# example of Riemann triple integral

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

 $(x^{2}\!+\!y^{2}\!+\!z^{2})^{3}\;=\;3a^{3}xyz$

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

 $D\;:=\;\{(x,\,y,\,z)\in\mathbb{R}^{3}\,\vdots\;\,x\geqq 0,\;\,y\geqq 0,\;\,z% \geqq 0,\;\,(x^{2}\!+\!y^{2}\!+\!z^{2})^{3}\leqq 3a^{3}xyz\}.$

By the definition

 $\mathbf{meas}(D)\;:=\>\int\chi_{D}(v)\,dv$

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

 $\displaystyle V\;=\;\int_{D}1\,dv\;=\;\iiint_{D}dx\,dy\,dz.$ (1)

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

 $\displaystyle\begin{cases}x\;=\;r\cos\varphi\cos\lambda\\ y\;=\;r\cos\varphi\sin\lambda\\ z\;=\;r\sin\varphi\end{cases}$

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

 $\displaystyle\frac{\partial(x,\,y,\,z)}{\partial(r,\,\varphi,\,\lambda)}\>=\;% \left|\begin{matrix}\frac{\partial x}{\partial r}&\frac{\partial y}{\partial r% }&\frac{\partial z}{\partial r}\\ \frac{\partial x}{\partial\varphi}&\frac{\partial y}{\partial\varphi}&\frac{% \partial z}{\partial\varphi}\\ \frac{\partial x}{\partial\lambda}&\frac{\partial y}{\partial\lambda}&\frac{% \partial z}{\partial\lambda}\end{matrix}\right|\;=\;\left|\begin{matrix}\cos% \varphi\cos\lambda&\cos\varphi\sin\lambda&\sin\varphi\\ -r\sin\varphi\cos\lambda&-r\sin\varphi\sin\lambda&r\cos\varphi\\ -r\cos\varphi\sin\lambda&r\cos\varphi\cos\lambda&0\end{matrix}\right|,$

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

 $r^{6}\;=\;3a^{3}r^{3}\cos^{2}\varphi\sin\varphi\cos\lambda\sin\lambda,$

or

 $r\;=\;\sqrt{3a^{3}\cos^{2}\varphi\sin\varphi\cos\lambda\sin\lambda}\;:=\;r(% \varphi,\,\lambda).$

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

 $r\;=\;0\quad\mbox{if only if}\;\;\cos^{2}\varphi\sin\varphi\cos\lambda\sin% \lambda\;=\;0.$

Thus we can write

 $V\;=\;\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{r(\varphi,\,% \lambda)}\!r^{2}\cos\varphi\;d\varphi\;d\lambda\;dr\;=\;\frac{1}{3}\int_{0}^{% \frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\left(\operatornamewithlimits{\Big{/}}_{% \!\!\!r=0}^{\,\quad r(\varphi,\,\lambda)}\!r^{3}\right)\cos\varphi\;d\varphi\;% d\lambda,$

getting then

 $V\;=\;a^{3}\int_{0}^{\frac{\pi}{2}}\!(\cos^{3}\varphi)(-\sin\varphi)\,d\varphi% \cdot\int_{0}^{\frac{\pi}{2}}\!(\cos\lambda)(-\sin\lambda)\,d\lambda\;=\;a^{3}% \!\operatornamewithlimits{\Big{/}}_{\!\!\!\varphi=0}^{\,\quad\frac{\pi}{2}}\!% \frac{\cos^{4}\varphi}{4}\cdot\!\operatornamewithlimits{\Big{/}}_{\!\!\!% \lambda=0}^{\,\quad\frac{\pi}{2}}\!\frac{\cos^{2}\lambda}{2}\;=\;\frac{a^{3}}{% 8}.$

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

 $\iiint_{D}f(x,\,y,\,z)\,dx\,dy\,dz\>=\;\iiint_{\Delta}\!f(x(\xi,\,\eta,\,\zeta% ),\,y(\xi,\,\eta,\,\zeta),\,z(\xi,\,\eta,\,\zeta))\left|\frac{\partial(x,\,y,% \,z)}{\partial(\xi,\,\eta,\,\zeta)}\right|\,d\xi\,d\eta\,d\zeta.$
 Title example of Riemann triple integral Canonical name ExampleOfRiemannTripleIntegral Date of creation 2013-03-22 19:10:59 Last modified on 2013-03-22 19:10:59 Owner pahio (2872) Last modified by pahio (2872) Numerical id 14 Author pahio (2872) Entry type Example Classification msc 26A42 Synonym volume as triple integral Related topic Volume2 Related topic VolumeAsIntegral Related topic SubstitutionNotation Related topic ChangeOfVariablesInIntegralOnMathbbRn Related topic ExampleOfRiemannDoubleIntegral