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) $\int_{-\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{1}^{3} \frac{1}{\sin (\theta)}+\frac{2}{\cos (\theta)} d r d \theta$

Question

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) $\int_{-\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{1}^{3} \frac{1}{\sin (\theta)}+\frac{2}{\cos (\theta)} d r d \theta$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To change the given function $f(x, y) = \frac{1}{x} + \frac{2}{y}$ into polar coordinates, we use the relations $x = r\cos( heta)$ and $y = r\sin( heta)$. ‖ ‖ step 2 ⋮ The function in polar coordinates becomes $f(r, heta) = \frac{1}{r\cos( heta)} + \frac{2}{r\sin( heta)}$. ‖ ‖ step 3 ⋮ The Jacobian of the transformation from Cartesian to polar coordinates is $r$, so the area element $dA$ becomes $r dr d heta$. ‖ ‖ step 4 ⋮ The limits for $ heta$ 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 $\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \left(\frac{1}{r\cos( heta)} + \frac{2}{r\sin( heta)} ight) r dr d heta$. ‖ ‖ step 7 ⋮ Simplifying the integrand, we get $\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \left(\frac{1}{\cos( heta)} + \frac{2}{\sin( heta)} ight) dr d heta$. ‖ ‖ 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 $ heta$ 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 $f(r, heta)$, and the area element $dA$ becomes $r dr d heta$. The limits of integration for $r$ and $ heta$ must correspond to the region of integration in polar coordinates.◊What is the formula for converting Cartesian coordinates to polar coordinates?⍭ Generate me a similar question◊⚹

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