Question

5. Express as a double integral the area enclosed by one loop of the curve 𝑟 = 3 cos 2𝜃 and

evaluate the integral

Answer 1

One loop of the curve is presented on the image below:

where $\theta\in[-\frac{\pi}{4},\frac{\pi}{4}]$. $r\in[0,3].$

Remind that for the polar coordinates: $x=r\,cos\,\theta$ and $y=r\,sin\,\theta$. $x\in[0,3\frac{\sqrt{2}}{2}]$, $y\in[-3\frac{\sqrt{2}}{2},3\frac{\sqrt{2}}{2}]$ It implies $x^2+y^2=r^2$. The trigonometric formula for $cos\,2\theta$ has the form: $cos\,2\theta=2\,cos^2\theta-1=\frac{2x^2}{r^2}-1=\frac{2x^2}{x^2+y^2}-1$. Thus, the formula of the curve takes the form: $\sqrt{x^2+y^2}=\frac{2x^2}{x^2+y^2}-1$. Simplifying the latter we receive: $(x^2+y^2)^{3/2}=x^2-y^2$. The latter implies $(x^2+y^2)^3=(x^2-y^2)^2$ . The double integral for the area takes the form: $S=\int\int_{D}dxdy$, where $D$ is the area enclosed by the curve on the picture. I.e., $(x^2+y^2)^3=(x^2-y^2)^2$, where $x\in[0,3\frac{\sqrt{2}}{2}]$, $y\in[-3\frac{\sqrt{2}}{2},3\frac{\sqrt{2}}{2}]$. However, it is easier to compute such an area in polar coordinates. The formula is: $S=\frac{1}{2}\int_{-\pi/4}^{\pi/4}r^2d\theta=\frac{1}{2}\int_{-\pi/4}^{\pi/4}9cos^2(2\theta)d\theta=\frac{9}{2}\int_{-\pi/4}^{\pi/4}\frac{1+cos(4\theta)}{2}d\theta=\frac{9}{4}(\theta+\frac{sin(4\theta)}{4})|_{-\pi/4}^{\pi/4}=\frac{9}{8}\pi$

The answer is $\frac{9}{8}\pi.$