Question #119856
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
2020-06-04T21:23:18-0400

As we havex2+y2+z2r2we getρ=0r,ϕ=0π and θ=02π So, we obtainDcdV=02π0π0rcρ2sin(ϕ)dρdϕdθ=cρ330rcos(ϕ)0πθ02π=cr33(cos(π)cos(0)(2π0))=cr33(2)(2π)=43πcr3Then, the correct answer is (b)\text{As 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{Then, the correct answer is (b)}


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