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Find the exact area of the surface obtained by rotating the curve about the x-axis.

$ y = \cos (\frac{1}{2} x) $ , $ 0 \le x \le \pi $

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$$\pi \sqrt{5}+4 \pi \ln \left(\frac{1+\sqrt{5}}{2}\right)$$

Calculus 2 / BC

Chapter 8

Further Applications of Integration

Section 2

Area of a Surface of Revolution

Applications of Integration

Campbell University

Oregon State University

University of Michigan - Ann Arbor

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14:56

Find the exact area of the…

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Mhm. In this video will be finding the surface area obtained by rotating some curve around the X axis. So we're given a curve and a certain range. And what we want to do first is we obviously want to find some derivatives so we can do. So that's the following. I'll do it in blue. We have if y is co sine of X over two then why prime would be negative one half times to sign of X over two. Yeah. And this is due to the chain rule. This is due to the chain rule making our life a little bit easier for us to deal with. So we can put this into our integration formula Along with the bounds of integration which is zero and 2 pi. So what happens is our surface area will be equal to bringing the two pi out. We're going to have to pi the integral from zero to pi of why? Which is just cosign of X over two. So it will be like that coastline of X over two times square of one plus now why? Prime squared squares both of the terms in the product. So this negative here will become a plus 1/4 times the sine squared of X over two. All of that will be integrated with respect to X at this point we can perform a small U substitution basically what that means. Is that we're going to have Mhm Yes, we're going to let all of this here or rather maybe not that maybe not all of that. Well let's just the sine term. We'll just let the sine term become you. This will make our calculations just a tiny bit easier. So what that means is that we're going to say let you yeah let it be equal to one half times the sine of X over two. Yeah. When I differentiate that deal will be just 1/4. The co sign of X over two dx. Uh huh. Now obviously I don't have this 1/4. This additional factor of 1/4 in. So I can multiply it and divide it So I can divide and multiply by 1/4 at the same time as follows. This also means that the bounds will change right as follows. When X zero you will be sine of zero which is zero and when X. S. Pie you will be 1/2 times the sine of pi and the sine of pi over two. That's one, that's one. So that will be just one half simplifying this little bit. Are integral will become four times 2 pi will be eight pi. The integral from 0 to 1/2 and we're going to have one plus U. Squared. Do you? Okay and at this point we need to make another substitution because we're doing some tricks up now. So we can do the following we can say that you will be equal to the tangent of tea. And then do you will be the second square. This is because we have a a squared plus U squared situation. Mhm. This also means the bands. Do you have to change once more? So When you is zero T. The tangent, the tangent of what angle is zero. That's obviously zero. And when you is 1/2 The tangent of what would be 1/2? Well, just some arbitrary angle. I don't know what that is. There is probably no exact value for it. In any case We can now substitute all of this again into the formula. So we would have eight pi the integral from zero to The Arc Tangent of 1/2. And we're going to have one Plus Tangent Squared T. Times to seek in squared of T DT. Mhm. So yeah, all of this here becomes second squared within the square root which is just going to be second And multiplying that with the second squared it's going to become a pie times the integral from syria To the arc tangent of 1/2. We're going to have second cubed of T DT. And yeah, we'll assume that we know how to integrate, seeking cubed already. It's going to involve some integration by parts and we'll just give the result right now. So it will be one half times the second. The tangent plus the log Using Basie of course of the absolute value of the second plus the tension. And that should be an absolute value bar. But it makes sense. So applying this anti derivative here will give us the following, this one has multiplies with the eight pi and we're going to have four pi Uh huh. And this would be yeah, the second of t times the tangent of tea plus the natural the natural log. Which All right. It's just the log. And since seeking since we're clearly positive, I can drop the absolute value bars within the log and just make it parentheses. So have the second plus the tangent. And this will be evaluated yeah. From T equals 02 T equals the arc tangent of one half. And plugging all of this in would give us we would have the second of the arc tangent of one half plus the tangent of the arc tangent of one half plus the natural log Of the 2nd of the Arc Tangent of one half plus the Plus the tangent of the arc tangent of 1/2. And I need one more parentheses here. Yeah. And one more there. And this will have to subtract of course, with four pi times the second of zero. Plus the tangent of zero. Mhm. Plus the log of the second of zero. Plus the tangent. 00 and one more parentheses there. Okay. Before we do anything we ought to figure out what the secret of the arc tangent of one half is. When we do. So we're going to find we'll find that this is just equal to. You can apply some trick to this. You'll find that this is going to be equal to the square root of 5/2. Yeah. So plugging all of the sudden also realized that the tangent and arc tangent will cancel each other out. Hello? And also the tangent of zero is 0 and seeking to zero is since second as one of the co sign, we have one over the Co signer zero, which is just one. So basically what happens, we can cancel some terms out. This goes away. This goes away. This second log goes away because the log of one is just zero. So this vanishes. So basically what this becomes is we're going to have more pie Times 5/2 plus one half plus the log us read 5/2 plus one half. And this will be all subtracted from four pi. Okay. Yeah. Mhm. And you can certainly distribute all of this for pie and stuff if you want to. Which we'll do when we do. So This will be equal to two pi route five plus two pi. Yeah. Plus four pi Log of root 5/2 plus one half minus four pi. Which would give us the following we'll have two pi route 5 -2 pi plus four pi log of root 5/2 plus one half. Yeah. Okay. Mm hmm. Okay. And I forgot this is a multiplication. Actually that's a multiply and that's a multiply. So actually what happens is this entire term vanishes. So this goes away. This is a month supply. This is a multiply. Okay. Let's let's redo that. Let's redo the evaluation part here. Okay. Mm hmm. Yeah. When these two multiply, we're going to get this is going to be equal to route five over four. So you can multiply this times this times that which will give us the final answer of pi group five plus four pi log. And we'll combine the fraction here too. So we'll have root five plus one. The whole thing divided by check which is the final answer to this question.

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