I have problem with verifying Green`s theorem for integral: $$\int _C (x^2+y^2+\cos(x))dx+(x^2+y^2+\sin(y))dy$$ C is the boundary of the semicircle: $$\{(x,y) \in R^2: x^2+y^2\leqslant4 \wedge x\geqslant0\}$$
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proving or verifying ? – rrr Sep 05 '14 at 10:13
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You`re right, verifying, sorry. – Robert Sep 05 '14 at 10:14
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its okay :) so basically you need to show that the closed loop line integral equals the double integral of curl of given vector field – rrr Sep 05 '14 at 10:16
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maybe start by parameterizing the path – rrr Sep 05 '14 at 10:17
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Yes, I know that I should parametrize path, but after parametrizing curve I have in integral something like sin(2cos(x)). – Robert Sep 05 '14 at 10:20
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Yeah I see that doesn't look pleasant to integrate >.< – rrr Sep 05 '14 at 10:28
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Parametrise the boundary by the line connecting (-2,0) and (2,0) (choose $x$ as parameter) (in that direction) and the circle by its angle $\theta$. The contourintegral then becomes the sum of the integrals over both the straight line and the circle. I assume the straight line does not pose any problems. The part of the circle becomes: $$ 2\int_0^\pi \left[ 4(\cos \theta - \sin \theta) -\sin(\theta)\cos(2\cos(\theta)) + \cos(\theta)\sin(2\sin(\theta)) \right]d\theta $$ Now take a look at the derivative of $\cos(\sin(\theta))$: $$ (\cos(\sin(\theta)))' = -\sin(\sin(\theta))\cos(\theta) $$ Appart from a constant factor this is exactly what we're asked to integrate. Therefor the integral becomes: $$ 2\int_0^\pi \left[ 4(\cos \theta - \sin \theta) +\frac{1}{2}\frac{d}{d\theta}\sin(2\cos(\theta)) + \frac{1}{2}\frac{d}{d\theta}\cos(2\sin(\theta)) \right]d\theta $$ For which you are now able to apply the first main theorem of calculus. (at least on the second part of the integral)
