Calculating Integral Using Greens Theorem

boneill3
Messages
126
Reaction score
0

Homework Statement



Use greens theorem to calculate.
\int_{c}(e^{x}+y^{2})dx+(e^{x}+y^{2})dy

Where c is the region between y=x2y=x

Homework Equations



Greens Theorem

\int_{c}f(x.y)dx+g(x,y)dy= \int_{R}\int (\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y})dA


The Attempt at a Solution



\frac{\partial g}{\partial x}= 2x
\frac{\partial g}{\partial x}= 2y
Calculate the integral

\int_{0}^{x}\int_{0}^{\sqrt{y}}2x-2y\text{ }dy dx

=\frac{x^2}{2}-\frac{4x^{5/2}}{5}

Does this look right?
regards
 
Physics news on Phys.org
with f(x,y)=g(x,y)=exp(x)+y*y, dg/dx=exp(x), the second dg/dx is a typo.

if you want the region bounded by y=x^2 and y=x, the inside integral must be from x^2 to x and the outside 0 to 1 with area element dydx, the result needs to be a value rather than a function, just something to get use to with multiple integrals.
 
Thanks

\int_{0}^{1}\int_{x}^{x^2}2x-2y\text{ }dy dx

=\frac{1}{30}

With the outside limits of double integrals eg 0 to 1 do they always have to be constants?
regards
 
If the result is supposed to be a constant, then, yes, the limits of the integral have to be numbers, not variables!
 
Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top