Question

g Evaluate the integral below by changing to spherical coordinates. 3 −3 9 − y2 − 9 − y2 9 − x2 − y2 x2z y2z z3 dz dx dy − 9 − x2 − y2

Answer 1

My best guess as to what the integral is supposed to be is

$\displaystyle\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_{-\sqrt{9-x^2-y^2}}^{\sqrt{9-x^2-y^2}}(x^2z+y^2z+z^3)\,\mathrm dz\,\mathrm dx\,\mathrm dy$

which is the integral over the sphere of radius 3 centered at (0, 0, 0). In spherical coordinates, the integral is equivalent to

$\displaystyle\int_0^{2\pi}\int_0^{\pi}\int_0^3(\rho^3\cos^2\theta\sin^2\varphi\cos\varphi+\rho^3\sin^2\theta\sin^2\varphi\cos\varphi+\rho^3\cos^3\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^{\pi}\int_0^3\rho^5\sin\varphi\cos\varphi\,\mathrm d\rho\,\mathrm d\varphi$

$=\displaystyle\pi\int_0^{\pi}\int_0^3\rho^5\sin(2\varphi)\,\mathrm d\rho\,\mathrm d\varphi$

$=\displaystyle\pi\left(\int_0^{\pi}\sin(2\varphi)\,\mathrm d\varphi\right)\left(\int_0^3\rho^5\,\mathrm d\rho\right)=\boxed0$