Dear all,
Questions IA,B, were testing knowledge of the big theorems useful for computing integrals and recognizing gradients.
IA,a: f(q) -f(p).
b. the path integral of F dot dr = the surface integral of (del cross F) dot n dsigma, page 906.
c. the surface integral of F dot n dsigma over S, equals the volume integral of del dot F dV, over R.
IB a) True, the curl of a gradient is always zero, by the equality of mixed partials.
i.e. the entries are the differences of second partials of f taken in the opposite order.
i.e. curl f = (fyz - fzy, fzx - fxz, fxy - fyx) = (0,0,0).
b) False, curl(F) = 0 only guarantees that F is locally a gradient, as we saw an example "dtheta", of a field wioth zero curl, but only a gradient in regions that do not wind around the origin.
c) True, here the region U is simply connected so curl(G) = 0 does guarantee that G is a gradient in U, so all closed curve path integrals are zero.
d) True, stokes theorem equates the flux integral of curl(F) over a surface, with the path integral of F itself over the bloundary curve.
so if two surfaces have the same boundary ciurve, then stokes equates both flux integrals to the same path integral.
we had explicitly answered this question, a homework problem from the book. page 913, problem 11.
e) this is true, by the divergence theorem, since every sphere is the boundary surface of a ball, and the divergence theorem
says to get the surface integral, we can just integrate the divergence, which is zero, over the ball.IIa) This is a simple path integral we did several times, for the area of the region inside the path, namely an ellipse of semi - axes a,b, the area is pi ab,
which here is 6pi.
IIb) here is one way to see it gives area, since by greens theorem, it equals the double integral of dxdy over the interior of the ellipse,
i.e. area., see problem 21, page 885.
IIIa) del cross F here i.e. curl(F), is just ( y, x, 1).
By the true statement IB d), we can replace the hemisphere H by any other simpler surface with the same oriented boundary,
such as the disc of radius 2, in the x,y plane.
then the normal vector to the disc is just (0,0,1), so in the flux integral, the dot product of curl(F) with n is just 1,
and the surface flux integral becomes just dxdy over the disc,
i.e. the area of the disc, or 4pi.
the path integral is not too hard either, and during the test i even did the surface flux integral over the hemisphere,
using spherical coords, and it was not too bad either. it finally came out as the integral from phi = 0 to phi = pi/2,
of 8pi sin(phi) cos(phi) which is again 4pi.
IV. div(F) here is just z.
using the divergence theorem, we are integrating z over the tetrahedron, so at each height z, if we integrate in the order z,x,y, we are
integrating z times the area of the triangular slice at height z, and that area is (1/2)(1-z)^2.
so we are integrating (1/2)z(1-z)^2 from z=0 to z=1, and get 1/24.
i also parametrized the faces of the tetrahedron and did the masochist's computation of the flux integral, and finally got the same thing.
there are three pieces to the surface integrand as usual, one each for dydz, dzdx, dxdy, and 4 faces for the tetrahedron, so potentially 12 parametrized area integrals to do, but 10 lf them are equal to zero,
because dzdx for instance is always zero in the x= 0 plane and z=0 plane, and ydzdx will be zero also in the y=0 plane.
and one of the two non zero integrals cancels part of the other one, for reasons of opposite orientation,
so we are left finally with an integral over the triangular base that also comes out 1/24.
Archimedes knew the value of this integral by the way because he knew the center of gravity of a tetrahedron is 1/4 of the way up from the base,
so at height 1/4, but the height of the center of gravity is the average z coordinate, which equals the integral ,of the z coordinates divided by the volume of the tetrahedron, as we know, (pages 817-818),
so the integral of z is the producto f the height of the centyer of gravity by the volume of the tetrahedron, i.e. (1/4) times 1/6 = 1/24.
recall the volume of a pyramid is 1/3 the product of the height by the area of the base.
actually archimedes computed centers of mass first and then deduced formulas for volume.best regards,
roy