




1. Let F = (3 + 3x^{2}y^{3})i + (3x^{3}y^{2
} 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 (3x^{3}y^{2 } 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 G_{k} 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.