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

The right answer is (b).

As D is the sphere with the radius r and the centre (0, 0, 0), we can do this integral in spherical coordinates.

This is the relationship between the Cartesian and spherical coordinate systems:

$x=\rho sin\phi cos\theta,$

$y=\rho sin\phi sin\theta,$

$z=\rho cos\phi,$

$x^2+y^2+z^2=\rho^2.$

We also have such restrictions on the coordinates in D:

$0\le\rho\le r, 0\le\phi\le \pi, 0\le\theta\le 2\pi.$

Then $dV=\rho^2 sin\phi d\rho d\theta d\phi.$

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

$=\int_0^\pi \int_0^{2\pi} (\frac{c\rho^3}{3} |_0^r) sin\phi d\theta d\phi=$

$=\int_0^\pi \int_0^{2\pi} \frac{cr^3}{3} sin\phi d\theta d\phi= \int_0^\pi \frac{cr^3}{3}(\theta|_0^{2\pi})sin\phi d\phi=$

$=\int_0^\pi \frac{2\pi cr^3}{3} sin\phi d\phi==-\frac{2\pi cr^3}{3} (cos\phi |_0^{\pi})=$

$=-\frac{2\pi cr^3}{3} (cos\pi-cos0)=-\frac{2\pi cr^3}{3}*(-2)=\frac{4\pi cr^3}{3}.$