TITLE: Volume of cos(x) using disc method with y-axis
QUESTION [0 upvotes]: 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.

REPLY [0 votes]: 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
$2\pi \int_0^{\frac \pi 2} x \cos x \;dx\\
2\pi (x \sin x + \cos x)|_0^{\frac \pi2}\\
2\pi (\frac \pi2 - 1)$
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