Let $S$ be the quarter of the cylinder with height 1 and radius 5 such that $\pi

Q: Let $S$ be the quarter of the cylinder with height 1 and radius 5 such that $\pi<\theta<\frac{3 \pi}{2}$, whose axis is parallel to the $z$-axis and whose lower base is centered at the origin. What is the triple integral of the scalar field $f(x, y, z)=y z$ over $S$ in cylindrical coordinates? Choose 1 answer: (A) $\int_{0}^{5} \int_{0}^{1} \int_{\pi}^{\frac{3 \pi}{2}} r^{2} z \sin (\theta) d \theta d r d z$ (B) $\int_{0}^{1} \int_{0}^{5} \int_{\pi}^{\frac{3 \pi}{2}} r^{2} z \sin (\theta) d \theta d r d z$ (C) $\int_{0}^{5} \int_{0}^{1} \int_{\pi}^{\frac{3 \pi}{2}} r z \cos (\theta) d \theta d r d z$ (D) $\int_{0}^{1} \int_{0}^{5} \int_{\pi}^{\frac{3 \pi}{2}} r z \cos (\theta) d \theta d r d z$

∻Solution by Steps∻ ‖ step 1 ⋮ Convert the scalar field $f(x, y, z) = yz$ to cylindrical coordinates. Since $y = r\sin(\theta)$, we have $f(r, \theta, z) = r\sin(\theta)z$. ‖ ‖ step 2 ⋮ Set up the triple integral in cylindrical coordinates with the given limits for $S$: $\int_{0}^{5} \int_{0}^{1} \int_{\pi}^{\frac{3\pi}{2}} r\sin(\theta)z \, d\theta \, dr \, dz$. ‖ ‖ step 3 ⋮ Integrate with respect to $\theta$ from $\pi$ to $\frac{3\pi}{2}$: $\int_{\pi}^{\frac{3\pi}{2}} \sin(\theta) \, d\theta = [-\cos(\theta)]_{\pi}^{\frac{3\pi}{2}} = -(-1) - (-1) = 0$. ‖ ‖ step 4 ⋮ Since the integral of $\sin(\theta)$ over the interval $[\pi, \frac{3\pi}{2}]$ is zero, the entire triple integral evaluates to zero. ‖ ***D*** ∻Key Concept∻ ⚹Evaluating the innermost integral first⚹ ∻Explanation∻ ⚹When the innermost integral of a triple integral evaluates to zero, the entire integral evaluates to zero regardless of the other limits of integration.⚹◊What are the advantages of using cylindrical coordinates for solving triple integrals in certain problems⍭ Generate me a similar question◊

Question

Let $S$ be the quarter of the cylinder with height 1 and radius 5 such that $\pi<	heta<\frac{3 \pi}{2}$, whose axis is parallel to the $z$-axis and whose lower base is centered at the origin.

What is the triple integral of the scalar field $f(x, y, z)=y z$ over $S$ in cylindrical coordinates?

Choose 1 answer:
(A) $\int_{0}^{5} \int_{0}^{1} \int_{\pi}^{\frac{3 \pi}{2}} r^{2} z \sin (	heta) d 	heta d r d z$
(B) $\int_{0}^{1} \int_{0}^{5} \int_{\pi}^{\frac{3 \pi}{2}} r^{2} z \sin (	heta) d 	heta d r d z$
(C) $\int_{0}^{5} \int_{0}^{1} \int_{\pi}^{\frac{3 \pi}{2}} r z \cos (	heta) d 	heta d r d z$
(D) $\int_{0}^{1} \int_{0}^{5} \int_{\pi}^{\frac{3 \pi}{2}} r z \cos (	heta) d 	heta d r d z$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Convert the scalar field $f(x, y, z) = yz$ to cylindrical coordinates. Since $y = r\sin( heta)$, we have $f(r, heta, z) = r\sin( heta)z$. ‖ ‖ step 2 ⋮ Set up the triple integral in cylindrical coordinates with the given limits for $S$: $\int_{0}^{5} \int_{0}^{1} \int_{\pi}^{\frac{3\pi}{2}} r\sin( heta)z \, d heta \, dr \, dz$. ‖ ‖ step 3 ⋮ Integrate with respect to $ heta$ from $\pi$ to $\frac{3\pi}{2}$: $\int_{\pi}^{\frac{3\pi}{2}} \sin( heta) \, d heta = [-\cos( heta)]_{\pi}^{\frac{3\pi}{2}} = -(-1) - (-1) = 0$. ‖ ‖ step 4 ⋮ Since the integral of $\sin( heta)$ over the interval $[\pi, \frac{3\pi}{2}]$ is zero, the entire triple integral evaluates to zero. ‖ ***D*** ∻Key Concept∻ ⚹Evaluating the innermost integral first⚹ ∻Explanation∻ ⚹When the innermost integral of a triple integral evaluates to zero, the entire integral evaluates to zero regardless of the other limits of integration.⚹◊What are the advantages of using cylindrical coordinates for solving triple integrals in certain problems⍭ Generate me a similar question◊

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