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I cannot recall how to calculate the volume of $y=\cos(x)$ bounded by $x=-\frac{\pi}{2} , x=\frac{\pi}{2}$ using the disc method around the $y$-axis.

It has been a long time since I have done this. Perhaps someone would help refresh my memory. Here is what I want to do:

$$V= \pi \int_{0}^1 g(y)^2 dy$$

Is $g(y)= \arccos(y)$? I'm not sure.

Kind regards.

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    You should speak in your title of the volume **of revolution** generated when revolving the curve of $y=\cos(x) $ around the y axis. – 2017-01-13
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    Are you sure you want do use disks? Shells will be much easier. – 2017-01-13

1 Answers 1

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By the disk method, around the $y$ axis.

$\pi \int_0^1 x^2 dy\\ x = \arccos y\\ \pi \int_0^1 (\arccos y)^2 dy$

Integration by parts:

$u = (\arccos y)^2; dv = dy\\ du = -2\frac {(\arccos y)}{\sqrt {1-y^2}} dy; v = y\\ \pi[y (\arccos y)^2 + \int \frac {2y}{\sqrt {1-y^2}} (\arccos y) dy]\\ y (\arccos y)^2|_0^1 = 0\\ \pi\int \frac {2y}{\sqrt {1-y^2}} (\arccos y) dy$

And parts again.

$u = \arccos y ; dv = \frac {2y}{\sqrt {1-y^2}} dy\\ du = -\frac {1}{\sqrt {1-y^2}} dy; v = -2\sqrt{1-y^2}\\ \pi[-2\sqrt{1-y^2}\arccos y - \int 2dy]\\\ \pi^2 - 2\pi$

Shells

$2\pi \int_0^{\frac \pi 2} x \cos x \;dx\\ 2\pi (x \sin x + \cos x)|_0^{\frac \pi2}\\ 2\pi (\frac \pi2 - 1)$