Question #120079
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
2020-06-09T18:17:06-0400

Since we havex2+y2+z2r2,we getρ=0rϕ=0π andθ=02πSo, we obtainDcdV=02π0π0rcρ2sin(ϕ)dρdϕdθ=cρ330rcos(ϕ)0πθ02π=cr33(cos(π)cos(0)(2π0))=cr33(2)(2π)=43πcr3So, the correct answer is   (b)\text{Since we have}\\ x^2+y^2+z^2\leq r^2,\\ we \ get\\ \rho=0 \rightarrow r\\ \phi=0 \rightarrow \pi \ \text{and}\\ \theta=0 \rightarrow 2 \pi\\ \text{So, we obtain}\\ \iiint\limits_{D} c d V=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{r} c \rho^{2} \sin (\phi) d \rho d \phi d \theta\\ =-\left.\left.\left.c \frac{\rho^{3}}{3}\right|_{0} ^{r} \cos (\phi)\right|_{0} ^{\pi} \theta\right|_{0} ^{2 \pi}\\ =-c \frac{r^{3}}{3}(\cos (\pi)-\cos (0)(2 \pi-0))\\ =-c \frac{r^{3}}{3}(-2)(2 \pi)\\ =\frac{4}{3} \pi c r^{3}\\ \text{So, the correct answer is }\ \ (b)


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