Evaluate the line integral

**m777** · June 7th 2007, 09:26 PM

Hello,
please try to solve this question.
Thanks

**ThePerfectHacker** · June 8th 2007, 06:05 AM

$\oint_C x^3 dx - 2xy dy = \iint_D \left( \frac{\partial (-2xy)}{\partial x} - \frac{\partial x^3}{\partial y} \right) \ dA = -2\iint_D y \ dA =$ $- 2\int_0^2 \int_0^{2-x} y \ dy \ dx$

**topsquark** · June 8th 2007, 06:24 AM

Evaluate the line integral $\int_C x^3 dx - 2xy dy$ where C comprises the three sides of a triangle joining O (0, 0) A (2, 0) and B (0, 2)

From O to A we have that y does not change so dy = 0.
$\int_{O to A} x^3 dx - 2xy dy = \int_0^2 x^3 dx$

$= \frac{1}{4}x^4|_0^2 = \frac{1}{4}2^4 = 4$

From A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so
$dy = -dx$

Thus
$\int_{(2,0) to (0,2)} x^3 dx - 2xy dy$

$= \int_2^0 x^3 dx + 2x(-x + 2)(-dx)$

$= \int_2^0 (x^3 + 2x^2 - 4x)dx$

$= \left ( \frac{1}{4}x^4 + \frac{2}{3}x^3 - 2x^2 \right ) _2^0$

$= - \left ( \frac{1}{4}2^4 + \frac{2}{3}2^3 - 2 \cdot 2^2 \right ) = - \left ( 4 + \frac{16}{3} - 8 \right )$

$= 4 - \frac{16}{3}$

So the overall integral will be:
$4 + 4 - \frac{16}{3} = 8 - \frac{16}{3} = \frac{8}{3}$

-Dan

**ThePerfectHacker** · June 8th 2007, 06:34 AM

Originally Posted by topsquark

From O to A we have that y does not change so dy = 0.
$\int_{O to A} x^3 dx - 2xy dy = \int_0^2 x^3 dx$

$= \frac{1}{4}x^4|_0^2 = \frac{1}{4}2^4 = 4$

From A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so
$dy = -dx$

Thus
$\int_{(2,0) to (0,2)} x^3 dx - 2xy dy$

$= \int_2^0 x^3 dx + 2x(-x + 2)(-dx)$

$= \int_2^0 (x^3 + 2x^2 - 4x)dx$

$= \left ( \frac{1}{4}x^4 + \frac{2}{3}x^3 - 2x^2 \right ) _2^0$

$= - \left ( \frac{1}{4}2^4 + \frac{2}{3}2^3 - 2 \cdot 2^2 \right ) = - \left ( 4 + \frac{16}{3} - 8 \right )$

$= 4 - \frac{16}{3}$

So the overall integral will be:
$4 + 4 - \frac{16}{3} = 8 - \frac{16}{3} = \frac{8}{3}$

-Dan

My answer is exactly negative of your answer if you do it out by Green's theorem! Maybe you did it clockwise.

**topsquark** · June 8th 2007, 08:15 AM

Originally Posted by ThePerfectHacker

My answer is exactly negative of your answer if you do it out by Green's theorem! Maybe you did it clockwise.

I suppose I should have suspected this, but I didn't realize it was an integral over a closed area. I only did the integral from O to A to B. I never connected B to O.

So from B (0, 2) to O (0, 0) dx = 0, with x = 0. Thus
$\int_{(0, 2) to (0, 0)} x^3 dx - 2xy dy = \int_2^0 (-2 \cdot 0 \cdot y) dy = \int_2^0 0 dy = 0$

So my final integral is STILL $\frac{8}{3}$ .

And if you look at the succession of points I did the integral in the counterclockwise sense. I can't argue with your result (though I was never very good at implementing Green's theorem anyway) but I can't find a mistake in my own work?

-Dan

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