Question

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

Answer 1

In this question we have to find the area under the trigonometric function $r=3\cos (\theta )$ therefore we will use double integration to get the required answer. We will the function $r$from its intervals $0$to $3\cos \theta$and then integrate that value from $0$ to $\pi$, to get the required solution.

Complete step-by-step answer:
We have the expression given to us as $r=3\cos (\theta )$ which means that $r$ in the between the values of $0$ to $3\cos \theta$and then integrate it from $0$ to $\pi$.
Therefore, in the integration form it can be written as:
$A=\int\limits_{o}^{\pi }{\int\limits_{0}^{3\cos \theta }{rdrd\theta }}$
We will first complete the internal integration. We know that $\int{r}dr=\dfrac{{{r}^{2}}}{2}$ therefore, on using the formula, we get:
$A=\int\limits_{0}^{\pi }{\left[ \dfrac{{{r}^{2}}}{2} \right]_{0}^{3\cos \theta }d\theta }$
On putting the values of the limits, we get:
$A=\int\limits_{0}^{\pi }{\dfrac{{{3}^{2}}{{\cos }^{2}}\theta }{2}-\dfrac{{{0}^{2}}}{2}d\theta }$
On simplifying the values, we get:
$A=\int\limits_{0}^{\pi }{\dfrac{9{{\cos }^{2}}\theta }{2}d\theta }$
Since the term $\dfrac{9}{2}$ in is multiplication, we can take it out of the integral as:
$A=\dfrac{9}{2}\int\limits_{0}^{\pi }{{{\cos }^{2}}\theta d\theta }$
On we will integrate the outer integral. We know that ${{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}$ therefore, on substituting it in the integral, we get:
$A=\dfrac{9}{2}\int\limits_{0}^{\pi }{\dfrac{1+\cos 2\theta }{2}d\theta }$
On taking the term $2$ which is in division out of the integral, we get:
$A=\dfrac{9}{4}\int\limits_{0}^{\pi }{(1+\cos 2\theta )d\theta }$
Now we know that $\int{1}d\theta =\theta$ and $\int{\cos (a\theta )d\theta =\dfrac{\sin (a\theta )}{a}}$
On using the formula and integrating, we get:
$A=\dfrac{9}{4}\int\limits_{0}^{\pi }{\left[ \theta +\dfrac{\sin 2\theta }{2} \right]_{0}^{\pi }}$
On splitting the integral values, we get:
$A=\dfrac{9}{4}\left[ \pi +\dfrac{\sin 2\left( \pi \right)}{2}-0+\dfrac{\sin 2\left( 0 \right)}{2} \right]$
Now we know that $\sin 0=0$ and $\sin 2\pi =0$ therefore, on substituting, we get:
$A=\dfrac{9}{4}\left[ \pi +0-0+0 \right]$
Which can be simplified as:
$A=\dfrac{9}{4}\pi$, which is the area under the polar curve $r=3\cos (\theta )$.

Note: It is to be remembered that the area under the polar graph is approximately the sum of all the skinny wedges which are under it. It is also to be remembered that the derivative is the inverse of integration. Integration helps in finding the area or volume while derivative finds the equation of a line or a curve.