How do you find the area of the region bounded by the polar curves $r = 3 + 2 cos (θ)$ $r=3+2cos(theta)$ and $r = 3 + 2 sin (θ)$ $r=3+2sin(theta)$ ?

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Answer 1 · 2014-11-08T11:06:33

Let us look at the region bounded by the polar curves, which looks like:

enter image source here

Red: $y = 3 + 2 cos θ$ $y=3+2cos theta$
Blue: $y = 3 + 2 sin θ$ $y=3+2sin theta$
Green: $y = x$ $y=x$

Using the symmetry, we will try to find the area of the region bounded by the red curve and the green line then double it.

$A = 2 \int_{\frac{π}{4}}^{\frac{5 π}{4}} \int_{0}^{3 + 2 cos θ} r d r d θ$ $A=2int_{pi/4}^{{5pi}/4}\int_0^{3+2cos theta}rdrd theta$

$= 2 \int_{\frac{π}{4}}^{\frac{5 π}{4}} {[\frac{r^{2}}{2}]}_{0}^{3 + 2 cos θ} d θ$ $=2int_{pi/4}^{{5pi}/4}[r^2/2]_0^{3+2cos theta} d theta$

$= \int_{\frac{π}{4}}^{\frac{5 π}{4}} (9 + 12 cos θ + 4 {cos}^{2} θ) d θ$ $=int_{pi/4}^{{5pi}/4}(9+12cos theta+4cos^2theta)d theta$

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

$= \int_{\frac{π}{4}}^{\frac{5 π}{4}} (11 + 12 cos θ + 2 cos 2 θ) d θ$ $=int_{pi/4}^{{5pi}/4}(11+12cos theta+2cos2theta)d theta$

$= {[11 θ + 12 sin θ + sin 2 θ]}_{\frac{π}{4}}^{\frac{5 π}{4}}$ $=[11theta+12sin theta+sin2theta]_{pi/4}^{{5pi}/4}$

$= \frac{55 π}{4} - 6 \sqrt{2} + 1 - (\frac{11 π}{4} + 6 \sqrt{2} + 1)$ $={55pi}/4-6sqrt{2}+1-({11pi}/4+6sqrt{2}+1)$

$= 11 π - 12 \sqrt{2}$ $=11pi-12sqrt{2]$

Hence, the area of the region is $11 π - 12 \sqrt{2}$ $11pi-12sqrt{2}$ .

I hope that this was helpful.

How do you find the area of the region bounded by the polar curves r=3+2cos(θ)r=3+2cos(theta) and r=3+2sin(θ)r=3+2sin(theta) ?