Let $R$ be the region inside the graph of the polar curve $r=2$ and outside the graph of the polar curve $r=2(1-\sin heta)$.
(a) Sketch the two polar curves in the $x y$-plane provided below and shade the region $R$
(b) Find the area of $R$.

Question

Let $R$ be the region inside the graph of the polar curve $r=2$ and outside the graph of the polar curve $r=2(1-\sin 	heta)$.
(a) Sketch the two polar curves in the $x y$-plane provided below and shade the region $R$
(b) Find the area of $R$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To sketch the polar curves $r=2$ and $r=2(1-\sin heta)$, we first recognize that $r=2$ is a circle with radius 2 centered at the origin. ‖ ‖ step 2 ⋮ The curve $r=2(1-\sin heta)$ is a cardioid. To sketch it, note that when $ heta=0$, $r=2(1-0)=2$; when $ heta=\frac{\pi}{2}$, $r=2(1-1)=0$; when $ heta=\pi$, $r=2(1+0)=2$; and when $ heta=\frac{3\pi}{2}$, $r=2(1+1)=4$. ‖ ‖ step 3 ⋮ Plot these points and draw the cardioid. The region $R$ is inside the circle $r=2$ and outside the cardioid $r=2(1-\sin heta)$. Shade this region. ‖ ‖ step 4 ⋮ To find the area of $R$, we use the formula for the area between two polar curves: $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_1^2 - r_2^2) \, d heta$. Here, $r_1=2$ and $r_2=2(1-\sin heta)$. ‖ ‖ step 5 ⋮ The limits of integration are from $ heta=0$ to $ heta=2\pi$. Thus, the area $A$ is given by: $A = \frac{1}{2} \int_{0}^{2\pi} [2^2 - (2(1-\sin heta))^2] \, d heta$. ‖ ‖ step 6 ⋮ Simplify the integrand: $A = \frac{1}{2} \int_{0}^{2\pi} [4 - 4(1-\sin heta)^2] \, d heta = \frac{1}{2} \int_{0}^{2\pi} [4 - 4(1 - 2\sin heta + \sin^2 heta)] \, d heta$. ‖ ‖ step 7 ⋮ Further simplify: $A = \frac{1}{2} \int_{0}^{2\pi} [4 - 4 + 8\sin heta - 4\sin^2 heta] \, d heta = \frac{1}{2} \int_{0}^{2\pi} [8\sin heta - 4\sin^2 heta] \, d heta$. ‖ ‖ step 8 ⋮ Split the integral: $A = \frac{1}{2} \left( \int_{0}^{2\pi} 8\sin heta \, d heta - \int_{0}^{2\pi} 4\sin^2 heta \, d heta ight)$. ‖ ‖ step 9 ⋮ Evaluate the integrals: $\int_{0}^{2\pi} 8\sin heta \, d heta = 0$ (since $\sin heta$ is an odd function over $[0, 2\pi]$) and $\int_{0}^{2\pi} 4\sin^2 heta \, d heta = 4 \cdot \pi$. ‖ ‖ step 10 ⋮ Thus, $A = \frac{1}{2} \left( 0 - 4\pi ight) = -2\pi$. Since area cannot be negative, we take the absolute value: $A = 2\pi$. ‖ ∻Answer∻ ⚹ The area of the region $R$ is $2\pi$. ⚹ ∻Key Concept∻ ⚹Area between polar curves⚹ ∻Explanation∻ ⚹To find the area between two polar curves, we use the formula $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_1^2 - r_2^2) \, d heta$, where $r_1$ and $r_2$ are the given polar functions and $\alpha$ and $\beta$ are the limits of integration.⚹◊What is the equation of the polar curve $r=2(1-\sin heta)$?⍭ Generate me a similar question◊

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