Green's Theorem

**Adebensjp05** · January 29th 2007, 05:36 PM

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.

**ThePerfectHacker** · January 29th 2007, 07:07 PM

Originally Posted by Adebensjp05

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,
$2\cdot \frac{1}{4} \cdot \pi \cdot a\cdot b=\frac{\pi ab}{2}$
Thus, my answer matches with thy answer.

**Adebensjp05** · January 29th 2007, 08:09 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.

**qpmathelp** · January 29th 2007, 10:04 PM

Originally Posted by Adebensjp05

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.

**ThePerfectHacker** · January 30th 2007, 09:37 AM

Originally Posted by Adebensjp05

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.

**Adebensjp05** · January 30th 2007, 09:45 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?

**ThePerfectHacker** · January 30th 2007, 09:52 AM

Originally Posted by Adebensjp05

Well, actually I think it should be:

double integral of (2x)dA.

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

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.

**Adebensjp05** · January 30th 2007, 10:01 AM

I did it quickly, but it looks like you get zero in the denominator.

**Adebensjp05** · January 30th 2007, 10:04 AM

Yeah I'm wrong. Thanks for the help. I really appreciate it.

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