Another line integral question....

**daskywalker** · November 20th 2009, 09:26 AM

Evaluate the line integral directly, WITHOUT using Green's theorem.

Integral of (x-y)dx+(x+y)dy where C is the circle with center the origin and radius 2.

I have to show that both methods give me the same answer. I know how to do it with Green's theorem...but not otherwise.

**qmech** · November 20th 2009, 09:48 AM

Rectangular and polar coordinates have the relation:
$ x = r cos \theta $
$ y = r sin \theta $
from which we get:
$ dx = cos \theta dr - r sin\theta d\theta $
$ dy = sin \theta dr + r cos\theta d\theta $
Substitute into your problem:
$(x-y)dx = (rcos\theta - rsin\theta)(cos\theta dr - r sin\theta d\theta ) $
$(x+y)dy = (rcos\theta + rsin\theta)(sin\theta dr + r cos\theta d\theta ) $
Your contour is a circle, so r=constant, so the integral over r is 0 - we can ignore the dr terms.
Multiply out the rest to get to get
$ (x-y)dx + (x+y)dy= -r^2 sin \theta cos \theta d \theta + r^2 sin^2 \theta d\theta + r^2 cos^2 \theta d\theta + r^2 sin \theta cos\theta d\theta $
The RHS collapses to $r^2 d\theta$
Your integral is for r = 2, and $\theta$ going from 0 to $2 \pi$
so it equals $2^2 * 2 \pi = 8 \pi$

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