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I'm working on a problem that is asking for the volume of $y=\cos x$ when rotated about the line $y=1$, with a restricted domain of $\bigl[0,\frac{\pi}{2}\bigr]$. The range is $\bigl[0,2\bigr]$.

There's a problem I'm working on that is asking for the volume of $y=\cos x$ and $y=\sin x$ when rotated about line $y=-1$ with a domain of $\bigl[0,\frac{\pi}{4}\bigr)$

I don't even know where to begin. First off; which method would be most effective in this case? A shell, washer or disc?

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  • $\begingroup$ You should draw a picture first to see what sort of shape you will get. $\endgroup$ – user164587 Sep 22 '14 at 1:34
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I assume that for the first problem, you are rotating the region below $y=\cos x$, above the $x$-axis, from $x=0$ to $x=\frac{\pi}{2}$.

Both Slicing and the Method of Shells work. Perhaps Slicing is a little easier.

Draw a picture. Make a slice perpendicular to th $x$-axis "at" $x$.

The cross-section is a "washer," that is, a circle with a circular hole in it. The radius of the outer circle is $1$, and the radius of the inner circle is $1-y$, that is, $1-\cos x$. Thus the area $A(x)$ of cross-section is given by $$A(x)=\pi\left(1^2-(1-\cos x)^2\right).$$ It follows that the volume of the solid is $$\int_0^{\pi/2} \pi\left(1^2-(1-\cos x)^2\right)\,dx.$$ Before integrating, it will be useful to simplify the integrand. You will need, among other things, to integrate $\cos^2 x$. There are several ways to do it. One standard one is to use the fact that $\cos^2 x=\frac{\cos(2x)+1}{2}$.

The second problem has a very similar structure.

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