Triple Integration in Spherical Coordinates - AP Calculus BC

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Question

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere .

Answer

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

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Question

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere .

Answer

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

Compare your answer with the correct one above

Question

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere .

Answer

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

Compare your answer with the correct one above

Question

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere .

Answer

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

Compare your answer with the correct one above

Question

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere .

Answer

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

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