Calculus – Multiple integrals – Triple integrals in spherical coordinates

Spherical coordinates are:

\[\left\{\begin{array}{l}x=\rho\sin(\phi)\cos(\theta)\\[2.5mm]y=\rho\sin(\phi)\sin(\theta)\\[2.5mm]z=\rho\cos(\phi)\end{array}\right. \quad\Longrightarrow\quad \rho^2=x^2+y^2+z^2.\]

The volume element equals: \(dV=\rho^2\sin(\phi)\,d\rho\,d\phi\,d\theta\).

The counterpart of a rectangular box in spherical coordinates is a spherical wedge given by \(a\leq\rho\leq b\), \(\alpha\leq\theta\leq\beta\) and \(\gamma\leq\phi\leq\delta\), where \(a\geq0\), \(\beta-\alpha\leq2\pi\) and \(\delta-\gamma\leq\pi\). The volume of such a wedge is approximately equal to that of a box with sides of length \(d\rho\), \(\rho\,d\phi\) and \(r\,d\theta\), where \(r=\rho\sin(\phi)\) denotes the distance to the \(z\)-axis, which is \(d\rho\cdot\rho\,d\phi\cdot\rho\sin(\phi)\,d\theta=\rho^2\sin(\phi)\,d\rho\,d\phi\,d\theta\).

Suppose that

\[E=\{(\rho\sin(\phi)\cos(\theta),\rho\sin(\phi)\sin(\theta),\rho\cos(\phi)\,|\,\alpha\leq\theta\leq\beta,\;\gamma\leq\phi\leq\delta,\;u_1(\phi,\theta))\leq\rho\leq u_2(\phi,\theta)\},\]

then we have:

\[\iiint\limits_Ef(x,y,z)\,dV=\int_{\alpha}^{\beta}\int_{\gamma}^{\delta}\int_{u_1(\phi,\theta)}^{u_2(\phi,\theta)} f(\rho\sin(\phi)\cos(\theta),\rho\sin(\phi)\sin(\theta),\rho\cos(\phi))\,\rho^2\sin(\phi)\,d\rho\,d\phi\,d\theta.\]

Stewart §15.8, Example 3
Evaluate \(\displaystyle\iiint\limits_Be^{(x^2+y^2+z^2)^{3/2}}\,dV\), where \(B\) is the unit ball \(B=\{(x,y,z)\,|\,x^2+y^2+z^2\leq1\}\).

Solution: In rectangular coordinates the integral would be

\[\iiint\limits_Be^{(x^2+y^2+z^2)^{3/2}}\,dV=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}}e^{(x^2+y^2+z^2)^{3/2}}\,dz,\,dy\,dx,\]

which would be extremely difficult to evaluate. However, in spherical coordinates we have \(x^2+y^2+z^2=\rho^2\) and

\[B=\{(\rho\sin(\phi)\cos(\theta),\rho\sin(\phi)\sin(\theta),\rho\cos(\phi)\,|\,0\leq\rho\leq1,\;0\leq\phi\leq\pi,\;0\leq\theta\leq2\pi\}.\]

Hence we obtain

\begin{align*} \iiint\limits_Be^{(x^2+y^2+z^2)^{3/2}}\,dV&=\int_0^{2\pi}\int_0^{\pi}\int_0^1e^{(\rho^2)^{3/2}}\rho^2\sin(\phi)\,d\rho\,d\phi\,d\theta =\int_0^{2\pi}d\theta\int_0^{\phi}\sin(\phi)\,d\phi\int_0^1\rho^2e^{\rho^3}\,d\rho\\[2.5mm] &=2\pi\bigg[-\cos(\phi)\bigg]_{\phi=0}^{\pi}\bigg[\frac{1}{3}e^{\rho^3}\bigg]_{\rho=0}^1=2\pi\cdot2\cdot\frac{1}{3}(e-1) =\frac{4}{3}\pi(e-1). \end{align*}

Stewart §15.8, Example 4
Use spherical coordinates to find the volume of the solid that lies above the cone \(z=\sqrt{x^2+y^2}\) and within the sphere \(x^2+y^2+z^2=z\).

Solution: Note that

\[x^2+y^2+z^2=z\quad\Longleftrightarrow\quad x^2+y^2+(z-\tfrac{1}{2})^2=\tfrac{1}{4},\]

which implies that it is the sphere with center \((0,0,\frac{1}{2})\) and radius \(\frac{1}{2}\). In spherical coordinates we have

\[x^2+y^2+z^2=z\quad\Longleftrightarrow\quad\rho^2=\rho\cos(\phi)\quad\Longrightarrow\quad\rho=\cos(\phi).\]

For the cone we obtain

\[z=\sqrt{x^2+y^2}\quad\Longleftrightarrow\quad\rho\cos(\phi)=r=\rho\sin(\phi)\quad\Longrightarrow\quad\cos(\phi)=\sin(\phi) \quad\Longrightarrow\quad\phi=\tfrac{1}{4}\pi.\]

Hence, the volume equals

\begin{align*} \iiint\limits_EdV&=\int_0^{2\pi}\int_0^{\frac{1}{4}\pi}\int_0^{\cos(\phi)}\rho^2\sin(\phi)\,d\rho\,d\phi\,d\theta =\int_0^{2\pi}d\theta\int_0^{\frac{1}{4}\pi}\sin(\phi)\bigg[\frac{1}{3}\rho^2\bigg]_{\rho=0}^{\cos(\phi)}\,d\phi =\frac{2}{3}\pi\int_0^{\frac{1}{4}\pi}\cos^3(\phi)\sin(\phi)\,d\phi\\[2.5mm] &=\frac{2}{3}\pi\bigg[-\frac{1}{4}\cos^4(\phi)\bigg]_{\phi=0}^{\frac{1}{4}\pi} =\frac{2}{3}\pi\cdot\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{2}{3}\pi\cdot\frac{3}{16}=\frac{1}{8}\pi. \end{align*}

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Last modified on October 4, 2021