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.

In a sense "Polar coordinates does not help much since the integrand likes them while the domain does not". 
It helps to know about partial integration, that is 
$$\int f'(x)g(x)dx = f(t)g(t)- \int f(x)g'(x)dx +C $$
In addition one should know about the derivatives of $\arcsin$ and $\arctan$.
Now to the integral, first we simplify 
\begin{eqnarray}\int_0^x\sqrt{a^2-x^2-y^2}dy &=& \int_0^x\sqrt{a^2-x^2}\sqrt{1-\left(\frac{y}{\sqrt{a^2-x^2}}\right)^2}dy \\ &=&\int_0^{x/\sqrt{a^2-x^2}}(a^2-x^2)\sqrt{1-t^2}dy\tag{1}\end{eqnarray}
Next we use partial integration
\begin{eqnarray}I=\int_0^\eta\sqrt{1-t^2}dt &=& \left[t\sqrt{1-t^2}\right]_0^\eta-\int_0^\eta t\cdot\left(\frac{-t}{\sqrt{1-t^2}}\right)dt\\&=& \eta\sqrt{1-\eta^2}+\int_0^\eta \frac{t^2}{\sqrt{1-t^2}}dt
\end{eqnarray}
and then $$\int_0^\eta \frac{t^2}{\sqrt{1-t^2}}dt = \int_0^\eta \frac{-(1-t^2)}{\sqrt{1-t^2}}dt+\int_0^\eta \frac{1}{\sqrt{1-t^2}}dt=-I +\arcsin{\eta}$$
Thus 
$$I=\frac12\left(\eta\sqrt{1-\eta^2}  +\arcsin{\eta} \right)\tag{2}$$
which we use in (1) to get 
\begin{eqnarray}
\int_0^{x/\sqrt{a^2-x^2}}(a^2-x^2)\sqrt{1-t^2}dy
 &=& \frac{(a^2-x^2)}2\Big(\frac{x}{\sqrt{a^2-x^2}}\sqrt{1-\frac{x^2}{a^2-x^2}}  \\
&&+\arcsin{\frac{x}{\sqrt{a^2-x^2}}} \Big)\\
&=&\frac{x}{2}\sqrt{a^2-2x^2}  
+\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}} \end{eqnarray}
It follows that 
$$\int_0^b \int_0^x\sqrt{a^2-x^2-y^2}dy\,dx= \int_0^b\frac{x}{2}\sqrt{a^2-2x^2}dx+ \int_0^b  
\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}}dx$$
and since 
$$\int_0^b\frac{x}{2}\sqrt{a^2-2x^2}dx= \frac{1}{12}(a^2-2b^2)^{3/2}$$
it suffice to calculate 
\begin{eqnarray}\int_0^b  
\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}}dx&=&\frac1{2a^2}\int_0^b  
\frac{(1-(x/a)^2)}2\arcsin{\frac{(x/a)}{\sqrt{1-(x/a)^2}}}dx\\
&=&\frac1{2a}\int_0^{b/a}  
(1-t^2)\arcsin{\frac{t}{\sqrt{1-t^2}}}dt\end{eqnarray}
which can be calculated in the same fashion as we did with $I$ above - I leave this fun part to you. 
