Evaluate the line integral $\int_C \mathbf F \cdot \, \mathrm d \mathbf r$ where, $\mathbf F = (e^x + xy) \mathbf i+ e^y - xy^2\mathbf j$

Question

Evaluate the line integral $\int_C \mathbf F \cdot \, \mathrm d \mathbf r$ where, $\mathbf F = (e^x + xy) \mathbf i+ e^y - xy^2\mathbf j$ and $C$ is the circle $x^2+y^2=4$ oriented clockwise.

My attempt:

$P(x,y)=e^x+xy$ and $Q(x,y)=e^y-xy^2$.

Hence, $\frac{\partial{Q}}{\partial{x}}=-y^2$ and $\frac{\partial{P}}{\partial{y}}=x$

Applying Green's Theorem where D is the interior of C, i.e. D is the disc such that $x^2+y^2\leq4$

$$\int_C(e^x + xy)dx + (e^y - xy^2)dy =\int\int_D(-y^2-x)dxdy$$

Not sure where to go from here or if this is currently correct.

Answer 1

You have gone through most half of the way! The remaining is to solve the integral you found: $$\int_{D}{-y^2-xdxdy}=\int_{D}{-r^2sin^2(\theta)-rcos(\theta)rd\theta dr}$$ $$=\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}{-r^3{{1-cos(2\theta)}\over2}-r^2cos(\theta)d\theta dr}$$ since $sin$ and $cos$ functions integrate to zero on a length of a multiple of their period ,say, $\int_{0}^{2\pi}cos(\theta)d\theta=0$ and $\int_{0}^{2\pi}cos(2\theta)d\theta=0$ then we have: $$\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}{-r^3{{1-cos(2\theta)}\over2}-r^2cos(\theta)d\theta dr}=\int_{r=0}^{r=2}{-\pi r^3dr}=-4\pi$$

Answer 2

Because the circle is oriented clockwise, I believe this makes the answer +4pi, rather than -4pi. The working is the same as in the answer above, but the sign is inverted due to the orientation.