How do you Evaluate the integral by changing to cylindrical coordinates?

$\setminus {\int}_{- 2}^{2} \setminus {\int}_{- \setminus \sqrt{4 - {y}^{2}}}^{\setminus \sqrt{4 - {y}^{2}}} {\int}_{\sqrt{{x}^{2} + {y}^{2}}}^{2} \left(x z\right) \mathrm{dz} \mathrm{dx} \mathrm{dy}$

Jul 19, 2018

$\setminus {\int}_{- 2}^{2} \setminus {\int}_{- \setminus \sqrt{4 - {y}^{2}}}^{\setminus \sqrt{4 - {y}^{2}}} {\int}_{\sqrt{{x}^{2} + {y}^{2}}}^{2} \left(x z\right) \setminus \mathrm{dz} \setminus \mathrm{dx} \setminus \mathrm{dy}$

The limit on the innermost integral defines the volume inside the cone, vertex at origin, concentric with z-axis, radius 2 at $z = 2$.

The other limits lie on or outside this cone and so they can be simplified as follows in cylindrical:

$= \setminus {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2} {\int}_{r}^{2} {\underbrace{r \cos \theta \setminus z}}_{= x z} \cdot {\underbrace{r \setminus \mathrm{dz} \setminus \mathrm{dr} \setminus d \theta}}_{\equiv \mathrm{dA}}$

• Note that for every positive $x z$ isnside the cone, there is a corresponding negative $x z$ value, so one would expect the integration to result in zero.

$= \setminus {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2} {\left[{r}^{2} \cos \theta {z}^{2} / 2\right]}_{r}^{2} \setminus \mathrm{dr} \setminus d \theta$

$= \setminus {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2} 2 {r}^{2} \cos \theta - {r}^{4} / 2 \cos \theta \setminus \mathrm{dr} \setminus d \theta$

$= \setminus {\int}_{0}^{2 \pi} {\left[\frac{2}{3} {r}^{3} \cos \theta - {r}^{5} / 10 \cos \theta\right]}_{0}^{2} \setminus d \theta$

$= \frac{32}{15} \setminus {\int}_{0}^{2 \pi} \cos \theta \setminus d \theta = 0$