Answer to Question #119690 in Calculus for Olivia

Question #119690
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
2020-06-03T19:24:31-0400

Correct option is (b).

Reason:



Given ,c,rc,r are constants.

Assume radius is RR instead of rr for the time being,then replace RR by rr in the final result.

DD is the region described as D={(x,y,z):x2+y2+z2r2}D=\{(x,y,z):x^2+y^2+z^2≤r^2\} ,clearly DD contains all the points lies inside and surface of the sphere whose radius is rr .

Thus,

I=DcdV=cDdVI=\int \int\int_{D}cdV=c\int \int\int_{D}dV

Let's draw the elemental volume


Hence,

I=cDdV=c0R0π02πr2sin(θ)dθdϕdr    I=c0R(r20π(sin(θ)02πdϕ)dθ)dr=c4πR33I=c\int \int\int_{D}dV=c\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}r^2\sin(\theta)d\theta d\phi dr\\ \implies I=c\int_{0}^{R}\bigg(r^2\int_{0}^{\pi}\bigg(\sin(\theta)\int_{0}^{2\pi}d\phi \bigg)d\theta \bigg)dr=c\frac{4\pi R^3}{3}

Therefore the final answer is

I=c4πr33I=c\frac{4\pi r^3}{3}


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