The question is as follows:
$f(x) = x^2$ and $g(x) = -x^2$, rotated about the y-axis. It is also bounded by $x = -2$ Find the volume by rotating the region around the given axis.
The first method I tried was "shell" method. My equation was:
$$2\pi\int_{-2}^0-x(2x^2)dx $$
The second method (according to my teacher) was to take the region in Quadrant II, rotate it and find volume, then multiply by 2 (using disk method):
$$ 2\times\pi\int_{-2}^0 x^4dx$$
And of course they ended up being different results. I'm not sure if I messed up in shell method equation, second method equation, or both, so I would appreciate it if someone offered correction.
Best Answer
The following diagram is a representation of the solid of revolution.
Because of the symmetry, you can integrate along the horizontal axis. from $0$ to $2$ rather than from $-2$ to $0$.
For the cylindrical method, the cylindrical sections have area
$$ A=2\pi rh $$
In the cylindrical shell method the cross-sections through the region are taken parallel to the axis of revolution. Thus the cross-sections are perpendicular to the $x$-axis. So we will integrate with respect to $x$ and find $r$ and $h$ as functions of $x$.
For this example, $r=x$ and $h=2x^2$ so the volume is
$$ V=\int_0^24\pi x^3\,dx=16\pi $$
For the annulus method, we take a cross-section of the region perpendicular to the axis of revolution. These cross-sections are perpendicular to the $y$-axis so we will integrate with respect to $y$ and find the variables of the area formula
$$A=\pi(R^2-r^2)$$
in terms of $y$. We see that the outer radius is $R=2$ and the inner radius is $r=\sqrt{y}$. So we get a volume of
$$ V=\int_{-4}^4\pi(2^2-(\sqrt{y})^2)\,dy $$
but we can use your teacher's suggestion and just double the volume of the top half:
$$ V=2\pi\int_{0}^4(4-y)\,dy=16\pi $$