One triple integral.

Relevant wiki: Double and Triple Integrals

A sphere reminds of the Spherical Co ordinates , so let the equation of a sphere be $x^2+y^2+z^2=a^2$

Let $x=r\sin\theta\cos\phi,y=r\sin\theta\sin\phi,z=r\cos\theta$ be the set of substitutions. The Jacobian of the substitution is given by,

$\displaystyle J=\dfrac{\partial(x,y,z)}{\partial(r,\theta,\phi)}=\begin{vmatrix} \dfrac{\partial{x}}{\partial{r}} & \dfrac{\partial{x}}{\partial{\theta}} & \dfrac{\partial{x}}{\partial{\phi}} \\ \dfrac{\partial{y}}{\partial{r}} & \dfrac{\partial{y}}{\partial{\theta}} & \dfrac{\partial{y}}{\partial{\phi}} \\ \dfrac{\partial{z}}{\partial{r}} & \dfrac{\partial{z}}{\partial{\theta}} & \dfrac{\partial{z}}{\partial{\phi}} \end{vmatrix}=r^2 \sin\theta$

Since we are integrating in the positive octant of the sphere so we have after substituing for $x,y,z$ and $dx\;ddy\;dz=J\;dr\;d\theta\;d\phi$ ,

$\displaystyle \begin{aligned} \iiint_{S}xyz\;dx\;dy\;dz&=\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{a} r^5 \sin^3\theta\cos\theta\sin\phi\cos\phi\;dr\;d\theta\;d\phi \\ &= \int_0^{\pi/2} \sin^3\theta\cos\theta d\theta\;\int_0^{\pi/2}\sin\phi\cos\phi\;d\phi\;\int_0^{a}r^5\;dr \\ &=\left(\dfrac{1}{4}\right)\times \left(\dfrac{2}{4}\right)\times \left(\dfrac{a^6}{6}\right) \\ &=\dfrac{a^6}{48} \end{aligned}$

Here $a=1$ which makes the answer $\displaystyle \boxed{1+48=49}$

The integral with the given boundary can be written as $\begin{aligned}&\int_0^1x\,dx\Bigg\{\int_0^{\beta(x)}y\,dy\bigg(\int_0^{\alpha (x,y)} z\,dz\bigg)\Bigg\}\quad \text{where } \alpha (x,y)=\sqrt{1-x^2-y^2} \text{ and } \beta(x)=\sqrt{1-x^2}&\\ &=\int_0^1x\,dx\Bigg\{\int_0^{\beta(x)}y\,dy\bigg(\frac{1-x^2-y^2}{2}\bigg)\Bigg\}&\\ &=\int_0^1x\,dx\Bigg\{\bigg(\frac{1-x^2}{2}\bigg)^2- \frac{(1-x^2)^2}{8}\Bigg\}&\\ &=\int_0^1x\,dx\Bigg\{\frac{(1-x^2)^2}{8}\Bigg\}&\\ &=\Bigg[-\frac{(1-x^2)^3}{48}\Bigg]_0^1=\frac{1}{48}&\end{aligned}$

$\implies \boxed{a+b=49}$