find the area a enclosed by the curve r = 2 + 2sin(θ)

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 cardioid $r = 2 + 2\sin\theta$, the limits of integration are from $0$ to $2\pi$. So, $A=\frac{1}{2}\int_{0}^{2\pi}(2 + 2\sin\theta)^{2}d\theta$.

Step2: Expand the integrand

Expand $(2 + 2\sin\theta)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. Here, $a = 2$ and $b = 2\sin\theta$, so $(2 + 2\sin\theta)^{2}=4+8\sin\theta + 4\sin^{2}\theta$. Then the integral becomes $A=\frac{1}{2}\int_{0}^{2\pi}(4 + 8\sin\theta+4\sin^{2}\theta)d\theta$.

Step3: Split the integral

Split the integral into three separate integrals: $A=\frac{1}{2}\left(\int_{0}^{2\pi}4d\theta+\int_{0}^{2\pi}8\sin\theta d\theta+\int_{0}^{2\pi}4\sin^{2}\theta d\theta\right)$.

Step4: Evaluate each integral

• $\int_{0}^{2\pi}4d\theta=4\theta\big|_{0}^{2\pi}=4(2\pi - 0)=8\pi$.
• $\int_{0}^{2\pi}8\sin\theta d\theta=- 8\cos\theta\big|_{0}^{2\pi}=-8(\cos(2\pi)-\cos(0))=-8(1 - 1)=0$.
• Use the double - angle formula $\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}$. Then $\int_{0}^{2\pi}4\sin^{2}\theta d\theta=\int_{0}^{2\pi}4\times\frac{1 - \cos(2\theta)}{2}d\theta=\int_{0}^{2\pi}(2-2\cos(2\theta))d\theta=(2\theta-\sin(2\theta))\big|_{0}^{2\pi}=2(2\pi)-\sin(4\pi)-(0 - \sin(0)) = 4\pi$.

Step5: Combine the results

$A=\frac{1}{2}(8\pi+0 + 4\pi)=\frac{1}{2}(12\pi)=6\pi$.

Answer:

$6\pi$