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Green's Theorem Problem
- Thread starter JerryCan
- Start date
Pretty typical Green's Theorem question but the fact that the circle is off-center is throwing me off!
Any help would be appreciated.
Suppose you were to just make two simple substitutions
u = x - 1
v = y - 5
What would be the equation of the circle in terms of u and v and what would the integral become?
Thanks for the response Ishuda! But I think I've realized this question is actually relatively simple; since the partial derivate of the y-component of the vector field in terms of x minus the partial derivative of the x-component of the vector field in terms of y is constant (equal to 1) - the value of the integral is the same regardless of where the circle is in the field.
So the value of the integral is just 1 times the area of the circle = pi.
Am I right here?
So the value of the integral is just 1 times the area of the circle = pi.
Am I right here?
Are you sure you have the sign correct? I think it isThanks for the response Ishuda! But I think I've realized this question is actually relatively simple; since the partial derivate of the y-component of the vector field in terms of x minus the partial derivative of the x-component of the vector field in terms of y is constant (equal to 1) - the value of the integral is the same regardless of where the circle is in the field.
So the value of the integral is just 1 times the area of the circle = pi.
Am I right here?
\(\displaystyle \oint\, (P\, dx\, +\, Q\, dy)\, =\, \int\, (\frac{\partial\, Q}{\partial\, x}\, -\, \frac{\partial\, P}{\partial\, y})\, dA\)