TITLE: Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$
QUESTION [9 upvotes]: What is the best method for evaluating the following double integral?

$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$

Is there exist an easy method?
My try:
$$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx=\int_0^{\frac{\pi}{4}}\int_0^{b\sec(\theta)}r\sqrt{a^2-r^2}dr\,d\theta$$
$$=\int_0^{\frac{\pi}{4}}\frac{-1}{3}\left[(a^2-r^2)\sqrt{a^2-r^2}\right]_0^{b\sec(\theta)}d\theta$$
$$=\frac{1}{3}\int_0^{\frac{\pi}{4}}\left[a^3-(a^2-b^2\sec^2(\theta))\sqrt{a^2-b^2\sec^2(\theta)}\right]d\theta$$
but evaluating above integral is very difficult and antiderivative is very complexity! see here.

REPLY [14 votes]: Of course this is an exercise in integration. But nevertheless it can be done using elementary geometry alone. We are told to compute the volume of a body $K$  which is bounded by several planar faces and a peace of a spherical surface. The following figure shows the situation as seen from the tip of the $x$-axis. I have put $a=1$ and written $p$ instead of $b$.

The body $K$ contains (a)  the pyramid $P$ with base the triangle with vertices $(0,0,0)$,$(p,0,0)$, $(p,p,0)$, and of height $\sqrt{1-2p^2}$. The volume of this pyramid is
$${\rm vol}(P)={1\over6}p^2\sqrt{1-2p^2}\ .$$
Furthermore $K$ contains (b) part of a sector $S$ of central angle $$\alpha:=\arctan{p\over\sqrt{1-2p^2}}$$ of a spherical segment. The two radii of this segment are $1$ and $\sqrt{1-p^2}$, and its thickness is $p$. The volume of the full sector $S$ is therefore given by
$${\rm vol}(S)={\alpha\over 2\pi}{\pi\over6} p\bigl(3+3(1-p^2) +p^2\bigr)={\alpha\over6}p(3-p^2)\ .$$
From the volume of $S$ we (c) have to deduct the volume of a triangular spherical sector $T$, whereby the angles of the spherical triangle in question (shaded in the figure) are ${\pi\over2}$, ${\pi\over4}$ and a certain $\beta$. One leg of this triangle is $\alpha$, and a standard formula for right spherical triangles then tells us that
$$\cos\beta=\sin{\pi\over4}\cos\alpha={1\over\sqrt{2}}\cos\alpha\ .$$
The spherical area of the triangle is then $\beta-{\pi\over 4}$, so that
$${\rm vol}(T)={1\over3}\bigl(\beta-{\pi\over 4}\bigr)\ .$$
Finally
$${\rm vol}(K)={\rm vol}(P)+{\rm vol}(S)-{\rm vol}(T)\ ,$$
which maybe can be simplified somewhat.