This is a homework question, but i am just checking if what i am saying is correct,
The question in the book states that
A sphere of radius $r$ is cut by a plane of $h$ ($h < r$) units above the equator. Find the volume of the solid (spherical segment) above the plane.
Now i know that the upper hemisphere of any circle can be mapped on a 2 dimensional graph as
$$ f(x) = \sqrt{r^2-x^2} $$ If this plane would be represented as a line on this graph that intersects with this function at 2 points, calling those points $a$ and $b$
$$ h = \sqrt{r^2-x^2} $$ The positive point of intersection would be $$ \sqrt{r^2-h^2} = a $$
If i were to revolve this around the x-axis the volume would be
$$ V = 4\pi \int_{0}^{a} (\sqrt{r^2-x^2}-h^2)^2dx $$
Due to symmetry, then i would continue with the integration
$$ V = 4\pi\int_0^{\sqrt{r^2-h^2}}(r^2-x^2 - 2h^2 \sqrt{r^2-x^2} - h^4)dx $$
Therefore $$ V = 4\pi [r^2\sqrt{r^2 - h^2} - \frac{(r^2-h^2)^{3/2}}{3} - h^4 \sqrt{r^2-h^2} + \int_{0}^{\sqrt{r^2-h^2}} 2h^2 \sqrt{r^2-x^2} dx] $$
$$ \int_0^\sqrt{r^2-h^2} 2h^2 \sqrt{r^2-x^2} dx = h^3 \sqrt{r^2-h^2} + r^2 \tan^{-1}(\frac{\sqrt{r^2-h^2}}{h}) $$
Therefore the volume of this spherical segment would be
$$ V = 4\pi [r^2\sqrt{r^2 - h^2} - \frac{(r^2-h^2)^{3/2}}{3} - h^4 \sqrt{r^2-h^2} + h^3 \sqrt{r^2-h^2} + h^2 r^2 \tan^{-1}(\frac{\sqrt{r^2-h^2}}{h})] $$
This was a problem for my calcus 2 class, and it seems like i over did this problem since i have never done a problem of this magnitude in the class thus far. Did i do this correctly? Was there an easier way to go about doing this?
EDIT: I know what i did wrong, i made a mistake with the washer method and did
$$ V = 2 \pi \int_a^b (R-r^2)^2 dx $$
Instead of
$$ V = 2 \pi \int_{a}^b (R^2-r^2)dx $$
I answered my own question down below.
It might help you to draw a labeled diagram. Figure out where the first disk is, where the last one is, and what is the radius of a disk somewhere between them.
The integration limits $\int_0^{\sqrt{r^2 - x^2}}$ look vaguely like something you could use to find the volume by cylindrical shells. But the integrand for that method would be different from what you used.
The integrand for the disk method is actually much simpler than what you tried.