Compute the line integral of v r cos2 r r cos sin 3r

Compute the line integral of v = (r cos2 ?) r – (r cos ? sin ?) ? + 3r ? around the path shown in Fig. 1.50 (the points are labeled by their Cartesian coordinates). Do it either in cylindrical or in spherical coordinates. Check your answer, using Stokes\' theorem.

Solution

Solution. Here’s a picture of the surface S. x y z To use Stokes’ Theorem, we need to first find the boundary C of S and figure out how it should be oriented. The boundary is where x 2 + y 2 + z 2 = 25 and z = 4. Substituting z = 4 into the first equation, we can also describe the boundary as where x 2 + y 2 = 9 and z = 4. To figure out how C should be oriented, we first need to understand the orientation of S. We are told that S is oriented so that the unit normal vector at (0, 0, 5) (which is the lowest point of the sphere) is h0, 0, 1i (which points down). This tells us that the blue side must be the “positive” side. We want to orient the boundary so that, if a penguin walks near the boundary of S on the “positive” side (which we’ve already decided is the blue side), he keeps the surface on his left. If we imagine looking down on the surface from a really high point like (0, 0, 100), then the penguin should walk clockwise (from our vantage point). So, using Stokes’ Theorem, we have changed the original problem into a new one: Evaluate the line integral Z C F~ · d~r, where C is the curve described by x 2 + y 2 = 9 and z = 4, oriented clockwise when viewed from above. Now, we just need to evaluate the line integral, using the definition of the line integral. (This is like #4(a) on the worksheet “Vector Fields and Line Integrals”.) We start by parameterizing C. One possible parameterization is ~r(t) = h3 cost, 3 sin t, 4i, 0 t < 2. (1) If we look at this from above, it is oriented counterclockwise, which is the wrong orientation. Therefore, Z C F~ · d~r = Z 2 0 F~ (~r(t)) · r~0(t) dt = Z 2 0 h3 sin t, 3 cost, 36 costsin ti · h3 sin t, 3 cost, 0i dt = Z 2 0 9 dt = 18