1. If a vector field  is given by , where M, N and P are functions of x, y and z, show that the field is conservative if , the subscripts indicating partial differentiation with respect to these variables. Using this show that the field  is conservative. Here .

The curl of the given field is . Thus the curl becomes zero when the quantities inside each bracket  vanishes. For the given vector field  Thus  and , we have used  etc.

 

2. Verify Stokes theorem for the vector  field  where the contour is a circle of radius R  in the xy plane centered at the origin. Take the surfaces to be (1) a disk (2) a hemisphere  and (3) a right circular cone radius R and height h.

Let us first calculate the line integral directly. We have . Since  the contour is in the xy plane (z=0), the second and the third integrals vanish. The first integral can be evaluated by polar parameterization  This gives the line integral to be .

       The curl of the given field is seen to be

1. For a disk, if the contour is in the xy plane and in counterclockwise direction, the outward normal is along the + direction. Thus
2. For the hemisphere, the outward normal is in the radially outward direction. The unit normal  is given by , so that the surface integral of the flux is . The integral is readily evaluated in a spherical polar coordinates . The flux integral is thus given by . The first two terms are zero because the azimuthal integral vanishes .The last term gives the flux to be .
3. (this is a hard problem)

The area element on the surface is  where is along the slant of the cone,  which gives . Thus the surface integral of the curl  is

Substitute
Using these the surface integral becomes
The first two terms vanish as the azimuthal integration gives zero, leaving us with the last term, which gives .

3. Evaluate  over a circle of radius 2 whose centre is located at (0,2,2).

The curl of the given field is .  Since the contour is in the yz plane, x=0 and the normal to the disk is in the x direction. . This leaves us with a single surface  integral The line integral is equally easy as  x=0 on the contour leaving us with a single integral  Since the center of the circle is at (2,2) in the yz plane, the polar parameterization is . The line integral becomes

4. Show that the line integral of  along any closed contour is zero.

The curl of the given field being zero, the line integral; on any closed contour which is equal to the surface integral of the curl would vanish.

5. Find the contour integral of  along a triangle whose vertices are at the points (1,0,0), (0,1,0) and (0,0,1).

(this is a hard problem) The line integral is easy to calculate directly. Since the corners of the triangle are at (1,0,0), (0,1,0) and (0,0,1), the equation to the plane is x+y+z=1. Defining the boundary in anticlockwise direction gives the first line in xy plane  second line in yz plane  and the  third line in xz plane  Because of the symmetry of the integrand we need to calculate any of the three line integrals and multiply by 3. Let us take the first line in xy plane with z=0. The integral is  . Note that on this line x=y=1 so that the integral can be written as   Thus the value of the integral over the boundary is -1.

To  get this result by application of Stoke’s theorem, we first calculate  the curl of the field which is  .  The unit normal to the surface is . Thus we have to evaluate . To evaluate this we need to take a projection of the surface into any convenient plane. Taking the projection onto xy plane for which the normal is along  direction, we can write . Thus the surface integral is . Since the equation to the plane is  we can eliminate z and write the integral as . The limits  are as follows : . The integral is