Q. 32 - Calculus

Question

Areas of regions bounded by polar functions: Find the areas of the following regions. 

The area bounded by the function $r = a \cos \theta$, where $a$ is a positive constant. 

Step-by-Step Solution

Answer

The area bounded by the function is $\frac{{\mathrm{\pi a}}^2}{2} {\mathrm{unit}}^2$

Step 1: Given information.

The given function is $r = a \cos \theta$,where a is a positive constant.

Step 2: Area

The area of the curve is given by,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}{\left(a\cos \theta \right)}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\left(a^2{\cos }^2\theta \right)d\theta \\ A=\frac{a^2}{2}{\int }_0^{2\mathrm{\pi }}\left(\frac{1+\cos 2\theta }{2}\right)d\theta \\ A=\frac{a^2}{4}{\int }_0^{2\mathrm{\pi }}\left(\frac{1+\cos 2\theta }{2}\right)d\theta \\ A=\frac{a^2}{4}{\left[\theta +\frac{\sin 2\theta }{2}\right]}_0^{2\mathrm{\pi }} \\ AA=\frac{a^2}{4}{\left[2\mathrm{\pi }+\frac{\sin 4\mathrm{\pi }}{2}\right]}_0^{2\mathrm{\pi }} \\ A=\frac{{\mathrm{\pi a}}^2}{2} {\mathrm{unit}}^2 \end{gathered}$$

The area of the function is $A=\frac{{\mathrm{\pi a}}^2}{2} {\mathrm{unit}}^2$