Question

Math

Posted 6 months ago

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Let $R$ be the region shown below.
Let $f(x, y)=\frac{1}{x}+\frac{2}{y}$.
What is $\iint_{R} f(x, y) d A$ after a change of variables into polar coordinates?
Choose 1 answer:
(A)
$\int_{-\frac{\pi}{3}}^{\frac{5 \pi}{6}} \int_{1}^{3} \frac{1}{\cos (\theta)}+\frac{2}{\sin (\theta)} d r d \theta$
(B) $\int_{-\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{1}^{3} \frac{1}{\cos (\theta)}+\frac{2}{\sin (\theta)} d r d \theta$
(C) $\int_{-\frac{\pi}{3}}^{\frac{5 \pi}{6}} \int_{1}^{3} \frac{1}{\sin (\theta)}+\frac{2}{\cos (\theta)} d r d \theta$
(D) $\int_{-\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{1}^{3} \frac{1}{\sin (\theta)}+\frac{2}{\cos (\theta)} d r d \theta$
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Answer from Sia

Posted 6 months ago

Solution by Steps

step 2

The function in polar coordinates becomes $f(r, \theta) = \frac{1}{r\cos(\theta)} + \frac{2}{r\sin(\theta)}$

step 3

The Jacobian of the transformation from Cartesian to polar coordinates is $r$, so the area element $dA$ becomes $r dr d\theta$

step 4

The limits for $\theta$ are from $0$ to $\frac{\pi}{2}$ since the region R is a quarter-circle in the first quadrant

step 5

The limits for $r$ are from $0$ to $3$ because the radius of the quarter-circle is $3$ units

step 6

The double integral in polar coordinates is $\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \left(\frac{1}{r\cos(\theta)} + \frac{2}{r\sin(\theta)}\right) r dr d\theta$

step 7

Simplifying the integrand, we get $\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \left(\frac{1}{\cos(\theta)} + \frac{2}{\sin(\theta)}\right) dr d\theta$

step 8

Comparing the simplified integrand with the given options, we find that none of the options match exactly because they all have incorrect limits for $\theta$ and/or the integrand is not correctly expressed

None of the options A, B, C, or D are correct.

Key Concept

Change of variables in double integrals

Explanation

When changing from Cartesian to polar coordinates, the function and the limits of integration must be correctly transformed. The function $f(x, y)$ becomes $f(r, \theta)$, and the area element $dA$ becomes $r dr d\theta$. The limits of integration for $r$ and $\theta$ must correspond to the region of integration in polar coordinates.

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