Green's Theorem — Question 6

Question 6 – Solution

Strategy. Combine Green’s area formula with polar boundary coordinates.

Step 1: Convert the differential With $x=r\cos\theta$ and $y=r\sin\theta$, $$dx=(r'\cos\theta-r\sin\theta)d\theta, \quad dy=(r'\sin\theta+r\cos\theta)d\theta.$$ The $rr'$ terms cancel, leaving $$x\,dy-y\,dx=r^2(\cos^2\theta+\sin^2\theta)d\theta=r^2d\theta.$$

Step 2: Evaluate the area $$\begin{align*} A&=\frac 12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta\\ &=\frac 12\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta\\ &=\frac 12(2\pi+0+\pi)=\boxed{\frac{3\pi}{2}}. \end{align*}$$

Verification The standard polar area formula is $\frac 12\int r^2d\theta$, exactly what Green’s Theorem produced. The nonnegative integrand confirms the positive area.