OBJECTIVES

When you finish this section, you should be able to:

1. Convert rectangular coordinates to spherical coordinates (p. 956)
2. Find a triple integral using spherical coordinates (p. 958)

If the spherical coordinates of a point $P$ are $\left( 4,\dfrac{\pi }{6}, \dfrac{2\pi }{3}\right)$, find the rectangular coordinates of $P$.

Solution We use the equations (1) to convert spherical coordinates to rectangular coordinates. Then $$\begin{eqnarray*} x& =&\rho \sin \phi \cos \theta =4\sin \dfrac{2\pi }{3}\cos \dfrac{\pi }{6} =4\!\left( \dfrac{\sqrt{3}}{2}\right) \left( \dfrac{\sqrt{3}}{2}\right) =3 \qquad\!\! {\color {#0066A7}{\hbox{\(\rho =4; \theta =\dfrac{\pi}{6}, \phi =\dfrac{2\pi }{3}\)}}}\\ y& =&\rho \sin \phi \sin \theta =4\sin \dfrac{2\pi }{3}\sin \dfrac{\pi }{6} =4\!\left( \dfrac{\sqrt{3}}{2}\right) \left( \dfrac{1}{2}\right) =\sqrt{3} \\[4pt] z& =&\rho \cos \phi =4\cos \dfrac{2\pi }{3}=4\!\left( -\dfrac{1}{2}\right) =-2 \end{eqnarray*}$$

The rectangular coordinates of $P$ are $( 3,\sqrt{3},-2)$.

NOW WORK

Problem 7.

If the rectangular coordinates of a point $P$ are $( 1,\sqrt{3} ,-2)$, find the spherical coordinates of $P$.

Solution We use the equations (2) to convert rectangular coordinates to spherical coordinates. Then $$\begin{eqnarray*} \rho &=&\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{1+3+4}=\sqrt{8}=2\sqrt{2} \\ \tan \theta &=&\dfrac{y}{x}=\dfrac{\sqrt{3}}{1}, \hbox{ so }\theta =\tan ^{-1}\sqrt{3}=\dfrac{\pi }{3}\qquad \\ \cos \phi &=&\dfrac{z}{\rho }=\dfrac{-2}{2\sqrt{2}}=-\dfrac{\sqrt{2}}{2}, \hbox{ so }\phi =\cos ^{-1}\!\left( -\dfrac{\sqrt{2}}{2}\right) =\dfrac{ 3\pi }{4} \end{eqnarray*}$$

The spherical coordinates of $P$ are $\!\left( 2\sqrt{2},\dfrac{\pi }{3},\dfrac{3\pi }{4}\right)$.

NOW WORK

Problem 9.

NOTE

Points $(\rho ,\theta ,\phi )$ are called spherical coordinates because the surface formed when $\rho =a$, where $a\gt 0$ is a constant, is a sphere.

NOTE

Notice that the integrand contains the factor $\rho ^{2}\sin \phi$ because in spherical coordinates the differential ${\it dV}$ of the volume $V$ is ${\it dV}=\rho^{2}\sin \phi \,d\rho \,d\theta \,d\phi$.

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 $$0\leq \sqrt{x^{2}+y^{2}}\leq z\qquad 0\leq x^{2}+y^{2}+z^{2}\leq 1$$

Figure 60

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.

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*}$$

NOW WORK

Problems 21 and 25.

Find the volume of the solid that is removed from a hemisphere of radius 1 when it is cut by a cone that makes an angle of $30^{\circ}$ with the positive $z$-axis.

Figure 61

Solution Figure 61 shows the part of the hemisphere cut by the cone, namely, the volume under the sphere and inside the cone. The upper surface is the hemisphere of radius $r=1$; the lower surface is the $xy$-plane $(z=0)$. The angle $\theta$ ranges from $0$ to $2\pi$ and $\phi$ ranges from $0$ to $\dfrac{\pi}{6}.$ $$0\leq \rho \leq 1\qquad 0\leq \theta \leq 2\pi \qquad 0\leq \phi \leq \dfrac{\pi }{6}$$

Then the volume $V$ of the solid is $$\begin{eqnarray*} V &=&\iiint\limits_{E}\,\rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi =\int_{0}^{\pi /6}\int_{0}^{2\pi}\int_{0}^{1}\rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi\\ &=&\int_{0}^{\pi /6}\int_{0}^{2\pi}\left[\int_{0}^{1}\rho ^{2}d\rho \right] \sin\, \phi\, d\theta \,d\phi \notag \\ &=&\int_{0}^{\pi /6}\int_{0}^{2\pi }\left[ \dfrac{\rho ^{3}}{3}\right] _{0}^{1}\,d\theta \sin \phi \,d\phi =\int_{0}^{\pi /6}\left[\int_{0}^{2\pi } \dfrac{1}{3}d\theta \right] \sin \phi \,d\phi \notag \\ &=&\dfrac{2\pi}{3}\int_{0}^{\pi /6}\sin \phi \,d\phi =\left. \dfrac{2\pi }{3 }(-\cos \phi) \right] _{0}^{\pi /6}=\dfrac{2\pi }{3}\left( 1- \dfrac{\sqrt{3}}{2}\right) \end{eqnarray*}$$

NOW WORK

Problem 33.

NOTE

Since $\rho$ is used as a spherical coordinate, we use $\delta$ to represent the mass density.

Find the mass $M$ of a sphere if its mass density $\delta$ is proportional to the distance from the center of the sphere.

Solution Let $a$ be the radius of the sphere, and position the sphere so that its center is at the origin. In spherical coordinates, the equation of the sphere is $\rho =a$. The mass density $\delta$ of the sphere is $\delta =k\rho,$ where $k$ is the constant of proportionality. The mass $M$ is $$\begin{eqnarray*} M &=&\iiint\limits_{E}\delta \,{dV}=\iiint\limits_{E}k\rho \rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta =k\iiint\limits_{E}\rho ^{3}\sin \phi \,d\rho \,d\phi \,d\theta\\ &=&k\int_{0}^{2\pi }\int_{0}^{\pi }\left[ \int_{0}^{a}\rho ^{3}\,d\rho \right] \sin \phi \,d\phi \,d\theta \notag \\ & =&k\int_{0}^{2\pi}\int_{0}^{\pi }\left[ \dfrac{\rho ^{4}}{4}\right] _{0}^{a}\,\sin \phi \,d\phi \,d\theta =\dfrac{ka^{4}}{4}\int_{0}^{2\pi }\left[ \int_{0}^{\pi }\sin \phi \,d\phi \right] d\theta\\ &=&\dfrac{ka^{4}}{4} \int_{0}^{2\pi }\big[ -\cos \phi \big] _{0}^{\pi }\,d\theta \notag \\ & =&\dfrac{ka^{4}}{2}\int_{0}^{2\pi }\,d\theta =k\pi a^{4} \end{eqnarray*}$$

NOW WORK

Problem 37.

Greenwich, England, has a longitude of $0^\circ$ and an approximate latitude of $51.5^\circ$ N. If $R_{E}=3960$ miles is used for the radius of the Earth and degrees are used to measure angles instead of radians, then the spherical coordinates for Greenwich, England, are $(3960,0^{\circ},38.5^\circ )$. See Figure 62(b), where the angle $ECG=51.5^\circ$.