Let $R$ be the region in the first quadrant that is inside the polar curve $r=4 \cos ( heta)$ and inside the polar curve $r=4 \sin (2 heta)$, as shown in the graph. The curves intersect at $ heta=\frac{\pi}{6}$.

What is the area of $R$ ?
Use a graphing calculator and round your answer to three decimal places.

Question

Let $R$ be the region in the first quadrant that is inside the polar curve $r=4 \cos (	heta)$ and inside the polar curve $r=4 \sin (2 	heta)$, as shown in the graph. The curves intersect at $	heta=\frac{\pi}{6}$.

What is the area of $R$ ?
Use a graphing calculator and round your answer to three decimal places.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the area of the region $R$ inside the polar curves $r=4 \cos(\theta)$ and $r=4 \sin(2 \theta)$, we need to set up the integrals for each curve and then subtract the second integral from the first. ‖ ‖ step 2 ⋮ First, we calculate the integral of $0.5 (4 \cos(\theta))^2$ from $0$ to $\frac{\pi}{6}$: $$ \int_0^{\pi/6} 0.5 (4 \cos(\theta))^2 \, d\theta = 3.82645 $$ ‖ ‖ step 3 ⋮ Next, we calculate the integral of $0.5 (4 \sin(2 \theta))^2$ from $0$ to $\frac{\pi}{6}$: $$ \int_0^{\pi/6} 0.5 (4 \sin(2 \theta))^2 \, d\theta = 1.22837 $$ ‖ ‖ step 4 ⋮ Finally, we subtract the result of the second integral from the first integral to find the area of $R$: $$ 3.82645 - 1.22837 = 2.59808 $$ ‖ ∻Answer∻ ⚹ The area of the region $R$ is approximately $2.598$ square units. ⚹ ∻Key Concept∻ ⚹Area of a region bounded by polar curves⚹ ∻Explanation∻ ⚹To find the area of a region bounded by two polar curves, we integrate the square of the radius function for each curve over the given interval and then subtract the smaller area from the larger area.⚹◊What is the formula to calculate the area bounded by two polar curves?⍭ Generate me a similar question◊

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