I'm a bit confused regarding questions involving solids of revolution.
For example, a question involves a polynomal, say, $x^2+2$ is rotated about the y-axis, find the volume for the solid of revolution between $x=1$ and $x=3$.
Suppose I use the shell method, and my equation becomes:
$V=2\pi\int^3_1(x)(x^2+2) dx$
This volume would include whatever is between the curve and the x-axis, rotated about the y-axis.
However, if I were to use the disk method, my equation becomes:
$V=\pi\int^{11}_{3}(\sqrt{(y-2)})^2 dy$
But, in order for both regions to have the same volume, the equations would need to be:
$V=2\pi\int^3_1(x)(11-(x^2+2)) dx$
and
$V=\pi\int^{11}_{3}(\sqrt{(y-2)})^2-(1)^2 dy$
I was wondering if someone could clarify the reasoning behind this. Also, if it's rotated about the y-axis, does that mean the volume is between the curve and the y-axis, and the same if it's rotated about the x-axis?
Thanks.
Note that the range for $x$ is from 1 to 3. So, the cross section evolving around $y$ is enclosed by $x>1$, $y<11$ and above the curve.
The shell integral is, then,
$$ V=2\pi \int_1^3 x[y(3)-y(x)] dx = 2\pi\int_1^3 x[11-(x^2+2)] dx$$
The integral for the disk, actually a disk ring with inner radius at 1, is
$$ V= \pi \int_3^{11} [x^2(y)-1] dy = \pi\int_3^{11} (y-3)dy$$
They come out the same at $32\pi$.