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 $x y$ -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,

r\;=\;\sqrt[3]{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