Answer to Question #143571 in Calculus for Promise Omiponle

Question

Use spherical coordinates to evaluate the triple integral ∫∫∫E x^2+y^2+z^2 dV, where E is the ball: x^2+y^2+z^2≤16.

Answer 1

let's move on to spherical coordinates:

$x = rcos\varphi sin\theta ,y = rsin\varphi sin\theta ,z = rcos\theta$

$0 < r < 4,\,\,0 < \varphi < 2\pi ,\,\,0 < \theta < \pi$

Then

${x^2} + {y^2} + {z^2} = {r^2}$

$\begin{array}{l} \int {\int {\int\limits_E {({x^2} + {y^2} + {z^2})dV = \int\limits_0^{2\pi } {d\varphi } } } } \int\limits_0^4 {{r^2}dr} \int\limits_0^\pi {{r^2}\sin \theta d\theta } = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^4 {{r^4}dr} \int\limits_0^\pi {\sin \theta d\theta } = \\ = \left. \varphi \right|_0^{2\pi } \cdot \left. {\frac{{{r^5}}}{5}} \right|_0^4 \cdot \left. {( - \cos \theta )} \right|_0^\pi = 2\pi \cdot \frac{{{4^5}}}{5}( - \cos \pi + \cos 0) = \frac{{2048}}{5}\pi \cdot (1 + 1) = \frac{{4096\pi }}{5} \end{array}$

Answer: $\frac{{4096\pi }}{5}$