Greens Theorem: Evaluate Line Integral of 6xy-y^2

[bugatti79]

Homework Statement

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.

Homework 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?

 

[vela]

Homework Statement

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.

Homework 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?

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

 

[HACR]

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

 

[bugatti79]

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

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?

 

[vela]

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?

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

 

[bugatti79]

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

Homework Statement

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.

Homework 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?

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?

 

[vela]

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

 

[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?

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

but how would you evaluate the LHS of above if we don't have the limits for the LHS?

 

[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

for more info see wiki
http://en.wikipedia.org/wiki/Green's_theorem

 

[bugatti79]

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

for more info see wiki
http://en.wikipedia.org/wiki/Green's_theorem

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}...?

 

[lanedance]

what do you think?

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

 

[bugatti79]

what do you think?

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

Part of me thinks that the answers should be the same because we are calculating the area of the bounded region so it doesn't matter what direction we integrate along...?