Triple Integrals in Spherical Coordinates

Introduction

Recall what we know about spherical coordinates:

Spherical coordinates diagram

x=ρcos⁡θsin⁡ϕy=ρsin⁡θsin⁡ϕz=ρcos⁡ϕρ2=x2+y2+y2\begin{aligned} x &= \rho \cos \theta \sin \phi\ y &= \rho \sin \theta \sin \phi\ z &= \rho \cos \phi\ \ \rho^2 &= x^2+y^2+y^2 \end{aligned}

Therefore, a triple integral in rectangular coordinates can be rewritten in terms of spherical coordinates: ∭Df(x,y,z) dV=∭Df(ρ,ϕ,θ) ρ2sin⁡ϕ dρ dϕ dθ\iiint_D f(x,y,z)\ dV = \iiint_D f(\rho, \phi, \theta)\ \rho^2 \sin \phi\ d\rho\ d\phi\ d\theta

We'll tend to use spherical coordinates when we encounter a triple integral with x2+y2+z2x^2+y^2+z^2 somewhere.

Examples

Convert the following integral to spherical coordinates and evaluate. ∭D(x2+y2+z2)−3/2 dV\iiint_D (x^2+y^2+z^2)^{-3/2}\ dV where DD is the region in the first octant between two spheres of radius 11 and 22 centered at the origin.

Solution

∭D(x2+y2+z2)−3/2 dV=∫0π/2∫0π/2∫12ρ−3⋅ρ2sin⁡ϕ dρ dϕ dθ=∫0π/2∫0π/2sin⁡ϕ(ln⁡ρ)∣12 dϕ dθ=∫0π/2∫0π/2sin⁡ϕ⋅ln⁡2 dϕ dθ=∫0π/2−ln⁡2⋅cos⁡ϕ∣0π/2 dθ=∫0π/2ln⁡2 dθ=π2⋅ln⁡2\begin{aligned} \iiint_D &(x^2+y^2+z^2)^{-3/2}\ dV\ &= \int_0^{\pi/2} \int_0^{\pi/2} \int_1^2 \rho^{-3} \cdot \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta\ &= \int_0^{\pi/2} \int_0^{\pi/2} \sin \phi(\ln \rho) \bigg|_1^2 \ d\phi\ d\theta\ &= \int_0^{\pi/2} \int_0^{\pi/2} \sin \phi \cdot \ln 2 \ d\phi\ d\theta\ &= \int_0^{\pi/2} -\ln 2 \cdot \cos \phi\bigg|_0^{\pi/2} \ d\theta\ &= \int_0^{\pi/2} \ln 2 \ d\theta = \ans{\dfrac{\pi}{2}\cdot \ln 2} \end{aligned}

Find the volume of the solid DD that lies inside the cone ϕ=π6\phi = \dfrac{\pi}{6} and inside the sphere ρ=4ρ=4.

Solution

Ice cream cone looking solid

V=∫02π∫0π/6∫04ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π/6643sin⁡ϕ dϕ dθ=∫02π−643(32−1) dθ=128π3−64π33\begin{aligned} V &= \int_0^{2\pi} \int_0^{\pi/6} \int_0^4 \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta\ &= \int_0^{2\pi} \int_0^{\pi/6} \dfrac{64}{3} \sin\phi\ d\phi\ d\theta\ &= \int_0^{2\pi} -\dfrac{64}{3}\left(\dfrac{\sqrt{3}}{2}-1\right)\ d\theta\ &= \ans{\dfrac{128\pi}{3}-\dfrac{64\pi\sqrt{3}}{3}} \end{aligned}

  1. Use spherical coordinates to evaluate ∭D(x2+y2+z2)5/2 dV\displaystyle\iiint_D (x^2+y^2+z^2)^{5/2}\ dV, where DD is the unit ball.

    2. ∫02π∫0π∫01ρ5⋅ρ2sin⁡ϕ dρ dϕ dθ=π2\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi} \displaystyle\int_0^{1} \rho^5\cdot\rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{\pi}{2}

  2. Use spherical coordinates to evaluate ∭D1(x2+y2+z2)3/2 dV\displaystyle\iiint_D \dfrac{1}{(x^2+y^2+z^2)^{3/2}}\ dV, where DD lies between spheres of radius 11 and 22 centered at the origin.

    3. ∫02π∫0π∫12ρ−3⋅ρ2sin⁡ϕ dρ dϕ dθ=4πln⁡2\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi} \displaystyle\int_1^{2} \rho^{-3}\cdot\rho^2\sin\phi\ d\rho\ d\phi\ d\theta = 4\pi \ln 2

  3. Use spherical coordinates to evaluate ∫0π∫0π/6∫2sec⁡ϕ4ρ2sin⁡ϕ dρ dϕ dθ.\displaystyle\int_{0}^{\pi} \displaystyle\int_{0}^{\pi/6} \displaystyle\int_{2\sec \phi}^{4} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta.

    4. π(1889−3233)\pi\left(\dfrac{188}{9}-\dfrac{32\sqrt{3}}{3}\right)

  4. Use spherical coordinates to find the volume of the solid bounded by the sphere ρ=2cos⁡ϕρ=2 \cos ϕ and the hemisphere ρ=1ρ=1, z≥0z≥0.

    5. ∫02π∫0π/3∫01ρ2sin⁡ϕ dρ dϕ dθ=5π12\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi/3} \displaystyle\int_0^{1} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{5\pi}{12}

  5. Use spherical coordinates to find the volume of the solid outside the cone ϕ=π4ϕ=\dfrac{\pi}{4} and inside the sphere ρ=4cos⁡ϕρ=4 \cos ϕ.

    6. ∫02π∫π/4π/2∫04cos⁡ϕρ2sin⁡ϕ dρ dϕ dθ=8π3\displaystyle\int_0^{2\pi} \displaystyle\int_{\pi/4}^{\pi/2} \displaystyle\int_0^{4\cos\phi} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{8\pi}{3}

  6. Use spherical coordinates to find the volume of the part of the ball ρ≤4ρ≤4 that lies between the planes z=2z=2 and z=23z=2\sqrt{3}.

    7. ∫02π∫0π/6∫023/cos⁡ϕρ2sin⁡ϕ dρ dϕ dθ=8π3(93−8)\displaystyle\int_0^{2\pi} \displaystyle\int_0^{\pi/6} \displaystyle\int_0^{2\sqrt{3}/\cos\phi} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta = \dfrac{8\pi}{3}(9\sqrt{3}-8)