find the area a enclosed by the curve\nr = 4 + sin(4θ)

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 a complete loop of the curve $r = 4+\sin(4\theta)$, the period of $\sin(4\theta)$ is $\frac{2\pi}{4}=\frac{\pi}{2}$, and we need to integrate over one full - period. So, $\alpha = 0$ and $\beta = 2\pi$. Then $A=\frac{1}{2}\int_{0}^{2\pi}(4 + \sin(4\theta))^{2}d\theta$.

Step2: Expand the integrand

Expand $(4+\sin(4\theta))^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. Here, $a = 4$ and $b=\sin(4\theta)$, so $(4+\sin(4\theta))^{2}=16 + 8\sin(4\theta)+\sin^{2}(4\theta)$.

Step3: Split the integral

$A=\frac{1}{2}\int_{0}^{2\pi}(16 + 8\sin(4\theta)+\sin^{2}(4\theta))d\theta=\frac{1}{2}\left(\int_{0}^{2\pi}16d\theta+8\int_{0}^{2\pi}\sin(4\theta)d\theta+\int_{0}^{2\pi}\sin^{2}(4\theta)d\theta\right)$.

Step4: Evaluate each integral

• For $\int_{0}^{2\pi}16d\theta$, using the power - rule $\int kd\theta=k\theta + C$ ($k = 16$), we have $\int_{0}^{2\pi}16d\theta=16\theta\big|_{0}^{2\pi}=16\times(2\pi-0)=32\pi$.
• For $\int_{0}^{2\pi}8\sin(4\theta)d\theta$, let $u = 4\theta$, then $du = 4d\theta$. When $\theta = 0$, $u = 0$; when $\theta = 2\pi$, $u = 8\pi$. So, $\int_{0}^{2\pi}8\sin(4\theta)d\theta=8\times\frac{1}{4}\int_{0}^{8\pi}\sin(u)du = 2[-\cos(u)]_{0}^{8\pi}=2(-\cos(8\pi)+\cos(0))=2(-1 + 1)=0$.
• For $\int_{0}^{2\pi}\sin^{2}(4\theta)d\theta$, use the identity $\sin^{2}x=\frac{1-\cos(2x)}{2}$. Here $x = 4\theta$, so $\sin^{2}(4\theta)=\frac{1-\cos(8\theta)}{2}$. Then $\int_{0}^{2\pi}\sin^{2}(4\theta)d\theta=\int_{0}^{2\pi}\frac{1-\cos(8\theta)}{2}d\theta=\frac{1}{2}\int_{0}^{2\pi}(1-\cos(8\theta))d\theta=\frac{1}{2}\left(\int_{0}^{2\pi}1d\theta-\int_{0}^{2\pi}\cos(8\theta)d\theta\right)$. Let $v = 8\theta$, $dv = 8d\theta$. $\int_{0}^{2\pi}1d\theta=\theta\big|{0}^{2\pi}=2\pi$, and $\int{0}^{2\pi}\cos(8\theta)d\theta=\frac{1}{8}\int_{0}^{16\pi}\cos(v)dv=\frac{1}{8}[\sin(v)]{0}^{16\pi}=0$. So, $\int{0}^{2\pi}\sin^{2}(4\theta)d\theta=\frac{1}{2}(2\pi-0)=\pi$.

Step5: Calculate the total area

$A=\frac{1}{2}(32\pi+0+\pi)=\frac{1}{2}(33\pi)=\frac{33\pi}{2}$.

Answer:

$\frac{33\pi}{2}$