Green's Theorem #1

First things first, if we walk along the path in the direction indicated in the diagram, then our left hand will be over the enclosed area. Therefore, this path has a positive orientation and we can use Green's Theorem to evaluate the integral.

Let's move on to the solution.

From the integral we have, $P=yx^2$ $Q=-x^2$ Remember that $P$ is multiplied by $x$ and $Q$ is multiplied by $y$ and don't forget to pay attention to signs.

Using Green's Theorem the line integral becomes, $\int_{C} yx^2 \,dx-x^2 \,dy=\iint_D -2x-x^2 \,dA$ D is the region enclosed by the curve.

Now, since D is just a half circle using polar coordinates is more suitable for this problem. Hence, the limits for D in polar coordinates are, $\frac{1}{2} \pi \leq t \leq \frac{3}{2} \pi$ $0 \leq r \leq 5$

Finally, all we need to do is evaluate the double integral, after converting to polar coordinates. The resulting steps are as follows:

$\int_{C} yx^2 \,dx-x^2 \,dy=\iint_D -2x-x^2 \,dA$ $=\int_{\frac{1}{2}\pi}^{\frac{3}{2}\pi} \int_{5}^{0} r(-2r\cos\theta -r^2 \cos^2\theta) \,dr \,d \theta$ $=\int_{\frac{1}{2}\pi}^{\frac{3}{2}\pi} \int_{5}^{0} -2r^2 \cos \theta - \frac{1}{2} r^3 (1 + \cos(2 \theta)) \,dr \,d \theta$ $=\int_{\frac{1}{2}\pi}^{\frac{3}{2}\pi} (- \frac{2}{3} r^3 \cos \theta - \frac{1}{8} r^4 (1 + \cos (2 \theta))) |^{5}_{0} \,d \theta$ $=\int_{\frac{1}{2}\pi}^{\frac{3}{2}\pi} - \frac{250}{3} \cos \theta - \frac{625}{8} (1 + \cos(2 \theta)) \,d \theta$ $= [- \frac{250}{3} \sin \theta - \frac{625}{8} ( \theta + \frac{1}{2} \sin(2 \theta))] |^{\frac{3}{2} \pi}_{\frac{1}{2} \pi}$ $= \frac{500}{3} - \frac{625}{8} \pi$ $= -78.7703$