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Apply Green's theorem in a plane to evaluate $\displaystyle\int(2x^2-y^2)dx + (x^2+y^2)dy$, where $C$ is the curve enclosed by the semi-circle $x^2+y^2=1$ and the $x$-axis.

I've done this already $$ \int Pdx + Qdy = \iint (dQ/dx - dP/dy) \text{ } dxdy $$

$P=(2x^2-y^2)$

$Q=(x^2+y^2)$

$\displaystyle\frac{dQ}{dx}=2x$

$\displaystyle\frac{dP}{dy}=-2y$

$$ \iint (2x+2y)dxdy $$

So how do I get the range for $x$ and $y$ to integrate with respect to?

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  • $\begingroup$ Welcome to MSE. You'll get a much more positive response (more help and fewer votes to close) if you show that you've made an effort to solve the problem yours elf. What have you done so far? Where are you having trouble? $\endgroup$
    – saulspatz
    Sep 5, 2018 at 15:21
  • $\begingroup$ @saulspatz Thanks for response check my edit $\endgroup$
    – Eric kioko
    Sep 5, 2018 at 17:01

1 Answer 1

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You've done fine so far. The boundary of the region is the semicircle described, so the region you need for the double is simply the interior of that semicircle. In polar coordinates $0\leq r\leq1,\ 0\leq\theta\leq\pi.$

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