Use Green's Theorem to evaluate the line integral along the positively oriented closed curve $C$. $\oint _ { C } ( 6 x y + 5 \ln ( 1 + x ) ) d x + 2 x ^ { 2 } d y$, where $C$ is the cardioid $r = 2 + 2 \cos \theta$.
A) $- 20 \pi$
B) $- 80 \pi$
C) $- 10 \pi$
D) $- 40 \pi$

Question

Quizplus · Accepted Answer

Using Green's Theorem, the line integral becomes$$\begin{split} \oint_C (6xy+5\ln(1+x))dx+2x^2dy & = \iint_R \left( \frac{\partial}{\partial x} [2x^2] - \frac{\partial}{\partial y} [6xy+5\ln(1+x)] ight) dA \\ & = \iint_R (4x-6x) dA \\ & = \iint_R (-2x) dA \\ & = -2 \int_0^{2\pi} \int_0^{2+2\cos	heta} r \ dr \ d	heta \\ & = -2 \int_0^{2\pi} \left[ \frac{r^2}{2} ight]_0^{2+2\cos	heta} d	heta \\ & = -2 \int_0^{2\pi} (2+4\cos	heta+\cos^2	heta) d	heta \\ & = -2 \int_0^{2\pi} (2+4\cos	heta+\frac{1+\cos2	heta}{2}) d	heta \\ & = -2 \int_0^{2\pi} \left( \frac{5}{2} + 4\cos	heta + \frac{1}{2} \cos2	heta ight) d	heta \\ & = -2 \left[ \frac{5}{2} 	heta + 4\sin	heta + \frac{1}{4} \sin2	heta ight]_0^{2\pi} \\ & = -2 \left( \frac{5\pi}{2} ight) \\ & = -5\pi \end{split}$$

Use Green's Theorem to Evaluate the Line Integral Along the Positively