15.8 triple integrals in spherical coordinates: problem 1 (1 point) evaluate the triple integral of…

Explanation:

Step1: Convert to spherical coordinates

In spherical coordinates, $x = \rho\sin\varphi\cos\theta$, $y=\rho\sin\varphi\sin\theta$, $z = \rho\cos\varphi$, and $dV=\rho^{2}\sin\varphi d\rho d\varphi d\theta$, and $x^{2}+y^{2}+z^{2}=\rho^{2}$. The function $f(x,y,z)=z(x^{2}+y^{2}+z^{2})^{-\frac{3}{2}}$ becomes $f(\rho,\varphi,\theta)=\rho\cos\varphi\cdot\rho^{- 3}=\frac{\cos\varphi}{\rho^{2}}$. The equation of the ball $x^{2}+y^{2}+z^{2}\leq64$ gives $\rho\leq8$, and $z\geq4$ gives $\rho\cos\varphi\geq4$, so $\rho\geq\frac{4}{\cos\varphi}$.

Step2: Determine the limits of integration

For the region, $0\leq\theta\leq2\pi$ (full - rotation around the $z$ - axis). To find the limits for $\varphi$, from $z = \rho\cos\varphi$ and $\rho = 8$, when $z = 4$, we have $8\cos\varphi=4$, so $\cos\varphi=\frac{1}{2}$ and $\varphi=\frac{\pi}{3}$. So $0\leq\varphi\leq\frac{\pi}{3}$. And $\frac{4}{\cos\varphi}\leq\rho\leq8$.

Step3: Set up the triple - integral

The triple - integral $\iiint_{W}f(x,y,z)dV$ in spherical coordinates is $\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos\varphi}}^{8}\frac{\cos\varphi}{\rho^{2}}\cdot\rho^{2}\sin\varphi d\rho d\varphi d\theta=\int_{0}^{2\pi}d\theta\int_{0}^{\frac{\pi}{3}}\cos\varphi\sin\varphi\left(\int_{\frac{4}{\cos\varphi}}^{8}d\rho\right)d\varphi$.

Step4: Integrate with respect to $\rho$

$\int_{\frac{4}{\cos\varphi}}^{8}d\rho=8 - \frac{4}{\cos\varphi}$. So the integral becomes $\int_{0}^{2\pi}d\theta\int_{0}^{\frac{\pi}{3}}\cos\varphi\sin\varphi\left(8-\frac{4}{\cos\varphi}\right)d\varphi=\int_{0}^{2\pi}d\theta\int_{0}^{\frac{\pi}{3}}(8\cos\varphi\sin\varphi - 4\sin\varphi)d\varphi$.

Step5: Integrate with respect to $\varphi$

Let $u=\sin\varphi$, then $du=\cos\varphi d\varphi$. $\int_{0}^{\frac{\pi}{3}}(8\cos\varphi\sin\varphi - 4\sin\varphi)d\varphi=\int_{0}^{\frac{\sqrt{3}}{2}}(8u - 4)du=\left[4u^{2}-4u\right]_{0}^{\frac{\sqrt{3}}{2}}=4\times\frac{3}{4}-4\times\frac{\sqrt{3}}{2}=3 - 2\sqrt{3}$.

Step6: Integrate with respect to $\theta$

$\int_{0}^{2\pi}d\theta=2\pi$. So the value of the triple - integral is $2\pi(3 - 2\sqrt{3})$.

Answer:

$2\pi(3 - 2\sqrt{3})$