Double Integral of e^(x^2) Using Green's Theorem

[mansi]

find the double integral of the function e^(x^2) over the region where
y/2 <= x<= 1 and 0<=y <=2 USING GREEN's THEOREM.

I can't imagine how we'd use green's theorem here...if F=(P,Q) is the function, are we supposed to find P and Q using green's theorem and then parametrize the boundary of the region. I've tried it...but it's too hard and besides,doesn't seem to work.
i

 

[Data]

well, to use Green's Theorem, you need P, Q s.t.

$$e^{(x^2)} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}.$$

What you need to realize is that you can take ANY P,Q satisfying this requirement. In this case, I would advise using

$$Q = 0, \ P = -ye^{(x^2)}.$$

By Green's Theorem, you then get

$$\int_0^2 \int_{y/2}^1 e^{(x^2)} \ dx \ dy = \oint_C -ye^{(x^2)} \ dx$$

where C is the boundary of the region you specified in the counter-clockwise direction. Now just parameterize it and see where you get!

 

[mansi]

thanks...but just to make it clearer,can you give me another set of of values of P and Q?

 

[HallsofIvy]

Do you have any reason to think that other values of P and Q will make this clearer?

Fine, take $$P(x,y)= cos(x)log(x)- ye^{x^2}$$ and Q(x,y)= $$e^{\frac{sin(y)}{e^{y^2}}$$!

Do you see why that would give exactly the same answer?
(What are $$\frac{\partial P}{\partial y}$$ and $$\frac{\partial Q}{\partial x}$$?

 

[mansi]

thanks...trust me,that DID help...