19. The first derivative of x = 4e^θ and y = e^-θ +...

19. The first derivative of $x = 4e^{\theta}$ and $y = e^{-\theta} + \frac{1}{16}e^{-5\theta}$ is $-rac{1}{16}e^{-5\theta}$. 20. The second derivative of $x = -10 \cos \theta$ and $y = -10 \sin \theta$ is $\frac{-1}{10 \sin^3 \theta} \theta$. 21. The slope of #19 above at $\theta = 0$ is zero. 22. The concavity for #20 above at $\theta = \frac{\pi}{2}$ is down. 23. The arc length of $x = \frac{1}{3}t^3$ and $y = 18t$ on $0 \le t \le 9$ is $81[\sqrt{2} - \ln\sqrt{2} + \ln(2 + \sqrt{2})]$. 24. The surface area for $x = t$, $y = 6t$, $0 \le t \le 2$ revolved about the x-axis is $24\sqrt{37}\pi$. 25. The interior area of $r = 3\cos\theta$ is $\frac{9\pi}{2}$.

Transcript

00:01 We're given a curve and we're asked to use either a computer algebra system or a table of integrals to find the exact area of the surface obtained by rotating this curve about the x -axis.

00:14 The curve is y equals the square root of x squared plus 1 defined on the interval 0 1 or sorry 0 3.

00:36 So it follows that d, y, d.

00:39 Y, dx is equal to, this is going to be 1 over 2 times the square root of x squared plus 1 times 2x, which is the same as x over the square root of x squared plus 1.

00:58 And therefore, the arc length element ds, which is equal to the square root of 1 plus d, d, x squared, times dx.

01:20 Is going to be equal to square root of 1 plus x squared over x squared plus 1 dx, which can also be written, well, we'll leave it as it is for now, and you'll see why.

01:43 So it follows that the surface area by formula from this section is the integral from x equals 0 to 3 of 2 pi y, which is square root of x squared plus 1 times d s which is the square root of 1 plus x squared over x squared plus 1 dx and then multiplying through the x squared plus 1 under the radical this becomes 2 pi times the integral from 0 to 3 of the square root of and then we have x squared plus 1 plus x squared which is 2x plus 1 dx, and then factoring out a root 2, because this is really root 2x squared, we factor out a root 2 and get, well, this is going to be 2 root 2 pi integral from 0 to 3 of, this will be x squared plus 1 1⁄2, and notice that 1⁄2 is the same.

03:36 Same as 1 over root 2 squared d x.

03:47 So this square root of x squared plus a constant squared, this actually is a familiar form that we'll see it in the back of the book and the table of integrals.

03:59 And so taking the antiderivative, or using a computer algebra system, we get 2 root 2 pi times 1 half x times the square root of x squared plus 1⁄2...

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