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 [0,pi/2]. The range is [0,2].
I need to setup equations using a) cylindrical shells and b) slicing. Slicing seems relatively easy to visualize and setup, but I'm not sure about the shell method. This is what I have:
a) Volume by Cylindrical Shells:
$$ 2π\int_0^{π/2} x(1-cosx) dx$$
b) Volume By Slicing
$$ π\int_0^1 (1-cosx)^2 dx$$
Is this the right way about doing it? I'm having a very hard time visualizing method a).
Best Answer
The problem is faulty: it gives you only one boundary of the region and leaves you to guess about the others. There are two possible regions. One, which I’ll call $R_1$, is bounded by $y=\cos x$, the segment of the $y$-axis between $y=0$ and $y=1$, and the segment of the $x$-axis between $x=0$ and $x=\frac{\pi}2$; it’s the region whose area you would be finding if you were asked to find the area under $y=\cos x$ between $x=0$ and $x=\frac{\pi}2$. The other, which I’ll call $R_2$, is bounded by $y=\cos x$, the segment of the line $x=\frac{\pi}2$ between $y=0$ and $y=1$, and the segment of the line $y=1$ between $x=0$ and $x=\frac{\pi}2$. My best guess is that you’re intended to use $R_1$, so that’s the one that I’ll use.
If you revolve $R_2$ about the line $y=1$, you get a roughly cone-shaped solid; its peak is to the left, at the point $\langle 0,1\rangle$, and its base is to the right. Its height, measured horizontally from peak to base, is $\frac{\pi}2$; its base has radius $1$. If you revolve $R_2$ about the line $y=1$, the resulting solid bears no resemblance to a cone.
If you chop $R_1$ into vertical slices, each slice produces an annulus. The outer radius of the annulus is the distance from the $x$-axis to the axis of revolution, which is the line $y=1$, so the outer radius is $1$. The inner radius is the distance from the top edge of $R_1$, the line $y=\cos x$, to the axis of revolution, so it’s $1-\cos x$. The area of the annulus is therefore
$$\pi\left(1^2-(1-\cos x)^2\right)=\pi\left(2\cos x-\cos^2x\right)\;,$$
and it contributes $$dV=\pi\left(2\cos x-\cos^2x\right)dx$$ to the volume. There is one such slice for each $x$ from $0$ to $\frac{\pi}2$, so the volume is
$$V=\pi\int_0^{\pi/2}\left(2\cos x-\cos^2x\right)dx\;.$$
Using cylindrical shells for this problem is a bad idea, but it can be done. Since you’re revolving the region about a horizontal axis, you must slice it up horizontally $-$ parallel to the axis of revolution $-$ in order to get shells. This means that you’ll have a shell for each value of $y$ from $0$ to $1$, and the infinitesimal thickness of each shell will be $dy$, not $dx$. (Infinitesimal here is informal, but it’s a perfectly good way to think about the problem in order to get it set up right.) The radius of the shell at a particular value of $y$ is the distance from $y$ to the axis of revolution at $y=1$, which is $1-y$. The height of the shell is the horizontal distance from the $y$-axis to the curve $y=\cos x$, which is $\cos^{-1}y$. The shell therefore contributes
$$dV=2\pi(1-y)\cos^{-1}y\,dy$$
to the volume of the solid of revolution, and the whole volume is
$$V=2\pi\int_0^1(1-y)\cos^{-1}y\,dy\;.$$