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

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Answer 1 · 2014-09-27T09:15:56

The area of the region is $\frac{9}{4} π$ $9/4pi$ .

Let us look at some details.

The region is a disk, which looks like this:
enter image source here

If you are allowed to use the formula for the area of a circle, then

$A = π r^{2} = π {(\frac{3}{2})}^{2} = \frac{9}{4} π$ $A=pir^2=pi(3/2)^2=9/4pi$

If you wish to use integration, then

$A = \int_{0}^{π} \int_{0}^{3 cos θ} r d r d θ$ $A=int_0^{pi}\int_0^{3cos theta}rdrd theta$

$= \int_{0}^{π} {[\frac{r^{2}}{2}]}_{0}^{3 cos θ} d θ$ $=int_0^{pi}[r^2/2]_0^{3cos theta}d theta$

$= \int_{0}^{π} \frac{9 {cos}^{2} θ}{2} d θ$ $=int_0^{pi}{9cos^2theta}/2 d theta$

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

$= \frac{9}{4} \int_{0}^{π} (1 + cos 2 θ) d θ$ $=9/4int_0^{pi}(1+cos2theta) d theta$

$= \frac{9}{4} {[θ + \frac{sin 2 θ}{2}]}_{0}^{π}$ $=9/4[theta+{sin2theta}/2]_0^{pi}$

$= \frac{9}{4} π$ $=9/4pi$

I hope that this was helpful.

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