Home | 18.022 | Chapter 11 | Section 11.5

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1. Let F = (3 + 3x2y3)i + (3x3y2 - 5)j
Evaluate the integral Fdl from left to right over the portion of the ellipse (x/2)2 + (y/5)2 = 1 in the first quadrant.

This integral goes from the point (0,5) to (2,0). If you integrate along the coordinate axes, the integrand is constant and easy. Going down 5 on the y axis integrating dl F you are integrating (3x3y2 - 5)j(-j) and the answer is 25; going up 2 on the x axis you get the integral of 3 or 6.
The curl of F vanishes, so that the integral is path independent.
The given integral along the original path is therefore 31.

2. Let G = F + yi
Evaluate the integral Gdl over the same path as above.

The integrals of F and G are identical on the axes, but now G = -k.
By Stokes' Theorem, the path integral of G down from (0, 5) to (2, 0) by the axes and back by the elliptical path is therefore the surface integral of Gk that is, of -1, over the enclosed area.
Changing the x scale downward by a factor of 2 and and the y scale by 5 makes the area into a quarter of a unit circle. The area is therefore 10/ 4 and the surface integral here is -5/2.
If we call the integral we want I, the we get that 31 - I = -5/ 2, and we conclude: I = 5/ 2 + 31.