# Green's Theorem

• Jan 29th 2007, 06:36 PM
Green's Theorem
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.
• Jan 29th 2007, 08:07 PM
ThePerfectHacker
Quote:

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.

I assume the path travels along the outside of the ellipse and the semi minor and major axes in the first quadrant. This is the region $D$.

For the first one,
$\oint_C (-ydx+xdy)=\int_D \int\left( \frac{\partial(x)}{\partial x}-\frac{\partial(-y)}{\partial y} \right) dA$
Thus,
$\int_D \int 2dA=2\int_D \int dA=2(\mbox{AREA})$
The area of ellipse (full) is $\pi ab$ in this case it is $1/4$ of that thus,
$2\cdot \frac{1}{4} \cdot \pi \cdot a\cdot b=\frac{\pi ab}{2}$
• Jan 29th 2007, 09:09 PM
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.
• Jan 29th 2007, 11:04 PM
qpmathelp
Quote:

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.

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.
• Jan 30th 2007, 10:37 AM
ThePerfectHacker
Quote:

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)?
.

Thus, you need to find,
$\oint_C (0dx +x^2 dy)=\int _D \int \left( \frac{\partial x^2}{\partial x}-\frac{\partial (0)}{\partial y} \right) dA=\int_D \int x^2 dA$

Now to find the integral use the transformation,
$x/a=u$ and $y/b=v$
This will simpify the region into a circle,
And the Jacobian is simplify a constant multiplier.
• Jan 30th 2007, 10:45 AM
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?
• Jan 30th 2007, 10:52 AM
ThePerfectHacker
Quote:

Well, actually I think it should be:

double integral of (2x)dA.

Correct, I made a mistake, forgot to take the partial derivative.
Quote:

I have no idea how to solve this. Can't I just parametrize it though?
Parametrize what? It is already in double integral form.
$\int_D \int (2x) dx$
The curve is,
$y=\frac{b}{a}\sqrt{a^2-x^2}$
$x=0$
$y=0$
$x=a$
Thus,
$\int_0^a \int_0^{(b/a)\sqrt{a^2-x^2}} 2x dy\, dx$

Integrate once,
$\int_0^a (b/a)(2x)\sqrt{a^2-x^2} dx=\frac{b}{a}\int_0^a 2x\sqrt{a^2 -x^2} dx$
This is easy to integrate.
The the substitution $u=a^2-x^2$ does the simplificiation.
• Jan 30th 2007, 11:01 AM
I did it quickly, but it looks like you get zero in the denominator.
• Jan 30th 2007, 11:04 AM