Triple Integrals in Spherical Coordinates — Question 1

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Question 1

Let EE be the part of the spherical shell 1≤x2+y2+z2≤31\le\sqrt{x^2+y^2+z^2}\le 3 in the first octant. Evaluate ∭E1x2+y2+z2dV\iiint_E\frac{1}{\sqrt{x^2+y^2+z^2}}\,dV using spherical coordinates.

Tasks

  1. Convert the region, integrand, and volume element.

  2. Evaluate the integral.

  3. Check the result using the integrand range and region volume.

Original worksheet page 1: question and worked solution for 4-7-001
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Question 1 – Solution

Strategy. A first-octant spherical shell has constant bounds in all three spherical variables, and the radial integrand cancels one power of the Jacobian.

Step 1: Convert Using x=ρsin⁡ϕcos⁡θ,y=ρsin⁡ϕsin⁡θ,z=ρcos⁡ϕ,x=\rho\sin\phi\cos\theta,\quad y=\rho\sin\phi\sin\theta, \quad z=\rho\cos\phi, the bounds are 0≤θ≤π2,0≤ϕ≤π2,1≤ρ≤3,0\le\theta\le\frac\pi 2,\qquad 0\le\phi\le\frac\pi 2, \qquad 1\le\rho\le 3, and dV=ρ2sin⁡ϕdρdϕdθdV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

Step 2: Evaluate I=∫0π/2∫0π/2∫131ρρ2sin⁡ϕdρdϕdθ=(π2)(1)[ρ22]13=2π.\begin{align*} I&=\int_0^{\pi/2}\int_0^{\pi/2}\int_1^3 \frac 1\rho\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\left(\frac\pi 2\right)(1)\left[\frac{\rho^2}{2}\right]_1^3 =\boxed{2\pi}. \end{align*}

Verification The shell-octant volume is 184π3(33−13)=13π/3\frac 18\frac{4\pi}{3}(3^3-1^3)=13\pi/3. Hence the average integrand is I/V=6/13I/V=6/13, which lies between its endpoint values 1/31/3 and 11.

Original worksheet page 2: question and worked solution for 4-7-001

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