Answer to Question #119890 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

$\text { The easiest way to do this is to make a switch to spherical coordinates.} \\[1 em] \text { There }\rho^{2}=x^{2}+y^{2}+z^{2} \text { and} d x d y d z=\rho^{2} \sin \phi d \rho d\theta d \varphi. \\[1 em] \iiint_{D} c dv=\iiint_{D} c d x d y d z =\iiint_{D} c \cdot \rho^{2} \sin \varphi d \rho d\theta d \varphi \\[1 em] \text { Now we are integrating over a region} D.\\[1 em] \text{What is} D\text{? It is a sphere of radius r centered at the origin.}\\[1 em] \begin{array}{l} \text { So } 0 \leq \rho \leq r,0 \leq \theta \leq 2 \pi \text { , and } 0 \leq \varphi \leq \pi \\[1 em] \qquad \iiint_{D}c \rho^{2} \sin \varphi d \rho d \theta d \varphi=\int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{r} c \rho^{2} \sin \varphi d \rho d \theta d \varphi \end{array} \\[1 em] \\[1 em] \begin{aligned} c\int_{0}^{2 \pi} d \theta \int_{0}^{\pi} \sin \varphi d \varphi \int_{0}^{r} \rho^{2} d \rho &= c (\theta\bigg| _{0}^{2 \pi} )(-\cos\varphi\bigg| _{0}^{\pi} )(\frac{ \rho^{3} }{3}\bigg| _{0}^{r}) \\ &=c(2 \pi)(-(-1-1))(\frac{ r^{3} }{3})\\ &=\frac{4\pi cr^3}{3} \end{aligned}\\ \text{The answer is } \fcolorbox{red}{aqua}{b}$