Finding a Triple Integral Using Spherical Coordinates

Use spherical coordinates to find the integral \[ \begin{equation*} \iiint\limits_{E}z\,{dx}\,{dy}\,{dz} \end{equation*} \]

where \(E\) is the solid defined by the inequalities \[ 0\leq \sqrt{x^{2}+y^{2}}\leq z\qquad 0\leq x^{2}+y^{2}+z^{2}\leq 1 \]

Solution We convert the rectangular coordinates to spherical coordinates to obtain \(E\) in spherical coordinates. Since \(\rho = \sqrt{ x^{2}+y^{2}+z^{2}}\) and \(x^{2}+y^{2}+z^{2}\leq 1,\) we have \(0\leq \rho \leq 1 \). This is the upper half of a spherical solid of radius 1. Next, the surface \(z= \sqrt{x^{2}+y^{2}}\) (which is a half-cone) can be written \( x^{2}+y^{2}=z^{2}\). The cross section of this cone with the \(xz\)-plane is the pair of lines \(z^{2}=x^{2}\) or \(z=\pm x\), each forming an angle \(\phi = \dfrac{\pi }{4}\) with the positive \(z\)-axis. See Figure 60.

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Since \(z=\rho \cos \phi\), the integral becomes \[ \begin{eqnarray*} \iiint\limits_{\kern-3ptE}z\,{dx}\,{dy}\,{dz}&=&\iiint\limits_{\kern-3ptE}\rho \cos \phi \cdot\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta \\ &=&\int_{0}^{2\pi }\int_{0}^{\pi/4}\int_{0}^{1}\rho ^{3}\sin \phi \cos \phi \,d\rho \,d\phi \,d\theta \\ &=&\int_{0}^{2\pi }\int_{0}^{\pi /4}\left[ \int_{0}^{1}\rho ^{3}d\rho \right] \sin \phi \cos \phi \,d\phi \,d\theta =\int_{0}^{2\pi }\int_{0}^{\pi /4}\dfrac{1}{4}\sin \phi \cos \phi \,d\phi \,d\theta\\ &=&\dfrac{1}{4}\int_{0}^{2\pi }\left[ \int_{0}^{\pi /4}\sin \phi \cos \phi\,d\phi \right] d\theta \underset{\underset{\underset{\color{#0066A7}{\hbox{when \(\phi =0,u=1\), when \(\phi =\dfrac{\pi }{4}\), \(u=\dfrac{\sqrt{2}}{2}\)}}}{\color{#0066A7}{\hbox{\(u=\cos \phi, du=-\sin \phi \,d\phi\)}}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{1}{4}\int_{0}^{2\pi }\left[ \int_{1}^{\sqrt{2}/2}-u\,{\it du}\right] d\theta \\ &=&\dfrac{1}{4}\int_{0}^{2\pi }\left[ -\dfrac{u^{2}}{2}\right] _{1}^{\sqrt{2}/2}d\theta =\dfrac{1}{4}\int_{0}^{2\pi }\left( -\dfrac{1}{4}+\dfrac{1}{2}\right) d\theta =\dfrac{1}{16}\int_{0}^{2\pi}\,d\theta =\dfrac{\pi}{8} \end{eqnarray*} \]