# Greens Theorem 2

1. Jan 9, 2012

### bugatti79

1. The problem statement, all variables and given/known data

Use greens theorem to evaluate this line integral $\oint_{C} 6xy- y^2$ assuming C is oriented counter clockwise. The region bounded by the curves y=x^2 and y=x.

2. Relevant equations

$\displaystyle \int\int_{R}\Bigl( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Biggr) = \oint_{C} Pdx + Qdy$

I calculate

$\displaystyle \oint_{C} 6xy- y^2 dx= \int_{0}^{1} \int_{x^2}^{ \sqrt x} (-6x+2y)dy dx$

This correct start?

2. Jan 9, 2012

### vela

Staff Emeritus
You need to specify the problem correctly. In other words, don't omit stuff. This expression doesn't make sense as written:
$$\oint_{C} 6xy- y^2$$

3. Jan 9, 2012

### HACR

If $$\int_{C} -3x^{2}dy-y^{2}dx$$ as the integral representation, then -6x+2y is totally valid.

4. Jan 10, 2012

### bugatti79

Sorry, that was a typo but I had it correct further on.

$\displaystyle \oint_{C} 6xy- y^2 dx$

Hence is my calculation in post 1 correct?

5. Jan 10, 2012

### vela

Staff Emeritus
You mean
$$\oint_{C} (6xy- y^2) dx.$$
The limits on your area integral are wrong.

6. Jan 10, 2012

### bugatti79

Sorry, that sqrt was a typo. It should read...

$\displaystyle \oint_{C} (6xy- y^2) dx= \int_{0}^{1} \int_{x^2}^{ x} (-6x+2y)dy dx$...good to go?

7. Jan 10, 2012

### vela

Staff Emeritus
Yes, that looks right. You could always check by evaluating the integrals both ways.

8. Jan 10, 2012

### bugatti79

but how would you evaluate the LHS of above if we dont have the limits for the LHS?

9. Jan 10, 2012

### lanedance

the LHS is a line intergal, and you are given the curves, so you should be able to subtitute y(x) into the integral and evaluate to the endpoint of the curve

In this case it will require splitting the integral into 2 parts, with one for each curve, and the endpoints will be where the curves intersect

http://en.wikipedia.org/wiki/Green's_theorem

10. Jan 10, 2012

### bugatti79

Ok. I calculated line integral (LHS) which involved 2 integrals and then summing them.

Now I am to evalute the following by changing dx to dy. Should I expect a different answer?

Now for $\displaystyle \oint_{C} (6xy- y^2) dy= \int_{0}^{1} \int_{x^2}^{ x} 6y dy dx=\frac{2}{5}$..........?

11. Jan 10, 2012

### lanedance

what do you think?

(PS not trying to be smart, but its worth you trying to think things through)

12. Jan 11, 2012

### bugatti79

Part of me thinks that the answers should be the same because we are calculating the area of the bounded region so it doesnt matter what direction we integrate along....?