I just posted something with this but now I want to use Green's Theorem:
Compute the following three line integrals directly around the boundary C of the part R of the interior ellipse (x^2/a^2)+(y^2/b^2)=1 where a>0 and b>0 that lies in the first quadrant:
(a) integral(xdy-ydx)
(b) integral((x^2)dy)
(c) integral((y^2)dx)
So for:
(a) integral(xdy-ydx)= double integral(2)dA what are the bounds?
(b) integral((x^2)dy)= double integral(-2x)dA what are the bounds?
(c) integral((y^2)dx)= double integral(2y)dA what are the bounds?
I know how to evaluate all the double integrals, I just don't know the bounds. I'm assuming I would use parameterization.
I think the answer for (a) is (pi/2)*ab, for (b) is (2/3)*(a^2)b, and for (c) is (-2/3)*a(b^2), but I solved them directly from a line integral not Green's Theorem.
Thank you.
Thank you for responding. That one seems easy enough.
How do you do the second one though (the third one appears to follow similarly from the second one)?
I can set up the integration:
double integral (-2acost) da dt where x=acos(t)
I just don't know the respective bounds to get an answer of (2/3)*(a^2)b.
for the positive quadrant of the ellipse
if you are using vertical strips,
from the eqn. of the ellipse y = (b/a) sqrt{a^2 - x^2}
so x varies from x=0 to x=a.
(origin to one end of the major axis)
while y varies from y =0 to y = (b/a) sqrt{a^2 - x^2}
(x-axis to the ellipse)
that was for direct integration without substitution
note that a is a constant not a variable
try sketching a rough figure.
Well, actually I think it should be:
double integral of (2x)dA.
I have no idea how to solve this. Can't I just parametrize it though?
That would make it double integral of (2acos(t)).
I don't know this transformation stuff, but why I can't I just find appropriate bounds for this double integral?
I also tried double integral of (2x)dydx where y goes from 0 to x and x goes from 0 to bsint. It doesn't equal (2/3)a^2*b.
Could anyone just clarify this?
Correct, I made a mistake, forgot to take the partial derivative.
Parametrize what? It is already in double integral form.I have no idea how to solve this. Can't I just parametrize it though?
The curve is,
Thus,
Integrate once,
This is easy to integrate.
The the substitution does the simplificiation.