15.8 triple integrals in spherical coordinates: problem 2 (1 point) evaluate, in spherical coordinates, the…

Explanation:

Step1: Recall triple - integral formula in spherical coordinates

The triple - integral in spherical coordinates is $\iiint_{E}f(\rho,\theta,\phi)\ dV=\int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2}\int_{\rho_1}^{\rho_2}f(\rho,\theta,\phi)\rho^{2}\sin\phi\ d\rho\ d\phi\ d\theta$. Here, $f(\rho,\theta,\phi)=\sin\phi$, $\theta_1 = 0$, $\theta_2=2\pi$, $\phi_1=\frac{\pi}{3}$, $\phi_2=\frac{\pi}{2}$, $\rho_1 = 1$, $\rho_2 = 5$. So the integral becomes $\int_{0}^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{1}^{5}\sin\phi\cdot\rho^{2}\sin\phi\ d\rho\ d\phi\ d\theta=\int_{0}^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{1}^{5}\rho^{2}\sin^{2}\phi\ d\rho\ d\phi\ d\theta$.

Step2: Integrate with respect to $\rho$

First, integrate $\int_{1}^{5}\rho^{2}d\rho$. Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int_{1}^{5}\rho^{2}d\rho=\left[\frac{\rho^{3}}{3}\right]_{1}^{5}=\frac{5^{3}}{3}-\frac{1^{3}}{3}=\frac{125 - 1}{3}=\frac{124}{3}$.

Step3: Integrate with respect to $\phi$

Next, integrate $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin^{2}\phi\ d\phi$. Since $\sin^{2}\phi=\frac{1-\cos(2\phi)}{2}$, then $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin^{2}\phi\ d\phi=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{1 - \cos(2\phi)}{2}d\phi=\frac{1}{2}\left[\phi-\frac{\sin(2\phi)}{2}\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}}=\frac{1}{2}\left[\left(\frac{\pi}{2}-0\right)-\left(\frac{\pi}{3}-\frac{\sin(\frac{2\pi}{3})}{2}\right)\right]=\frac{1}{2}\left[\frac{\pi}{2}-\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right]=\frac{1}{2}\left[\frac{\pi}{6}+\frac{\sqrt{3}}{4}\right]$.

Step4: Integrate with respect to $\theta$

Finally, integrate $\int_{0}^{2\pi}d\theta$. We know that $\int_{0}^{2\pi}d\theta=\left[\theta\right]_{0}^{2\pi}=2\pi$.

Step5: Calculate the final result

Multiply the results of the three integrations together: [ \begin{align*} &\frac{124}{3}\times\frac{1}{2}\left(\frac{\pi}{6}+\frac{\sqrt{3}}{4}\right)\times2\pi\ =&\frac{124\pi}{3}\left(\frac{\pi}{6}+\frac{\sqrt{3}}{4}\right)\ =&\frac{124\pi^{2}}{18}+\frac{124\pi\sqrt{3}}{12}\ =&\frac{62\pi^{2}}{9}+\frac{31\pi\sqrt{3}}{3} \end{align*} ]

Answer:

$\frac{62\pi^{2}}{9}+\frac{31\pi\sqrt{3}}{3}$