# Thread: Evaluate the line integral

1. ## Evaluate the line integral

Hello,
please try to solve this question.
Thanks

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

3. 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

4. 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.

5. 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