Find $\int_{\left(C\right)}xy{\rm d}x+y^{2}{\rm d}y$ with $\left(C\right)$ bound by $y\geq 0,x^{2}+y^{2}=4\left({\rm clockwise}\right).$

Question

Prob. Find $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y$ with $\left ( C \right )$ closed by the path $y\geq 0, x^{2}+ y^{2}= 4\left ( {\rm clockwise} \right ).$

My attempt: $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y= \int_{\left ( C \right )}xy{\rm d}x+ \frac{x^{2}-4}{2}{\rm d}y+ \int_{0\leq y\leq 2}\frac{3y^{2}}{2}{\rm d}y,$ let $P\left ( x, y \right )= xy, Q\left ( x, y \right )= \frac{x^{2}- 4}{2},$ so we have $\frac{\partial}{\partial y}P\left ( x, y \right )= \frac{\partial}{\partial x}Q\left ( x, y \right )= x$ on whole plane $\left ( Oxy \right )$ then the integral isn't up to the route. Replace $x= 2\cos t, y= 2\sin t\geq 0, 0\leq t\leq \pi,$ solve $\int_{\left ( C \right )}xy{\rm d}x+ y^{2}{\rm d}y= 0.$

This is my 1st time using Green's theorem, and I'm not sure of my opinion, I replaced $\int_{\left ( C \right )}\frac{3y^{2}}{2}{\rm d}y$ by $\int_{0\leq y\leq 2}\frac{3y^{2}}{2}{\rm d}y,$ it's no use. What is your thought on that ? Thanks a real lot.

Answer 1

If it is a closed path,

$\int_{\left(C\right)} \ xy \ {\rm d}x+y^{2} \ {\rm d}y$

$Q_x = 0, P_y = x$

Given it is clockwise, the line integral is equivalent to

$\int_C (P_y - Q_x) \ dx \ dy = \int_C x \ dx \ dy$

As $x$ is an odd function and we have semicircular region symmetric to $y-$axis, the integral is zero.

Answer 2

The $y^2dy$ piece is conservative so it will equal

$$\frac{1}{3}y^3\Bigr|_{(-2,0)}^{(2,0 )} = 0$$

by the fundamental theorem of line integrals. The first integral is

$$\int_{-2}^2x\sqrt{4-x^2}dx = 0$$

because the function is odd. Hence the whole thing is $0$.