21. Multiple Integrals in Curvilinear Coordinates

c. Integrating in Spherical Coordinates

2. Integral over a Spherical Region

We now explore integration in spherical coordinates with constant and non-constant limits.

Integral over a Spherical Box

We now want to integrate over the spherical box: ρ1ρρ2ϕ1ϕϕ2θ1θθ2 \rho_1 \le \rho \le \rho_2 \qquad \phi_1 \le \phi \le \phi_2 \qquad \theta_1 \le \theta \le \theta_2 Using the Riemann Sum definition with ΔVijk=ρi2sinϕjΔρiΔϕjΔθk\Delta V_{ijk}={\rho_i^*}^2\sin\phi_j^*\,\Delta\rho_i\,\Delta\phi_j\,\Delta\theta_k, the integral is: Rf(ρ,ϕ,θ)dV=limplimqlimri=1pj=1qk=1sf(ρi,ϕj,θk)ρi2sinϕjΔρiΔϕjΔθk \iiint\limits_R f(\rho,\phi,\theta)\,dV =\lim_{p\rightarrow\infty} \lim_{q\rightarrow\infty} \lim_{r\rightarrow\infty} \sum_{i=1}^p\sum_{j=1}^q\sum_{k=1}^s f(\rho_i^*,\phi_j^*,\theta_k^*) {\rho_i^*}^2\sin\phi_j^*\,\Delta\rho_i\,\Delta\phi_j\,\Delta\theta_k We recognize this as the triple iterated integral: Rf(ρ,ϕ,θ)dV=θ1θ2ϕ1ϕ2ρ1ρ2f(ρ,ϕ,θ)ρ2sinϕdρdϕdθ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\theta_1}^{\theta_2}\int_{\phi_1}^{\phi_2}\int_{\rho_1}^{\rho_2} f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta By Fubini's Theorem, there are five other orders of integrations obtained by reordering the differentials and the corresponding limits.

Other Integrals in Spherical Coordinates

In addition to integrating over spherical boxes, spherical coordinates can be used to integrate over other types of regions with appropriate definitions of the regions. The resulting integrals have the forms:

Type ρθ\rho\theta: Rf(ρ,ϕ,θ)dV=ρ1ρ2g(ρ)h(ρ)p(ρ,θ)q(ρ,θ)f(ρ,ϕ,θ)ρ2sinϕdϕdθdρ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\rho_1}^{\rho_2}\int_{g(\rho)}^{h(\rho)}\int_{p(\rho,\theta)}^{q(\rho,\theta)} f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\phi\,d\theta\,d\rho Type θρ\theta\rho: Rf(ρ,ϕ,θ)dV=θ1θ2g(θ)h(θ)p(ρ,θ)q(ρ,θ)f(ρ,ϕ,θ)ρ2sinϕdϕdρdθ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\theta_1}^{\theta_2}\int_{g(\theta)}^{h(\theta)} \int_{p(\rho,\theta)}^{q(\rho,\theta)} f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\phi\,d\rho\,d\theta Type ρϕ\rho\phi: Rf(ρ,ϕ,θ)dV=ρ1ρ2g(ρ)h(ρ)p(ρ,ϕ)q(ρ,ϕ)f(ρ,ϕ,θ)ρ2sinϕdθdϕdρ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\rho_1}^{\rho_2}\int_{g(\rho)}^{h(\rho)}\int_{p(\rho,\phi)}^{q(\rho,\phi)} f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\theta\,d\phi\,d\rho Type ϕρ\phi\rho: Rf(ρ,ϕ,θ)dV=ϕ1ϕ2g(ϕ)h(ϕ)p(ρ,ϕ)q(ρ,ϕ)f(ρ,ϕ,θ)ρ2sinϕdθdρdϕ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\phi_1}^{\phi_2}\int_{g(\phi)}^{h(\phi)} \int_{p(\rho,\phi)}^{q(\rho,\phi)} f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\theta\,d\rho\,d\phi Type θϕ\theta\phi: Rf(ρ,ϕ,θ)dV=θ1θ2g(θ)h(θ)p(θ,ϕ)q(θ,ϕ)f(ρ,ϕ,θ)ρ2sinϕdρdϕdθ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\theta_1}^{\theta_2}\int_{g(\theta)}^{h(\theta) }\int_{p(\theta,\phi)}^{q(\theta,\phi) }f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta Type ϕθ\phi\theta: Rf(ρ,ϕ,θ)dV=ϕ1ϕ2g(ϕ)h(ϕ)p(θ,ϕ)q(θ,ϕ)f(ρ,ϕ,θ)ρ2sinϕdρdθdϕ \iiint\limits_R f(\rho,\phi,\theta)\,dV =\int_{\phi_1}^{\phi_2}\int_{g(\phi)}^{h(\phi)} \int_{p(\theta,\phi)}^{q(\theta,\phi) }f(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\rho\,d\theta\,d\phi

Notice that each integral can have limits which depend only on the variables outside that integral.

Find the volume of the orange wedge given in spherical coordinates by 0ρ50 \le \rho \le 5, 0θπ60 \le \theta \le \dfrac{\pi}{6}, and 0ϕπ0 \le \phi \le \pi.

This is how the wedge looks:

wedge

This is a straightforward application of integration in spherical coordinates. Our integral with limits is as follows: V=0π/60π05ρ2sinϕdρdϕdθ=0π/60π[ρ33]ρ=05sinϕdϕdθ=12530π/60πsinϕdϕdθ=12530π/60πsinϕdϕdθ=12530π/6[cosϕ]ϕ=0πdθ=1253(2)0π/6dθ=2503π6=1259π\begin{aligned} V&=\int_0^{\pi/6}\int_0^\pi\int_0^5 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta =\int_0^{\pi/6}\int_0^\pi \left[\dfrac{\rho^3}{3}\right]_{\rho=0}^5\sin\phi\,d\phi\,d\theta \\ &=\dfrac{125}{3}\int_0^{\pi/6}\int_0^\pi \sin\phi\,d\phi\,d\theta =\dfrac{125}{3}\int_0^{\pi/6}\int_0^\pi \sin\phi\,d\phi\,d\theta \\ &=\dfrac{125}{3}\int_0^{\pi/6} \left[-\cos\phi\right]_{\phi=0}^\pi\,d\theta =\dfrac{125}{3}(2)\int_0^{\pi/6} \,d\theta \\ &=\dfrac{250}{3}\cdot\dfrac{\pi}{6} =\dfrac{125}{9}\pi \end{aligned}

We should have known this answer. Since π6\dfrac{\pi}{6} is 112\dfrac{1}{12} of a full circle 2π2\pi, the wedge is 112\dfrac{1}{12} of a sphere and its volume is 112\dfrac{1}{12} of the volume of a sphere of radius 55: V=11243π(5)3=1259πV= \dfrac{1}{12}\dfrac{4}{3}\pi(5)^3=\dfrac{125}{9}\pi. However, it is useful to write down the triple integral for the volume so we can use it to find the integrals for average value, centroid, mass, and center of mass.

Find the mass of the wedge in the previous example given that its density is δ=ρ\delta=\rho

Answer

M=62512πM=\dfrac{625}{12}\pi

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Solution

M=RδdV=Rδρ2sinϕdρdϕdθ=0π/60π05ρ3sinϕdρdϕdθ=[θ]0π/6[cosϕ]0π[ρ44]05=π6(2)6254=62512π\begin{aligned} M&=\iiint\limits_R \delta\,dV =\iiint\limits_R \delta\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=\int_0^{\pi/6}\int_0^\pi\int_0^5 \rho^3\sin\phi\,d\rho\,d\phi\,d\theta \\ &=\left[\theta\dfrac{}{}\right]_0^{\pi/6}\left[-\cos\phi\dfrac{}{}\right]_0^\pi\left[\dfrac{\rho^4}{4}\right]_0^5 =\dfrac{\pi}{6}(2)\dfrac{625}{4} =\dfrac{625}{12}\pi \end{aligned}

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