<s>[INST] Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ 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.
 [/INST] (The following is basically obsolete because already answered in
https://math.stackexchange.com/questions/571967/integral-inta2-b2-sec2x-sqrta2-b2-sec2xdx)
The first integral of Harry Peter is easy with the substitution $x^2=t$.
The second, attacked with partial integration, requires to integrate
$(a^2-x^2)/2$ which is $x(3a^2-x^2)/6$ and to differentiate the $\arcsin$
which gives $a^2/\sqrt{a^2-2x^2}/(a^2-x^2)$. So we need besides the
preintegrated term essentially
$$\int dx \frac{x(3a^2-x^2)}{6}\frac{a^2}{\sqrt{a^2-2x^2}(a^2-x^2)}$$
which turns with $x^2=t$ in something proportional to
$$\int dx \frac{(3a^2-t)}{6}\frac{a^2}{\sqrt{a^2-2t}(a^2-t)}$$.
By expansion of $(3a^2-6)/(a^2-t)$ into partial fractions only
the format $1/[\sqrt{a^2-2t}(a^2-t)]$ is required which seems with $y=a^2-2t$
to be covered by $$\int \frac{dy}{(a+by)\surd y} =\frac{2}{\sqrt{ab}}\arctan\sqrt{\frac{by}{a}}$$
</s>