# Verify greens for the integral

• joemama69

## 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, shouldn't it equal o.

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

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

Then I strongly suggest you revies Green's theorem!

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

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