Section 15.8: Triple Integration in Spherical Coordinates

SOLUTION We plot the point in Figure 6. From Equations 1 we have \[ x = \rho \sin \phi \cos \theta = 2 \sin \frac{\pi}{3} \cos \frac{\pi}{4} = 2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right) = \sqrt{\frac{3}{2}} \] \[ y = \rho \sin \phi \sin \theta = 2 \sin \frac{\pi}{3} \sin \frac{\pi}{4} = 2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right) = \sqrt{\frac{3}{2}} \] \[ z = \rho \cos \phi = 2 \cos \frac{\pi}{3} = 2\left(\frac{1}{2}\right) = 1 \] Thus the point \((2, \pi/4, \pi/3)\) is \((\sqrt{3/2}, \sqrt{3/2}, 1)\) in rectangular coordinates.

SOLUTION From Equation 2 we have \[ \rho = \sqrt{x^2+y^2+z^2} = \sqrt{0 + 12 + 4} = 4 \] and so Equations 1 give \[ \cos \phi = \frac{z}{\rho} = \frac{-2}{4} = -\frac{1}{2} \qquad \phi = \frac{2\pi}{3} \] \[ \cos \theta = \frac{x}{\rho \sin \phi} = 0 \qquad \theta = \frac{\pi}{2} \] (Note that \(\theta \neq 3\pi/2\) because \(y = 2\sqrt{3} > 0\).) Therefore spherical coordinates of the given point are \((4, \pi/2, 2\pi/3)\).