Answer to Question #119364 in Calculus for Olivia

Question

Let c,r be constants, and D={(x,y,z):x^2+y^2+z^2≤r^2}. The answer to ∭D cdV

is
Select one:
a. (π^2cr^3)/3


b. (4πcr^3)/3

c. 4πcr^3

d. (πcr^4)/3


e. (4πcr^2)/2

f. (4r^3)/3


g. (πcr^3)/3

Answer 1

Consider the region $D=\{ (x,y,z): x^2+y^2+z^2\leq r^2\}$

The region $D$ is inside the sphere of radius $r$ , so the region in spherical coordinates is given by,

$D=\{ (\rho,\phi,\theta)|0\leq\rho\leq r,0\leq\phi\leq \pi,0\leq\theta\leq 2\pi\}$

Now, evaluate the triple integral in spherical coordinates as,

$\iiint_DcdV=\int_0^{2\pi}\int_0^{\pi}\int_0^{r}c\rho^2\sin\phi d\rho d\phi d\theta$

$=\int_0^{2\pi}\int_0^{\pi}c[\frac{\rho^3}{3}]_0^{r}\sin\phi d\phi d\theta$

$=\frac{cr^3}{3}\int_0^{2\pi}\int_0^{\pi}\sin\phi d\phi d\theta$

$=\frac{cr^3}{3}\int_0^{2\pi}[ -\cos\phi]_0^{\pi}d\theta$

$=\frac{cr^3}{3}\int_0^{2\pi}(2)d\theta$

$=\frac{2cr^3}{3}\int_0^{2\pi}d\theta$

$=\frac{2cr^3}{3}[\theta ]_0^{2\pi}$

$=\frac{2cr^3}{3}(2\pi)$

$=\frac{4\pi cr^3}{3}$

Therefore, the triple integral is $\iiint_DcdV=\frac{4\pi cr^3}{3}$

Hence, option (b) is correct.