How do you find the area of the region bounded by the polar curves $r = 1 + cos (θ)$ $r=1+cos(theta)$ and $r = 1 - cos (θ)$ $r=1-cos(theta)$ ?

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How do you find the area of the region bounded by the polar curves $r = 1 + cos (θ)$ $r=1+cos(theta)$ and $r = 1 - cos (θ)$ $r=1-cos(theta)$ ?

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Answer 1 · 2014-11-09T16:57:56

The region bounded by the polar curves looks like:

enter image source here

Since the region consists of two identical leaves that are symmetric about the $y$ $y$ -axis, I will try to find a half of one leaf then multiply it by $4$ $4$ .

$A = 4 \int_{0}^{\frac{π}{2}} \int_{0}^{1 - cos θ} r d r d θ$ $A=4int_0^{pi/2}int_0^{1-cos theta}rdrd theta$

$= 4 \int_{0}^{\frac{π}{2}} {[\frac{r^{2}}{2}]}_{0}^{1 - cos θ} d θ$ $=4int_0^{pi/2}[r^2/2]_0^{1-cos theta}d theta$

$= 2 \int_{0}^{\frac{π}{2}} (1 - 2 cos θ + {cos}^{2} θ) d θ$ $=2int_0^{pi/2}(1-2cos theta+cos^2theta)d theta$

by ${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$ ,

$= \int_{0}^{\frac{π}{2}} (3 - 4 cos θ + cos 2 θ) d θ$ $=int_0^{pi/2}(3-4cos theta+cos2theta)d theta$

$= {[3 θ - 4 sin θ + \frac{1}{2} sin 2 θ]}_{0}^{\frac{π}{2}}$ $=[3theta-4sintheta+1/2sin2theta]_0^{pi/2}$

$= \frac{3 π}{2} - 4$ $={3pi}/2-4$

I hope that this was helpful.

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How do you find the area of the region bounded by the polar curves $r = 1 + cos (θ)$ $r=1+cos(theta)$ and $r = 1 - cos (θ)$ $r=1-cos(theta)$ ?

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How do you find the area of the region bounded by the polar curves r=1+cos(θ)r=1+cos(theta) and r=1−cos(θ)r=1-cos(theta) ?

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How do you find the area of the region bounded by the polar curves $r = 1 + cos (θ)$ $r=1+cos(theta)$ and $r = 1 - cos (θ)$ $r=1-cos(theta)$ ?