Question #119642
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
2020-06-02T17:44:06-0400

Since we havex2+y2+z2r2,we obtain ρ=0r,ϕ=0π andθ=02π.So,wegetDc dV =02π0π0rc ρ2 sin(ϕ) dρ dϕ dθ =c ρ330r cos(ϕ)0πθ02π  =c r33( cos(π)cos(0)(2π0)  =c r33(2)(2π)  =43πcr3So, the correct answer is   bSince \ we \ have \\ x^2+y^2+z^2\leq r^2, \\ we \ obtain\ \\ \rho=0\rightarrow r ,\\ \phi=0\rightarrow \pi\ and \\ \theta=0\rightarrow 2\pi. \\ So, we \, get\\ \iiint\limits_D c\ dV\ =\int\limits_0^{2\pi}\int\limits_0^{\pi}\int\limits_0^{r} c\ \rho^2\ sin(\phi ) \ d\rho \ d\phi\ d\theta\\ \\ \\ \qquad \qquad\ = -c\ \frac{\rho^3}{3}|_0^{r}\ cos(\phi )\large|_0^{\pi}\theta|_0^{2\pi}\\ \qquad \quad\ \ =-c\ \frac{r^3}{3}(\ cos(\pi )-cos(0)(2\pi-0)\\ \qquad \quad\ \ =-c\ \frac{r^3}{3}(-2)(2\pi)\\ \qquad \quad\ \ = \frac{4}{3} \pi c r^3\\ \text{So, the correct answer is} \ \ \ b


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