Question

use double integral to calculate area of region that is inside r=6-4Cos Θ

Answer 1

1 STEP: Let's draw a graph of this function

2 STEP: Define integration boundaries

$$\left\{\begin{array}{l} 0\le r\le 6-4\cdot\cos\theta\\ 0\le\theta\le\pi-\text{due to the symmetry of the graph} \end{array}\right.$$

Symmetry is indicated in order to calculate only the upper part of the area for which $0\le\theta\le\pi$ , and then multiply the resulting number by 2.

3 STEP: area calculation

Since the graph is given in polar coordinates, we will carry out calculations in polar coordinates.

To do this, we immediately indicate how the area element changes $dS=dxdy=rdrd\theta$

( More information: https://en.wikipedia.org/wiki/Polar_coordinate_system )

Then,

$$S=\iint_M1dxdy=2\cdot\int\limits_0^\pi d\theta\cdot\left(\int\limits_0^{6-4\cos\theta}rdr\right)=\\[0.3cm] =2\cdot\int\limits_0^\pi d\theta\cdot\left(\left.\frac{r^2}{2}\right|_0^{6-4\cos\theta}\right)= 2\cdot\frac{1}{2}\cdot\int\limits_0^\pi \left(6-4\cos\theta\right)^2d\theta=\\[0.3cm] =\int\limits_0^\pi \left(36-48\cos\theta+16\cos^2\theta\right)d\theta=\\[0.3cm] =\int\limits_0^\pi \left(36-48\cos\theta+16\cdot\frac{1+\cos2\theta}{2}\right)d\theta=\\[0.3cm] =\int\limits_0^\pi \left(36-48\cos\theta+8\cdot\left(1+\cos2\theta\right)\right)d\theta=\\[0.3cm] =\int\limits_0^\pi \left(44-48\cos\theta+8\cos2\theta\right)d\theta=\\[0.3cm] =\left.\left(44\theta-48\sin\theta+8\cdot\frac{\sin2\theta}{2}\right)\right|_0^\pi=\\[0.3cm] =\left(44\pi-48\sin\pi+4\sin2\pi\right)-\left(44\cdot0-48\sin0+4\sin2\cdot0\right)=\\[0.3cm] =44\pi-48\cdot0+4\cdot0-0+48\cdot0-4\cdot0=44\pi$$

ANSWER

$$\boxed{S=48\pi}$$