Question

Math

Posted 3 months ago

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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) $\int_{0}^{1} \int_{0}^{5} \int_{\pi}^{\frac{3 \pi}{2}} r z \cos (\theta) d \theta d r d z$
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

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.

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