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

• mansi
In summary, to 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, we can use any P and Q satisfying the condition e^(x^2) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}. One possible set of values is P(x,y)= cos(x)log(x)- ye^(x^2) and Q(x,y)= e^{\frac{sin(y)}{e^{y^2}}}!. This will give the same answer as using Q = 0 and P = -ye^(x^2), which is \int
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 boundry of the region. I've tried it...but it's too hard and besides,doesn't seem to work.
i

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!

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

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 $$\frac{\partial Q}{\partial x}$$?

Last edited by a moderator:
thanks...trust me,that DID help...

## 1. What is Green's Theorem and how is it used in double integration?

Green's Theorem is a mathematical tool that relates the line integral of a function around a closed curve to the double integral of the function over the region enclosed by the curve. In double integration, it allows us to evaluate the integral of a two-variable function by converting it into a line integral of a related function.

## 2. Why is it important to use Green's Theorem in this specific case of e^(x^2)?

In this case, the function e^(x^2) is not easily integrable using traditional methods. However, Green's Theorem allows us to convert the double integral of this function into a simpler line integral, making it easier to evaluate.

## 3. What is the general process for using Green's Theorem in double integration?

The general process for using Green's Theorem in double integration involves finding the partial derivatives of the function with respect to x and y, setting up the line integral using these derivatives, and then evaluating the integral using standard techniques.

## 4. Are there any limitations or conditions that must be met in order to use Green's Theorem for double integration?

Yes, Green's Theorem can only be applied to regions that are simply connected, meaning that there are no holes or intersecting boundaries. Additionally, the function being integrated must have continuous first-order partial derivatives over the region.

## 5. Can Green's Theorem be used for any type of double integration, or only specific cases?

Green's Theorem can only be used for certain types of double integration, specifically when the function being integrated is in the form of a two-variable function. It cannot be used for triple integration or for functions with more than two variables.

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