# Verify Green's Theorem in the plane for...

## Homework Statement

Use Green''s Theorem in the plane to check:
$$\oint_C (xy+y^2) \> dx + x^2 \> dy$$
Where C is the closed curveof the region bound between the curve of $y=x^2$ and the line $y=x$

## Homework Equations

$$\oint_C u \> dx + v \> dy = \int \int_A (\partial_x v - \partial_y u) \> dx \> dy$$

## The Attempt at a Solution

Right, this is my first time trying to solve a problem using Greens Theorem in the plane, and am having trouble. I think both the LHS and RHS are wrong, never mind having them the same haha.

First I did the RHS, the surface area part.
First I deduced that $v=x^2$ and $u=xy+y^2$, so that $\partial_x v = 2x$ and $\partial_y u = x+2y$ then tried setting the problem up...

$$I_1 = \int \int_A (\partial_x v - \partial_y u) \> dx \> dy \\ I_1 = \int \int_A (2x-x-2y) \> dx \> dy \\ I_1 = \int_0^1 dx \> \int_{y=x^2}^{y=x} x-2y \\ I_1 = \int_0^1 dx [xy-y^2]_{x^2}^{x} \\ I_1 = \int_0^1 dx \> ([x^2 - x^2]-[x^3 -x^4]) \\ I_1 = \int_0^1 (x^4 - x^3) \> dx = [\frac{x^5}{5}-\frac{x^4}{4}]_0^1 = -\frac{1}{20}$$

And for the contour integral I split it into two parts, with the first part $I_{x^2}$ being the integral going up the curve $y=x^2$ and the second part $I_x$ being the bit going back down the line $y=x$

For $I_{x^2}$ :we have $y=x^2$ therefore $dy=2x \> dx$ and I substitute in for $y=x^2$ to get everything in terms of x
$$I_{x^2} =\int (xy+y^2) \> dx + x^2 \> dy \\ I_{x^2} =\int_0^1 (x^3 + x^4 + x^2)2x \> dx \\ I_{x^2} = 2 \int_0^1 (x^5 + x^4 + x^3) \> dx = 2[\frac{x^6}{6}+\frac{x^5}{5}+\frac{x^4}{4}]_0^1 = \frac{37}{30} \\$$
Then for $I_{x}$ we have $y=x$ therefore $dy=dx \> dx$ and I substitute in for $y=x$ to get everything in terms of x
$$I_{x} =\int (xy+y^2) \> dx + x^2 \> dy \\ I_{x} =\int_1^0 (x^2+x^2+x^2) \> dx \\ I_{x} = -3 \int_0^1 x^2 = -3[\frac{x^3}{3}]_0^1 = -1$$
and then $I_2 = I_{x^2} + I_{x} = \frac{37}{30} - 1 = \frac{7}{30}$

I was bound to get it wrong on my first go at using the theorem, any help/advice is much appreciated! Thanks :)

EDIT:

I think I realised a mistake, in this bit
$$I_{x^2} =\int (xy+y^2) \> dx + x^2 \> dy \\ I_{x^2} =\int_0^1 (x^3 + x^4 + x^2)2x \> dx \\$$
Where it should only be the one term multiplied by 2x?
$$I_{x^2} =\int (xy+y^2) \> dx + x^2 \> dy \\ I_{x^2} =\int_0^1 ((x^3 + x^4) + (x^2)2x) \> dx \\ I_{x^2} =\int_0^1 (x^3 + x^4) + (2x^3) = [\frac{x^4}{4}+\frac{x^5}{5}+\frac{2x^4}{4}]_0^1 = \frac{1}{4}+\frac{1}{5}+\frac{2}{4}=\frac{19}{20}$$

Which gives $I_{x^2}+I_x = \frac{19}{20} - 1 =-\frac{1}{20}$

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I only did this really quick but I get ##-1/20## for both methods so I would guess your second one is wrong. Specifically redo the ##x^2## integral (it's wrong) and you're also integrating in the wrong direction. counter-clockwise is positive.

I only did this really quick but I get ##-1/20## for both methods so I would guess your second one is wrong. Specifically redo the ##x^2## integral (it's wrong) and you're also integrating in the wrong direction. counter-clockwise is positive.

Yeah did you see my edit? I redone it but still doesn't match. With regards to the direction, I assumed it was counter clockwise as all the examples we have had were counter clockwise. Obviously that doesn't necessarily mean this would be too, but how can you tell which way you have to go? I assume the negative answer for the surface bit gives it away yeah but if you did not know that, is there a way to tell?

Thanks for your help! Very much appreciated, at least I know I got the one bit correct, and will know when I have succeeded. Thanks!

Counter-clockwise is positive by definition (if you taken mechanics you have the same rule there for a right handed coordinate system.) The thing is that on ##x\in[0, 1]## ##x^2## is smaller (or "below") than ##x## so if we go counter clockwise we have to the ##y=x## curve first from 0 to 1 and then back again from 1 to 0 along the x^2 curve.

Look closely at the integral. You already have an ##dx## for the first part so you don't need to parameterizate that since it's already done. It's only for the second part you have ##dy = \frac{dy}{dx}dx##.

Edit: you got it! posted just after me!

• Noticed another mistake, not sure where the extra x factor came from!

$$I_{x^2} =\int (xy+y^2) \> dx + x^2 \> dy \\ I_{x^2} =\int_0^1 ((x^3 + x^4) + (x^2)2x) \> dx \\ I_{x^2} =\int_0^1 (x^3 + x^4) + (2x^3) = [\frac{x^4}{4}+\frac{x^5}{5}+\frac{2x^4}{4}]_0^1 = \frac{1}{4}+\frac{1}{5}+\frac{2}{4}=\frac{19}{20}$$

Which gives $I_{x^2}+I_x = \frac{19}{20} - 1 =-\frac{1}{20}$

Yay!

vela
Staff Emeritus
Homework Helper
$$I_{x^2} =\int (xy+y^2) \> dx + x^2 \> dy \\ I_{x^2} =\int_0^1 ((x^3 + x^4) + (x^2)2x) \> dx \\ I_{x^2} =\int_0^1 (x^3 + x^4) + (2x^3) = [\frac{x^6}{6}+\frac{x^5}{5}+\frac{2x^4}{4}]_0^1 = \frac{1}{6}+\frac{1}{5}+\frac{2}{4}=\frac{13}{15}$$
Where did the ##x^6## (and some of the other terms) come from?

Where did the ##x^6## (and some of the other terms) come from?
Yeah I just realised that, see my post directly above, I think I have solved it now. Thanks!

Edit: you got it! posted just after me!

Yeah thank's for your help and for taking a look :)

STEMucator
Homework Helper
Yeah did you see my edit? I redone it but still doesn't match. With regards to the direction, I assumed it was counter clockwise as all the examples we have had were counter clockwise. Obviously that doesn't necessarily mean this would be too, but how can you tell which way you have to go? I assume the negative answer for the surface bit gives it away yeah but if you did not know that, is there a way to tell?

Thanks for your help! Very much appreciated, at least I know I got the one bit correct, and will know when I have succeeded. Thanks!

It is conventional to assume counter-clockwise is the positive direction in a right handed co-ordinate system. Whenever you perform an integration over a simple region like the one in your problem statement, you should assume counter-clockwise integration over the contour.

What do you think would happen to the final answer if you integrated clockwise instead?

Now suppose you have a curve ##C_1 := x^2 + y^2 = 16##, which bounds a region ##D_1 := x^2 + y^2 \leq 16##. Let ##\vec F(x, y) = \frac{-y \hat i + x \hat j}{x^2 + y^2}##.

Imagine you wanted to perform a line integration for ##\vec F## over the curve ##C_1##, which by Green's theorem is equivalent to double integrating the difference of partial derivatives over an area that is bounded by ##C_1##. The problem is, the area bounded by ##C_1## contains the origin, so applying Green's theorem in its most general form is not possible. The reason being is ##\vec F## has a discontinuity at the origin, and so you will not be able to apply Green's theorem for this particular setup.

Applying Green's theorem so you can see this problem:

$$\oint_{C_1} \vec F \cdot d \vec r = \oint_{C_1} P \space dx + Q \space dy = \iint_{D_1} Q_x - P_y \space dA$$

How do you solve the problem?

Start by placing a smaller curve inside of the bigger one, lets say ##C_2 := x^2 + y^2 = 1##. The area bounded between ##C_1## and ##C_2## can now be used to solve the problem. Now a question for you:

How do we orient the two curves such that the way you orient them bounds the area between them?

If you orient the curves incorrectly, the inner curve will contain the origin, and the outer one will also contain the origin as a result because the smaller one contains the origin. If you orient the curves correctly, the origin will not be contained.