatever to his neck to 1/2 X and the limits of integration are from negative 1 to 1. So we have now is equal to pi times changing in worse off one minus 1/2 times one and then minus 10. Geant in worse off negative one minus 1/2 times Negative. What So that V is now equal to high times 10 years worth of oneness pi over four negative one times one is next to water with two minus. Changing yourself native to one, is negative power for negative 1/2 times. Next one is plus 1/2. So now if we just expand the expression we have Easy 4 to 5 times hide over four minus 1/2 plus I over four minus one of two. So that me is now equal to high times y over two minus one and that simplifies to musical two. Pi squared over two minus pi

So we are given why is equal to one over square look affects and the area is bounded by. Why is equal to zero. X is equal to two and excessive to six, and we are going to protect this area about via about the exact so use equal to in a cruel from 2 to 65 times Y squared DX. You can put the pie in the front so we can say musical comply kind integral from 2 to 6 and Y squared is one over x DX. The integral a one over X T X is equal to natural law effects of these equal to five times the natural law defects, with the limits off any creation going from 2 to 6. So these equal to pi times the natural log off six minus the natural off to on that is equal to read is equal to five times the natural off. Six divided by two, which would be threes or final answer. Volume is equal to five times the natural off. Three

So we are told that f of X is equal to dysfunction. And the interval with is this. And finally it is revolved about, um, the line. Why equals negative one. So it'll be awful to illustrate this situation to see, uh, what the discs for this, uh, solid of revolution would look like. So I will sketch out. Ah, the an estimation of this graph we will so noticed some from for native oneto one, uh, this value always be positive. And also, since there's a negative exponents, it'll be decreasing. So it could look something like this. And, um uh, there would be a nascent tote at X equals negative one. So I will draw that like this. Um So finally, I it's important enough that line that it's being revolved around. So let's say that's negative. Like, was native native one there, Um and ah, let's see. Okay, so but just look at what? What a disk in this case would be a disc of that sort of revolution look like, and I would draw that in green. So let's just taken a random base such as here, and it will Look, I think this, um, so this this length here would represent the radius. And that would be the value of FX plus one. Since there's divine is one below the middle of the disc, the center is one below the X axis. So we're going to say that our our vics, our meaning radius, would be equal Teoh f of x or this, uh, plus one. So that means our area forgiven. Expel you? Um, the area of a disc forgiven expert. I would be high times that our value that our function squared. So this So what this means is we will basically be taking We will have our volume be the following. The is equal to, um the integral from for the interval. Uh, negative 1 to 1 of, uh, that area function. So pie times off that and then course DX. So, um, firstly, no. This weekend, um, we can It will be easier to just take this constant city outside of the integral and ah, now to evaluate this in general, it will be more helpful to, uh, expand this binomial squared. Oh, by the way, I did forget to mention there definitely needs to be a square here because the area is pyre squared, so the same would apply here anyway. So, um, to make this to more easily evaluate this into growth will need to expand this binomial squared, Um, as follows. Using the usual A plus B squared is equal for just two a b plus B squared. So this would be this first all square this term to get, um, Let's see, the the two denominator in the exponents would just get cancelled out. So we have this and then, um we have Let's see, then it will be two times the two terms. So I was going by one, won't do anything and then, ah, just square the second terms, which is just gonna be one. And now Ah, we have this. So to find this definite integral first we will need to find what the indefinite integral is. So do that We're looking for this integral um de X there. So we will just evaluate thes three into girls separately and then add them. And no, we will not be putting an arbitrary costing at the end because we will be finding a definite integral anyway. So the integral of X was one to our native three. Using the power rule would be, um, this then Ah, for the next one, we would have, well, negative again by powerful we have. Let's see, he would end up being this because the exponents would become negative half and then ah, of course, I'm degree when we'll just give us X. So that is our indefinite into girl. And ah, remember what you're looking for this integral and ah, that's why I will move its down so we can work with it. So let's see, um, notice that this function is undefined. The reason that this is an improper integral because it's undefined When, um, X is equal to negative one. Because this inside the brackets here and here become zero. But there's a negative exposure, Which means did the nominator. It's in the denominator, so that would result in a zero denominator. So it's under fire. So that's why we need toe rewrite this in to grow as a limit like this. So we're gonna let a represent the lower bound. And since it's the lower bound, it'll you say from the positive side, the approach is negative. One from the positive side. And then, um You're right. A is the lower bound and the rest of the integral can remain the same. Uh, so to evaluate this, integral from the beta one. Um, we just need to input these about these bounds into the under indefinite integral Azzawi will do in the next line. So is equal to this. And in place of that in general, we will first input one in toothy on indefinite integral. Um, like this. So let's see. Oh, by the way, there should be a sign there. Green power rule. So yeah. No, you have one plus one. And then all that negative too. Then we have Ah, this 71 as we have this and then, Ah, one for X. So there's what we get when we sub in one. Now on the next line. Since run out of space, I will continue what I've been writing and now subtract. And then I will sub in the low, the lower bound A. So we have according to the indefinite in general, Negative and then in brackets A plus one and into the power of negative two. That's all over too. Then we have negative minus four over a squared of a plus one and then lastly, in place of one will write a So is what we have. And we're just going to evaluate this limit. Um, firstly, he will be helpful to simplify this constant this whole constant expression here. So we have, um this becomes too. So we have, uh, negative to to the negatives to the power of negative, too. Which would mean this is all just negative 1/2 times for what you say. So essentially, this is negative. 1/8, and this year be too. So we have 4/2 with a minus. Sign there, and that one stays the same. So we have that constant. We can just take that to the outside of the limits, since the limit doesn't affect a constant. So I will do that on the next line. Eso first will put little constant expression. So negative 1/8, minus four over to plus one. And now we will take the limit and we will take this negative sign to the outside of the limit. So we have this minus limit. Ah, because they approach the native one positive of this whole expression. No, from here. Uh, let's see what happens. So, firstly, we're gonna be evaluating this limit. So of course, for this a year, we can just ah sub in the negative one. So that's just negative one. And however, we're gonna look at the limits of this part and notice. Um, well, im going Teoh, rewrite this. So that's easier. We're going toe put the negative exponents part. We're just gonna be right this as one in the do Not in the numerator. And then, denominator, we have a plus one squared. Since that the exploding was negative two. And from here would see what happens. Well, as a approaches negative one from the positive side. This term, a post once weird will, um, will be approaching infinity. Now it will be approaching zero from the positive side. So this Tim will be approaching infinity. So this is all gonna be negative, Finney. And then, as for this part, the minus four over Ah, a plus one. Well, that denominator, as they approach, is negative one from the positive side that the nominee will also be approaching zero from the positive side. So this Ah, whole part minus four over route April's one will also be approaching native affinity and, uh, this whole part. So on that, I'm gonna rewrite everything on the next line, and we have high times, that whole constant expression, and then we have the minus sign, and that whole limit part was equal to negativity and then minus infinity minus one. Well, we haven't minus infinity than minus infinity. That becomes just minus infinity. And, um, we can expand this negative to this part. So we have that whole costs and expression, and then we have plus infinity and then expanding the negative to the minus one We give plus one. No, um, notice that this part, we have a bunch of Constance and we're adding them to infinity. Well, that just makes this whole that whole part equal to infinity. So we have with this sequel to kind many, and that's just able to infinity. So, technically, the answer to this integral ism, it diverges. So basically, the volume of this solid of revolution does not exist, which is going to be our final answer does not exist


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