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Problem 66 Hard Difficulty

Find the volume of the resulting solid if the region under the curve $ y = \frac{1}{(x^2 + 3x + 2)} $ from $ x = 0 $ to $ x = 1 $ is rotated about (a) the x-axis and (b) the y-axis.

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

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Video Transcript

Let's find the volume of the solid obtained after rotating the region underneath his curve from zero one. So here's a rough sketch of the graph when you plug in X equals zero to get one over two. As Ex increases, the denominator is larger, so the fraction is a whole good smaller. That's why it's decreasing. So for part A, we wrote it about the X axis, so basically we could see that a cross section is a disc. So we used the volume, so we know the disk will have radius pi r squared. That's a circle right there. So the radius is just the distance from Y equals zero Up until the curve that was in a straight line. Let me draw back here. So there's the radius from center to the end point. So our is just equal toe Why, which is given by this? So that's the radius of the disc. So for part, a pull off the pie that's coming from a circle area and then here's R R squared. So that's the formula for the volume using the disk method. So we'LL just have to evaluate this. Let's go ahead and factor that denominator before we do partial freshen And now we'LL just spend a few moments simplifying the inter grand so we can write this over explicit one. Be a restless once where c over X plus two and then be over X plus two squares. But so I'm running out of room here. I'Ll need to go to the next page, But before I do, let me note that for part B, when we rotate this thing about the Weisses, it may be easier to use cylindrical shells then to use this method. So for part b of hunger or use the shaman okay, simplifying our pre use equation. So we multiply both sides by the denominator on the left, clear out the denominator and then multiply out the right hand side and we'll have a system of equations coming up. Surely so here we can expand the whole thing on if we want to find a, B, C and D, or you could just go ahead and plug into Mike's values. For example, if you plug in X equals one, the left hand side is just one. We're on the right hand side. We should get in this case. I shouldn't be plugging in one. Let me plug in negative one. So if we plug in that we have zero over here, a zero over here and also zero over here. So then we just have be on the right. So that's one value. If we try, X equals negative too. We'Ll get d equals once. Now we could use some reasoning to find the other variables here. So, for example, if we look at the coefficient of that's cute on the right hand side, we see that there's two exes here, another X here. So that's one for the a term and that also see, we'LL have an excuse to access here Another meds here. So a plus e must be zero that is, this equals negative c and then looking at the constant term on the left, we have won on the right we have for a plus four b plus two C plus the yeah, plug in our values for being D and then also replaced, See with minus a and then So go ahead and solve this and we get that a equals two. Excuse me is minus Who so that c equals two. Now it's God and plug in these values for a, B, C and D and Tara. Partial fraction to composition. So we're still in party here, but we're basically finished. We've simplified our into grand, and now it's going to be simpler. Okay, there's a partial fraction to composition that they're plugging in are constants Now. We could go ahead and integrate all of these. The first general, if that plus one, is bothering you. Feel free to deal use up here. Same thing for the Senate next in a role as well, but here and could use the power rule. It's plus one. Similarly, here you have natural log and then power once more and then zero to one. So then which is good and plugging those end points? And I should have wrote. After that, there was a pie appear. A part of me should be a pile here the whole time. Now let's just go ahead and plug in those in points. So that's the plugging in the one. And then now we plug in zero Ellen of one zero. So that's what we get after we plug in the second and point and then just got and simplify this. Use your properties of the lot of them and we get in. Answer that looks like this. That's for answer for party now, as I mentioned before, you shells for part B after rotating it about the y axis. When I do so, my formula using shells too high in a girl zero one Then I have X and then on the denominator. We've already factored this. So let me write. This is X Plus one explodes too. This is the original denominated from the original function. Now let's do a partial fraction to composition. Here we can rewrite this using what the author calls case one distinct linear factors. Then you could multiply both sides of this equation here. By the denominator on the left, X equals a X plus two plus B. It's plus one. So here the partial fractions does seem a little simpler than party. Last time we had four coefficients. This time will we have to. So we see this has to be one that has to be zero. So go ahead and solve this for A and B. When will plug in these values of Andy into our partial fraction and then we'LL go ahead. And Mina Marie. So this all becomes plugging in negative one slugging. You're being equals two. Go ahead and evaluate these intervals where those will give us natural logs using you. Slip if you need to. And then our one point zero one. Now go ahead and simplify this. And then let's go ahead and finally break This's to pie. Natural log nine over eight And that will be our final answer for part B using show mother.

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Calculus: Early Transcendentals

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kristen Karbon

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals