.Line Integral in the First Quadrant

[kidia]

Question:
Starting from anyone of the three corners of the path described here under,evaluate the line integral \intc (2xy^3)dx + (4x^2y^2)dy where C is the closed path forming the boundary of region in the first quadrant enclosed by x-axis,the line x=1 and the curve y=x^3

 

[Galileo]

Well, what have you got so far?

 

[saltydog]

Question:
Starting from anyone of the three corners of the path described here under,evaluate the line integral \intc (2xy^3)dx + (4x^2y^2)dy where C is the closed path forming the boundary of region in the first quadrant enclosed by x-axis,the line x=1 and the curve y=x^3

Any time you have a nice closed path with the line integral in that form, try and remember to use Green's Theorem:

\oint_C Mdx+Ndy=\iint_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dA

Just plug it in right?

 

[cronxeh]

Not so fast saltydog - I don't think he is allowed to use Green or Stokes for now.

just parametrize the curves, find derivative of that curve and then you can find individual line integrals of int(F dot dr) for all segments

 

[dextercioby]

It's easier to compute the double integral than the line integral...So i vote for Green as well.

Daniel.

 

[snoble]

I might be embarrassing myself here but isn't there some sort of requirement that the function be analytic before you use Green's Theorem? And not all polynomials are analytic.

 

[mathwonk]

i don't think green has much hypotheses? maybe smoothness? try proving it and see what you need.

it follows from fubini (repeated integration) plus ftc.

of course all polynomials are analytic, but i am asuming you meant that you thought the form had to be closed, which is not necessary.

 

[dextercioby]

That domain should be described by x running from 0 to 1 and y from 0 to x^{3}...

Daniel.

 

[HallsofIvy]

I might be embarrassing myself here but isn't there some sort of requirement that the function be analytic before you use Green's Theorem? And not all polynomials are analytic.

All the polynomials I know are analytic!

(And you don't need the function to be analytic for Green's theorem- "continuously differentiable" is enough.)

The only question is whether kidia has had "Green's theorem" in Calculus class yet.

It's not that hard to actually integrate along the path:
1) from (0,0) to (1, 0) along the x- axis: take x= t, y= 0 so that dx= dt, dy= 0. Since y= 0 along that path, the integral on it is 0.

2) from (1,0) to (1,1) along the line x= 1:take x= 1, y= t so that dx= 0, dy= dt. integrate
4t²dt from 0 to 1.

3) from (1,1) to (0,0) along the line y= x: take x= t, y= t so that dx= dt, dy= dt. integrate from t= 1 down to t= 0.

 

[snoble]

I knew I was setting myself up for embarresment.