Question

Math

Posted 8 days ago

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

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Answer from Sia

Posted 8 days ago

Solution by Steps

step 1

To sketch the polar curves $r=2$ and $r=2(1-\sin \theta)$, we first recognize that $r=2$ is a circle with radius 2 centered at the origin

step 2

The curve $r=2(1-\sin \theta)$ is a cardioid. To sketch it, note that when $\theta=0$, $r=2(1-0)=2$; when $\theta=\frac{\pi}{2}$, $r=2(1-1)=0$; when $\theta=\pi$, $r=2(1+0)=2$; and when $\theta=\frac{3\pi}{2}$, $r=2(1+1)=4$

step 3

Plot these points and draw the cardioid. The region $R$ is inside the circle $r=2$ and outside the cardioid $r=2(1-\sin \theta)$. Shade this region

step 4

To find the area of $R$, we use the formula for the area between two polar curves: $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_1^2 - r_2^2) \, d\theta$. Here, $r_1=2$ and $r_2=2(1-\sin \theta)$

step 5

The limits of integration are from $\theta=0$ to $\theta=2\pi$. Thus, the area $A$ is given by: $A = \frac{1}{2} \int_{0}^{2\pi} [2^2 - (2(1-\sin \theta))^2] \, d\theta$

step 6

Simplify the integrand: $A = \frac{1}{2} \int_{0}^{2\pi} [4 - 4(1-\sin \theta)^2] \, d\theta = \frac{1}{2} \int_{0}^{2\pi} [4 - 4(1 - 2\sin \theta + \sin^2 \theta)] \, d\theta$

step 7

Further simplify: $A = \frac{1}{2} \int_{0}^{2\pi} [4 - 4 + 8\sin \theta - 4\sin^2 \theta] \, d\theta = \frac{1}{2} \int_{0}^{2\pi} [8\sin \theta - 4\sin^2 \theta] \, d\theta$

step 8

Split the integral: $A = \frac{1}{2} \left( \int_{0}^{2\pi} 8\sin \theta \, d\theta - \int_{0}^{2\pi} 4\sin^2 \theta \, d\theta \right)$

step 9

Evaluate the integrals: $\int_{0}^{2\pi} 8\sin \theta \, d\theta = 0$ (since $\sin \theta$ is an odd function over $[0, 2\pi]$) and $\int_{0}^{2\pi} 4\sin^2 \theta \, d\theta = 4 \cdot \pi$

step 10

Thus, $A = \frac{1}{2} \left( 0 - 4\pi \right) = -2\pi$. 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 $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_1^2 - r_2^2) \, d\theta$, where $r_1$ and $r_2$ are the given polar functions and $\alpha$ and $\beta$ are the limits of integration.

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