# Verify greens for the integral

## Homework Statement

verify greens for the integral (xy^2 i - x^2y j) dot dr over c, where c = y = x^2 from (-1,1) to (1,1). Evaluate both sides independantly to achieve the same answer

## The Attempt at a Solution

so i took the partials and got

-2xy - 2yx = -4xy, shouldnt it equal o.

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dx
Homework Helper
Gold Member

Why should it equal zero? What exactly is the question asking you to check?

HallsofIvy
Homework Helper

I'm confused by this. Green's theorem relates the integral around a closed path to the integral of the expression you give over the region bounded by the closed path.

"c = y = x^2 from (-1,1) to (1,1)" is not a closed path and so does not bound a region. Did you intend to include the line from (1, 1) back to (-1, 1)?

In any case that expression you give would be 0 only if the integral around the closed path were 0.

ya it does say "from -1,1 to 1,1 and the line segment from 1,1 to -1,1. sorry i didnt realize that was important

HallsofIvy
Homework Helper

Then I strongly suggest you revies Green's theorem!

benorin
Homework Helper

Forgot $$d\vec{r}$$.

Use $$\oint_{C}\left( xy^2\hat{i} -x^2y\hat{j}\right)\cdot d\vec{r} = \oint_{C}\left( xy^2\hat{i} -x^2y\hat{j}\right)\cdot \left( dx\hat{i} +2x\, dy\hat{j}\right) = \oint_{C}xy^2\, dx -2x^3y\, dy$$

oook check this

$$\oint$$xy2dx - 2x3dy

P = xy2 Q = 2x3y

$$\int$$$$\int$$6x2y - 2xy dy dx

-1 < x < 1, x2<y<1

$$\int$$-3x6 + x3 + 3x2 - x dx

= 8/7

benorin
Homework Helper

$$\oint_{C}xy^2\, dx -2x^3y\, dy = \int_{-1}^{1}\int_{x^2}^{1}\left( \frac{\partial }{\partial x}\left( -2x^3y\right) -\frac{\partial }{\partial y}\left( xy^2\right)\right) dydx$$​
$$= \int_{-1}^{1}\int_{x^2}^{1}\left(-6x^2y-2xy\right) dydx = \int_{-1}^{1}\left[ y^2\left(3x^2+x\right)\right|_{y=1}^{x^2} dx$$​
$$= \int_{-1}^{1}\left(3x^6+x^5-3x^2-x\right) dx=-{\frac{8}{7}}$$​