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## Homework Statement

Use Green''s Theorem in the plane to check:

[tex]

\oint_C (xy+y^2) \> dx + x^2 \> dy

[/tex]

Where C is the closed curveof the region bound between the curve of [itex]y=x^2[/itex] and the line [itex]y=x[/itex]

## Homework Equations

[tex]

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

[/tex]

## 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 [itex]v=x^2[/itex] and [itex]u=xy+y^2[/itex], so that [itex]\partial_x v = 2x[/itex] and [itex]\partial_y u = x+2y[/itex] then tried setting the problem up...

[tex]

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}

[/tex]

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

For [itex]I_{x^2}[/itex] :we have [itex]y=x^2[/itex] therefore [itex]dy=2x \> dx[/itex] and I substitute in for [itex]y=x^2[/itex] to get everything in terms of x

[tex]

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} \\

[/tex]

Then for [itex]I_{x}[/itex] we have [itex]y=x[/itex] therefore [itex]dy=dx \> dx[/itex] and I substitute in for [itex]y=x[/itex] to get everything in terms of x

[tex]

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

[/tex]

and then [itex]I_2 = I_{x^2} + I_{x} = \frac{37}{30} - 1 = \frac{7}{30}[/itex]

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

[tex]

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 \\

[/tex]

Where it should only be the one term multiplied by 2x?

[tex]

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}

[/tex]

Which gives [itex]I_{x^2}+I_x = \frac{19}{20} - 1 =-\frac{1}{20}[/itex]

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