<s>[INST] Disk/Washer method proof checking 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.
 [/INST] I found my mistake,
$$
V = 2 \pi \int_a^b (R^2 - r^2)dx
$$
Therefore
$$
V = 4 \pi \int_0^\sqrt{r^2-h^2} (r^2-x^2-h^2) dx
$$
$$
V = 4 \pi [r^2 \sqrt{r^2-h^2} - \frac{(r^2-h^2)^{3/2}}{3}
 - h^2 \sqrt{r^2-h^2}]$$
So much easier
</s>