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Find the area of the shaded region A shown in the figure below:

Question:

Find the area of the shaded region {eq}A {/eq} shown in the figure below:

Area:

Suppose that {eq}D {/eq} is a region of the plane. The area is then {eq}\iint_D \, dA {/eq}. If the region is bounded by polar curves, then we can convert the integral to polar coordinates. Recall that to do this we use {eq}dA=r\, dr\, d\theta {/eq}. If {eq}D=\{(r,\theta)\mid r_1(\theta)\leq r\leq r_2(\theta),\, \alpha \leq \theta \leq \beta\} {/eq}, then we have {eq}\text{Area}=\int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} r\, dr\, d\theta {/eq}.

Answer and Explanation:

The two curves are {eq}r=1 {/eq} and {eq}r=4\cos \theta {/eq}. Setting them equal we have

{eq}4\cos \theta=1\\ \cos \theta=\frac{1}{4}\\ \theta =\cos^{-1} \frac{1}{4} {/eq}

By symmetry we can see that the second point of intersection is

{eq}\theta=-\cos^{-1} \frac{1}{4} {/eq}

The region can then be described by

{eq}D=\{(r,\theta)\mid 1\leq r\leq 4\cos \theta,\, -\cos^{-1} \frac{1}{4}\leq \theta \leq \cos^{-1} \frac{1}{4}\} {/eq}. We then have

{eq}\begin{align} \text{Area}&=\iint_D \, dA & [\text{Converting to polar coordiantes.} ]\\ &=\int_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4} \int_1^{4\cos \theta} r\, dr\, d\theta\\ &=\int_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4} \frac{1}{2}r^2|_1^{4\cos \theta}\, d\theta\\ &=\int_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4} \frac{1}{2}[16\cos^2 \theta-1]\, d\theta\\ &=\int_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4} \frac{1}{2}[16(\frac{1+\cos (2\theta)}{2}-1]\, d\theta\\ &=\int_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4} \frac{1}{2}[8+8\cos (2\theta)-1]\, d\theta\\ &=\int_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4} \frac{1}{2}[7+8\cos (2\theta)]\, d\theta\\ &=\frac{1}{2}[7\theta+4\sin (2\theta)]|_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4}\\ &=\frac{1}{2}[7\theta+8\sin \theta\cos \theta]|_{-\cos^{-1} 1/4}^{\cos^{-1} 1/4}\\ &=\frac{1}{2}[7[\cos^{-1} 1/4-(-\cos^{-1} 1/4)+8\sin (\cos^{-1} 1/4)\cos (\cos^{-1} 1/4)-8\sin (-\cos^{-1} 1/4)\cos (-\cos^{-1} 1/4)]\\ &=\frac{1}{2}[14\cos^{-1} 1/4+8\frac{\sqrt{15}}{4}\frac{1}{4}+8\frac{\sqrt{15}}{4}\frac{1}{4} ]\\ &=\frac{1}{2}[14\cos^{-1} 1/4+\sqrt{15} ] \end{align} {/eq}


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