example of Riemann triple integral

Determine the volume of the solid in ${ℝ}^3$ by the part of the surface

$${(x^2+y^2+z^2)}^3=\mathrm{\hspace{0.33em}3}{a}^{3}xyz$$

being in the first octant ($a>0$).

Since $x^2+y^2+z^2$ is the squared distance of the point  $(x,y,z)$  from the origin, the solid is apparently defined by

$$D:=\{(x,y,z)\in {ℝ}^3\mathrm{⋮}x\geqq 0,y\geqq 0,z\geqq 0,{(x^2+y^2+z^2)}^3\leqq 3a^{3}xyz\}.$$

By the definition

$$\mathrm{𝐦𝐞𝐚𝐬}(D):=\int {\chi }_D(v)𝑑v$$

in the parent entry (http://planetmath.org/RiemannMultipleIntegral), the volume in the questionis

$V={\displaystyle {\int }_D}1𝑑v={\displaystyle {\iiint }_D}𝑑x𝑑y𝑑z.$(1)

For calculating the integral (1) we express it by the (geographic) spherical coordinates through

$\{\begin{array}{cc}x=r\cos \phi \cos \lambda \hfill & \\ y=r\cos \phi \sin \lambda \hfill & \\ z=r\sin \phi \hfill & \end{array}$

where the latitude angle $\phi$ of the position vector $\overrightarrow{r}$ is measured from the $xy$-plane (not as the colatitude $\varphi$ from the positive $z$-axis); $\lambda$ is the longitude.  For the change of coordinates, we need the Jacobian determinant

which is simplified to $r^2\cos \phi$.  The equation of the surface attains the form

$$r^6=\mathrm{\hspace{0.33em}3}{a}^3{r}^3{\cos }^2\phi \sin \phi \cos \lambda \sin \lambda ,$$

or

$$r=\sqrt[3]{3a^3{\cos }^2\phi \sin \phi \cos \lambda \sin \lambda }:=r(\phi ,\lambda ).$$

In the solid, we have  $0\leqq r\leqq r(\phi ,\lambda )$  and

$$r=\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{if only if}{\cos }^2\phi \sin \phi \cos \lambda \sin \lambda =\mathrm{\hspace{0.33em}0}.$$

Thus we can write

$$V={\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}}{\int }_0^{r(\phi ,\lambda )}{r}^2\cos \phi d\phi d\lambda dr=\frac{1}{3}{\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}}\left(\underset{r=0}{\overset{r(\phi ,\lambda )}{/}}{r}^3\right)\cos \phi d\phi d\lambda ,$$

getting then

$$V=a^3{\int }_0^{\frac{\pi }{2}}({\cos }^3\phi )(-\sin \phi )𝑑\phi \cdot {\int }_0^{\frac{\pi }{2}}(\cos \lambda )(-\sin \lambda )𝑑\lambda =a^3\underset{\phi =0}{\overset{\frac{\pi }{2}}{/}}\frac{{\cos }^4\phi }{4}\cdot \underset{\lambda =0}{\overset{\frac{\pi }{2}}{/}}\frac{{\cos }^2\lambda }{2}=\frac{a^3}{8}.$$

Remark.  The general for variable changing in a triple integral is

$${\iiint }_{D}f(x,y,z)𝑑x𝑑y𝑑z={\iiint }_{\mathrm{\Delta }}f(x(\xi ,\eta ,\zeta ),y(\xi ,\eta ,\zeta ),z(\xi ,\eta ,\zeta ))\left|\frac{\partial (x,y,z)}{\partial (\xi ,\eta ,\zeta )}\right|𝑑\xi 𝑑\eta 𝑑\zeta .$$