Answer to Question #119473 in Calculus for Olivia

Question #119473
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
1
Expert's answer
2020-06-07T16:19:00-0400

 The easiest way to do this is to make a switch to spherical coordinates. There ρ2=x2+y2+z2  anddxdydz=ρ2sinϕdρdθdφ.Dcdv=Dcdxdydz=Dcρ2sinφdρdθdφ Now we are integrating over a regionD.What isD? It is a sphere of radius r centered at the origin. So 0ρr,0θ2π , and 0φπDcρ2sinφdρdθdφ=0π02π0rcρ2sinφdρdθdφc02πdθ0πsinφdφ0rρ2dρ=c(θ02π)(cosφ0π)(ρ330r)=c(2π)((11))(r33)=4πcr33The answer is b\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}


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