Greens Theorem: Evaluating Integral of x^2 y dx +(y+x y^2)dy

[bugatti79]

Homework Statement

Use greens theorem to evaluate the integral

Homework Equations

\int x^2 y dx +(y+x y^2)dy where c is the boundary of the region enclosed by y=x^2 and x=y^2.

The Attempt at a Solution

The integral is \displaystyle \int_{0}^{1} \int_{x^2}^{\sqrt {x}} y^2+x^2 dy dx

1) How where the outer limits determined?

2) For the inner limits, why is x^2 on the bottom

3) Green's Theorem is based on partial integration right? ie we integrate one variable while keeping the other fixed.

Thanks

 

[TachyonRunner]

Green's Theorem states that
\int\int_{R}\Bigl( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Biggr) = \oint_{C} Pdx + Qdy,
where C is the boundary curve of the region R.
(1) Your outer limits are determined by the region begin integrated over. You were given the curve which encloses R, from that you can determine R. (Draw a picture)
(2) This goes along with (1). Draw a picture, and you will see why x^{2} is the lower bound. Remember, that first integral is integrating over y, that is to say you are integrating as y ranges from x^{2} to \sqrt{x}.
(3) Green's Theorem is not based on partial integration. It instead relates the integral of a vector field through a region to its integral around the boundary. Hence it is very useful as it can simplify integrals.

What you are describing in (3) is a method of evaluating double integrals, in which case you are right. In this scenario you are integrating over y first, then over x. Conversely you could integrate over x first and then y, except that you would then need to vary x from y^{2} to \sqrt{y} and x from 0 to 1 (a picture will allow you to see why this is so).

I also believe that there is a sign error in one of the two formulas which you posted so you should double check that.

 

[bugatti79]

Green's Theorem states that
\int\int_{R}\Bigl( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Biggr) = \oint_{C} Pdx + Qdy,
where C is the boundary curve of the region R.
(1) Your outer limits are determined by the region begin integrated over. You were given the curve which encloses R, from that you can determine R. (Draw a picture)
(2) This goes along with (1). Draw a picture, and you will see why x^{2} is the lower bound. Remember, that first integral is integrating over y, that is to say you are integrating as y ranges from x^{2} to \sqrt{x}.
(3) Green's Theorem is not based on partial integration. It instead relates the integral of a vector field through a region to its integral around the boundary. Hence it is very useful as it can simplify integrals.

What you are describing in (3) is a method of evaluating double integrals, in which case you are right. In this scenario you are integrating over y first, then over x. Conversely you could integrate over x first and then y, except that you would then need to vary x from y^{2} to \sqrt{y} and x from 0 to 1 (a picture will allow you to see why this is so).

I also believe that there is a sign error in one of the two formulas which you posted so you should double check that.

Ah...very good thank you. I get your 3 points.

Although there is no error as is on my worksheet...?

 

[TachyonRunner]

Well from Green's theorem we can see that the two integrals you have written down do not agree. Indeed from the first integral P = x^{2}y and Q = y+xy^{2} and hence \frac{\partial P}{\partial y} = x^{2} and \frac{\partial Q}{\partial x} = y^{2} so that \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x^{2} = y^{2} - x^{2}. From here you can see where the sign error in the expressions you have given is.

In fact, I would encourage you to explicitly compute the line integral and check that the line integral equals the double integral (check Green's theorem for this case). You will see that indeed the double integral which you have written down does not give the correct answer.

 

[bugatti79]

Well from Green's theorem we can see that the two integrals you have written down do not agree. Indeed from the first integral P = x^{2}y and Q = y+xy^{2} and hence \frac{\partial P}{\partial y} = x^{2} and \frac{\partial Q}{\partial x} = y^{2} so that \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x^{2} = y^{2} - x^{2}. From here you can see where the sign error in the expressions you have given is.

In fact, I would encourage you to explicitly compute the line integral and check that the line integral equals the double integral (check Green's theorem for this case). You will see that indeed the double integral which you have written down does not give the correct answer.

Ok, just bad computing on my part. Thanks

 

[bugatti79]

Well from Green's theorem we can see that the two integrals you have written down do not agree. Indeed from the first integral P = x^{2}y and Q = y+xy^{2} and hence \frac{\partial P}{\partial y} = x^{2} and \frac{\partial Q}{\partial x} = y^{2} so that \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x^{2} = y^{2} - x^{2}. From here you can see where the sign error in the expressions you have given is.

In fact, I would encourage you to explicitly compute the line integral and check that the line integral equals the double integral (check Green's theorem for this case). You will see that indeed the double integral which you have written down does not give the correct answer.

On second thoughts, how would one compute the line integral for comparison against the double integral when we don thave specific limits for the former?

Thanks.

 

[lanedance]

see your other post