Changing the interval of integration

Amaelle · Dec 19, 2021

Greetings Dear community!

Typeset textbook exercise 1: vector field F(x, y, z) = (-z, x, y) and curve gamma from the plane z = y intersecting the paraboloid z = x^2 + y^2, with projection gamma_1 onto z = 0 counterclockwise; compute work along gamma.

Here is the solutions using two different methods: the first method is the Green theorem and the second is the simple path integration method:

My question is why they integrate over [0.2pi] in the path integration method while they integrate within [0. pi] in the green method (I do agree with it)?

Diagram of a paraboloid cup opening upward with a red tilted planar cap cut by the plane z = y, a normal vector N, boundary curve gamma, and the yellow base circle C in the xy-plane with axes x, y, z.

Handwritten-style typeset math solution showing Stokes theorem setup: surface Sigma as graph of g(x,y)=y, normal vector N=(0,-1,1), curl of F equal to (1,-1,1), and integral of rotF dot N over the region.

Typeset worked solution: Stokes or Green step giving the integral of F dP equals the double integral of (1,-1,1)·(0,-1,1) dx dy, which is 2 times the area of C equal to pi/2; remark parametrizing the curve with cartesian equations and gamma(t) = (cos t/2, (1+sin t)/2, (1+sin t)/2) for t in [0,2pi], leading to the integral equal to pi/2.

thank you!

Orodruin · Dec 19, 2021

They don’t seem to be using any parametrization at all when applying Stokes’ theorem.

The value of an integral does not depend on the parametrization. You could have picked a parameter taking values in ##[-200.5,\pi]## and you would get the same result (although a bit more annoying expressions for the differentials).

Amaelle · Dec 19, 2021

for the stock theroem they calculate the surface by integrating from [0 ,pi], this is why i don't understand why they didn't do the same for the path integral

Changing the interval of integration

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