00:01
In this problem, we compute the surface area of x squared from 0 to 4 when it's revolved about the x -axis.
00:09
So recall that surface area is the integral from 0 to 4 of 2 pi, y, which in this case is x squared, times the square root of 1 plus y prime, which is 2x squared dx.
00:33
And then from here we can let 2x be tangent theta, which means that dx is one half times secant squared theta d theta.
00:57
And now we get that the surface area is the integral from zero to four.
01:03
Actually, let's go ahead and change our limits of integration.
01:12
We get that this is from zero to the inverse tangent of eight of ten, of two.
01:23
2 pi times x squared which in this case is 1 half times tangent theta squared times the square root of 1 plus tangent squared which is secant squared or that's the square root of secan squared theta which is just secant theta and then d x which is one half times secant squared theta d theta now we can pull our constants out in front so we have two pi times one -fourth times one -half.
02:19
So when we pull that out in front, we get that this is pi on four times the integral from zero to inverse tangent of 8 of tangent squared theta times secant cubed theta, d -theta.
02:51
And now from here we can rewrite tangent squared as secant squared minus one.
02:58
So this is pi on 4 times the integral from 0 to inverse tangent of 8 of secant to the 5th power of theta minus secant cubed theta d theta.
03:33
And now from here we can break this apart.
03:38
So this is piom 4 times the integral from 0 to inverse tangent of 8 of secant to the 5th theta d theta.
03:58
Minus pi on 4 times the integral from 0 to inverse tangent of 8 of secant cubed theta d theta.
04:21
So now that we have this, let's evaluate these integrals separately.
04:27
We can call this i1 and we can call this i2.
04:31
And we've seen i2 before.
04:35
So what we get when we integrate secant cubed theta, is the following.
04:46
So we get that this is equal to minus pi on 4.
04:58
Let's write i2 right here.
05:01
But we get that this is minus pi on 4 times 1⁄2cant of inverse tangent of 8 times tangent of inverse tangent of 8, which is 8, plus 1 half, times the natural log.
05:33
Again, of secret.
05:34
Of inverse tangent of 8 plus tangent of the inverse tangent of 8, which is just 8.
05:54
And then when we evaluate this at 0, that's just 0.
05:59
So we can stop there.
06:02
And let's write this as minus pi on 4 times 4.
06:11
We also should evaluate secant of inverse tangent of 8.
06:16
So to do so, let's construct a right triangle.
06:23
That angle is theta...