Using Green’s Theorem the line integral becomes,

\[\int\limits_{C}{{y{x^2}\,dx - {x^2}\,dy}} = \iint\limits_{D}{{ - 2x - {x^2}\,dA}}\]

\(D\) is the region enclosed by the curve.

Show Step 3

Since \(D\) is just a half circle it makes sense to use polar coordinates for this problem. The limits for \(D\) in polar coordinates are,

\[\begin{array}{c}\displaystyle \frac{1}{2}\pi \le \theta \le \frac{3}{2}\pi \\ 0 \le r \le 5\end{array}\] Show Step 4

Now all we need to do is evaluate the double integral, after first converting to polar coordinates of course.

Here is the evaluation work.

\[\begin{align*}\int\limits_{C}{{y{x^2}\,dx - {x^2}\,dy}} & = \iint\limits_{D}{{ - 2x - {x^2}\,dA}}\\ & = \int_{{\frac{1}{2}\pi }}^{{\frac{3}{2}\pi }}{{\int_{0}^{5}{{r\left( { - 2r\cos \theta - {r^2}{{\cos }^2}\theta } \right)dr}}\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{{\frac{3}{2}\pi }}{{\int_{0}^{5}{{ - 2{r^2}\cos \theta - \frac{1}{2}{r^3}\left( {1 + \cos \left( {2\theta } \right)} \right)dr}}\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{{\frac{3}{2}\pi }}{{\left. {\left( { - \frac{2}{3}{r^3}\cos \theta - \frac{1}{8}{r^4}\left( {1 + \cos \left( {2\theta } \right)} \right)} \right)} \right|_0^5\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{{\frac{3}{2}\pi }}{{ - \frac{{250}}{3}\cos \theta - \frac{{625}}{8}\left( {1 + \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\left[ { - \frac{{250}}{3}\sin \theta - \frac{{625}}{8}\left( {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right]} \right|_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{500}}{3} - \frac{{625}}{8}\pi = - 78.7703}}\end{align*}\]

Don’t forget the extra \(r\) from converting the \(A\) to polar coordinates and make sure you recall all the various tirg identities we need to deal with many of the various trig functions that show up in the integrals.