Question

I have to apply the Green's theorem to a triangle. I've got the integral of e^xcosy dx-e^xsiny dy. A=(ln2,0), C=(0,1) and B(-ln2,0).

Can someone help me please?
Thank you.

Answer 1

\(F(x,y)=(e^x\cos y, -e^x\sin y)=(P,Q) \)
Green:
\[
\text{your integral} = \int\int_\text{surface of triangle} (\frac{\partial Q}{\partial y}-\frac{\partial P}{\partial x}) dxdy
\]
You use Fubuni to write correctly \(\int\int_\text{surface of triangle}\), that is, you can choose the order of integration.
Here it is best to integrate on two different triangles: the two that you get when you cut along the \(y\)-axis.

Answer 2

your integral = \[\int_0^{\ln 2}(\int_0^{1-x} H(x,y) dy)dx +
\int_{-\ln 2}^0(\int_0^{x+1} H(x,y) dy)dx,\] where \(H(x,y)=(∂Q/∂y−∂P/∂x)\)

Answer 3

sorry I made a mistake with the integrals..

Answer 4

first integral:
\[
\int_0^{\ln 2}(\int_a^b H(x,y) \,dy)dx
\]
obviously, \(a=0\), but \(b=?\)

Answer 5

okay this is right:
\[
\int_{\ln2}^0 (\int_0^{b} H(x,y) \:dy)dx
\]where \(b\) is y_max. the equation of the line is \(y=(1/(-\ln2))x+1\). That is \(b\).

Answer 6

@reemii is it that? is it not line integral?

Answer 7

The statement is an integral over a path in \(\mathbb{R}^2\), but the Green formula enables to write an integral on a surface (the delimited by the closed path).

Answer 8

hold on, look at the function, it's .....dx - ..... dy, is it not the form of line integral? Please, don't confuse me. I need that part for my final, too.

Answer 9

I'm sure. what he wrote : "e^xcosy dx-e^xsiny dy on a triangle ABC"
is to be taken as a notation. I think the word for that topic is "flow integral".
Stated differently, the starting point is: \(\int_\text{triangle ABC} \text{flow} F \cdot \mathrm ds\). do you recognize this?

Answer 10

the question is unclear about the triangle. He disappeared , we cannot confirm the question but to me, it looks like line integral with parametric equation from A to B and then from B to C . Moreover, the form of function is not dA =dx dy but ....dx - ....dy, separately.

Answer 11

this is a notation. the fact that i'm pretty sure i'm right is beecause he mentions Green, this means the path is closed.
The notation "e^xcosy dx-e^xsiny dy" means that he has to compute the flow of the vector field (e^xcosy , -e^xsiny) along the path described be the triangle.

Answer 12

thanks for being here, the Asker,

Answer 13

oh, sorry i'm not aware of the vocabulary.
It seems the starting point was indeed : line integral of a vector field.

Answer 14

okk guys. My question is: integralo f C e^xcosydx-e^xsinydy where Cis the broken line from A=(ln2,0) to D=(0,1) and then from D to B=(-ln2,0). Hint: Apply Green's theorem to the integral around the closed curve ADBA.

Answer 15

ok, I am waiting for other's solution, no idea

Answer 16

The answer that I need to get is -3/2 . But something that I am doing is wrong cause I 'm finding other answers.

Answer 17

ok maybe don't cut the triangle.
Integrate like this: \(\int (\int ... dx )dy\).