Put spherical coordinates to use. Consider the triple integral&nbsp;(x2+y2)dV, where E lies between the spheres&nbsp;x2+y2+z2=4&nbsp;and&nbsp;x2+y2+z2=9

So,x2+y2=(ρsin⁡ϕ)2+(ρsin⁡ϕsin⁡θ)2&nbsp;=ρ2sin2ϕ(cos2θ+sin2θ)&nbsp;=ρ2sin2ϕ∫0π∫02π∫23(ρ2sin2ϕ)p2sin⁡ϕdpdθdϕ&nbsp;∫0π∫02π∫23ρ4sin3ϕdpdθdϕ&nbsp;∫0π∫02π15ρ5sin3ϕdθdϕ&nbsp;∫0π2115⋅2πsin3ϕdϕ&nbsp;Rewrite&nbsp;sin2=1−cos2&nbsp;422π5∫0πsin⁡ϕ(1−cos2ϕ)dϕ&nbsp;u=cos⁡ϕ,&nbsp;du=−sin⁡ϕdϕ&nbsp;422π5∫1−1−(1−u2)du&nbsp;422π5∫−111−u2du&nbsp;422π5(u−13u3)∣−11=1688π15&nbsp;Result:&nbsp;1688π15

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usagirl007A

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2021-10-07

Put spherical coordinates to use. Consider the triple integral $(x^{2} + y^{2}) d V$ , where E lies between the spheres $x^{2} + y^{2} + z^{2} = 4$ and $x^{2} + y^{2} + z^{2} = 9$

Answer & Explanation

Nathanael Webber

Skilled2021-10-08Added 117 answers

So,
$x^{2} + y^{2} = {(ρ \sin ϕ)}^{2} + {(ρ \sin ϕ \sin θ)}^{2}$
$= ρ^{2} \sin^{2} ϕ (\cos^{2} θ + \sin^{2} θ)$
$= ρ^{2} \sin^{2} ϕ$
$\int_{0}^{π} \int_{0}^{2 π} \int_{2}^{3} (ρ^{2} \sin^{2} ϕ) p^{2} \sin ϕ d p d θ d ϕ$
$\int_{0}^{π} \int_{0}^{2 π} \int_{2}^{3} ρ^{4} \sin^{3} ϕ d p d θ d ϕ$
$\int_{0}^{π} \int_{0}^{2 π} \frac{1}{5} ρ^{5} \sin^{3} ϕ d θ d ϕ$
$\int_{0}^{π} \frac{211}{5} \cdot 2 π \sin^{3} ϕ d ϕ$
Rewrite $\sin^{2} = 1 - \cos^{2}$
$\frac{422 π}{5} \int_{0}^{π} \sin ϕ (1 - \cos^{2} ϕ) d ϕ$
$u = \cos ϕ, d u = - \sin ϕ d ϕ$
$\frac{422 π}{5} \int_{1}^{- 1} - (1 - u^{2}) d u$
$\frac{422 π}{5} \int_{- 1}^{1} 1 - u^{2} d u$
$\frac{422 π}{5} (u - \frac{1}{3} u^{3}) ∣_{- 1}^{1} = \frac{1688 π}{15}$
Result: $\frac{1688 π}{15}$

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