If $0 < h < a$, then the plane $z = a - h$ cuts off a spherical segment of height h and radius b from the sphere $x^2 + y^2 + z^2 = a^2$.
a) Show that $b^2 = 2ah - h^2$
b) Show that the volume of the spherical segment is $V = \frac{1}{6}\pi h(3b^2+h^2)$
Part a is pretty straightforward but I'm looking for how to approach (b). I've seen it done with triple integration. Is it possible to do it for 2 integrals?
What I have so far is $\int_{0}^{2\pi}\int_{0}^{b} (\sqrt{1-r^2} - (a-h))rdrd\theta$.
What you are talking about is spherical cap. Yes you can find with single / double integral. There seems to be some mistake in how you have written it.
You know radius of the circle at any given $z$ is $\sqrt{a^2-z^2}$. You use the knowledge of perimeter of the circle or the area of the circle at a given $z$ to make this a double or single integral.
$\displaystyle 2\pi \int_{(a-h)}^{a} \int_0^{\sqrt{a^2-z^2}} r \, dr \, dz$
or you can write it as
$\displaystyle \pi \int_{(a-h)}^{a} (a^2 - z^2) \, dz$