11. -/6.66 points details my notes larcalc11 10.5 find the area of the region. interior of r² = 36 cos(2θ)

Explanation:

Step1: Recall area formula in polar coordinates

The area $A$ of a polar - curve $r = f(\theta)$ is given by $A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta$. For the curve $r^{2}=36\cos(2\theta)$, we first need to find the limits of integration. We know that $r^{2}\geq0$, so $\cos(2\theta)\geq0$. The solutions of $\cos(2\theta)\geq0$ are $-\frac{\pi}{4}\leq\theta\leq\frac{\pi}{4}$ and $\frac{3\pi}{4}\leq\theta\leq\frac{5\pi}{4}$. Due to the symmetry of the lemniscate (the curve $r^{2}=a\cos(2\theta)$ is a lemniscate), we can find the area of one - half of the curve and then double it. We will use the limits $\alpha=-\frac{\pi}{4}$ and $\beta = \frac{\pi}{4}$.

Step2: Set up the integral

Substitute $r^{2}=36\cos(2\theta)$ into the area formula: $A = 2\times\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}r^{2}d\theta=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}36\cos(2\theta)d\theta$.

Step3: Integrate $\cos(2\theta)$

The antiderivative of $\cos(2\theta)$ is $\frac{1}{2}\sin(2\theta)$. So, $\int36\cos(2\theta)d\theta=36\times\frac{1}{2}\sin(2\theta)+C = 18\sin(2\theta)+C$.

Step4: Evaluate the definite integral

$A=\left[18\sin(2\theta)\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=18\sin\left(2\times\frac{\pi}{4}\right)-18\sin\left(2\times\left(-\frac{\pi}{4}\right)\right)=18\sin\left(\frac{\pi}{2}\right)-18\sin\left(-\frac{\pi}{2}\right)$. Since $\sin\left(\frac{\pi}{2}\right)=1$ and $\sin\left(-\frac{\pi}{2}\right)= - 1$, we have $A = 18\times(1)-18\times(-1)=36$.

Answer:

$36$