Answer to Question #119364 in Calculus for Olivia

Question #119364
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
2020-06-04T18:27:28-0400

Consider the region D={(x,y,z):x2+y2+z2r2}D=\{ (x,y,z): x^2+y^2+z^2\leq r^2\}


The region DD is inside the sphere of radius rr , so the region in spherical coordinates is given by,


D={(ρ,ϕ,θ)0ρr,0ϕπ,0θ2π}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,


DcdV=02π0π0rcρ2sinϕdρdϕdθ\iiint_DcdV=\int_0^{2\pi}\int_0^{\pi}\int_0^{r}c\rho^2\sin\phi d\rho d\phi d\theta


=02π0πc[ρ33]0rsinϕdϕdθ=\int_0^{2\pi}\int_0^{\pi}c[\frac{\rho^3}{3}]_0^{r}\sin\phi d\phi d\theta


=cr3302π0πsinϕdϕdθ=\frac{cr^3}{3}\int_0^{2\pi}\int_0^{\pi}\sin\phi d\phi d\theta


=cr3302π[cosϕ]0πdθ=\frac{cr^3}{3}\int_0^{2\pi}[ -\cos\phi]_0^{\pi}d\theta


=cr3302π(2)dθ=\frac{cr^3}{3}\int_0^{2\pi}(2)d\theta


=2cr3302πdθ=\frac{2cr^3}{3}\int_0^{2\pi}d\theta


=2cr33[θ]02π=\frac{2cr^3}{3}[\theta ]_0^{2\pi}


=2cr33(2π)=\frac{2cr^3}{3}(2\pi)


=4πcr33=\frac{4\pi cr^3}{3}


Therefore, the triple integral is DcdV=4πcr33\iiint_DcdV=\frac{4\pi cr^3}{3}


Hence, option (b) is correct.

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