How do you find the area of the region bounded by the polar curve $r = 3 (1 + cos (θ))$ $r=3(1+cos(theta))$ ?

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How do you find the area of the region bounded by the polar curve $r = 3 (1 + cos (θ))$ $r=3(1+cos(theta))$ ?

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Answer 1 · 2014-09-07T18:01:45

The area of the region enclosed by the curve $r = 3 (1 = cos θ)$ $r=3(1=cos theta)$ can be found by the double integral
$\int_{0}^{2 π} \int_{0}^{3 (1 + cos θ)} r d r d θ = \frac{27}{2} π$ $int_0^{2pi}int_0^{3(1+cos theta)}rdrd theta=27/2pi$

Let us evaluate the double integral.
$\int_{0}^{2 π} \int_{0}^{3 (1 + cos θ)} r d r d θ$ $int_0^{2pi}int_0^{3(1+cos theta)}rdrd theta$
by Power Rule,
$= \int_{0}^{2 π} {[\frac{r^{2}}{2}]}_{0}^{3 (1 + cos θ)} d θ$ $=int_0^{2pi}[r^2/2]_0^{3(1+cos theta)}d theta$
$= \int_{0}^{2 π} \frac{{[3 (1 + cos θ)]}^{2}}{2} d θ$ $=int_0^{2pi}{[3(1+cos theta)]^2}/2d theta$
$= \frac{9}{2} \int_{0}^{2 π} (1 + 2 cos θ + {cos}^{2} θ) d θ$ $=9/2int_0^{2pi}(1+2cos theta+cos^2 theta)d theta$
by ${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2 theta=1/2(1+cos 2theta)$ ,
$= \frac{9}{2} \int_{0}^{2 π} (\frac{3}{2} + 2 cos θ + \frac{1}{2} cos 2 θ) d θ$ $=9/2int_0^{2pi}(3/2+2cos theta+1/2cos2theta)d theta$
$= \frac{9}{2} {[\frac{3}{2} θ + 2 sin θ + \frac{1}{4} sin 2 θ]}_{0}^{2 π}$ $=9/2[3/2theta+2sin theta+1/4sin2theta]_0^{2pi}$
$= \frac{9}{2} \cdot \frac{3}{2} (2 π) = \frac{27}{2} π$ $=9/2cdot3/2(2pi)=27/2pi$

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How do you find the area of the region bounded by the polar curve $r = 3 (1 + cos (θ))$ $r=3(1+cos(theta))$ ?

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How do you find the area of the region bounded by the polar curve r=3(1+cos(θ))r=3(1+cos(theta)) ?

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How do you find the area of the region bounded by the polar curve $r = 3 (1 + cos (θ))$ $r=3(1+cos(theta))$ ?