5.1
Vector Fields
299
Example 5 Show that the divergence of the inverse-square field
F(x, y, z) =
(.xi + yj + zk)
(r2 + y2 + 2)/2
is zero.
Answers
Answer:
please mark as brainlist please
Step-by-step explanation:
Let F be an inverse square field, that is F(r) = cr/|r|3, for some constant c, where r = x,y,z. Show that the flux of F across a sphere centered at the origin is independent of the radius of the sphere. Solution: Let S be a sphere of radius R centered at the origin. We will compute the flux and show that it doesn’t depend on R. The sphere is parametrized by r(φ,θ) = Rsinφcosθ,Rsinφsinθ,Rcosφ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π and the outward normal is rφ×rθ = (Rsinφ)r, as we computed in class on more than one occasion. Therefore F· ndS = cr |r|3 · (Rsinφ)r = cRsinφ|r|2 |r|3 So the flux is 2π F· ndS = π = cRsinφ |r| = cRsinφ R =csinφ csinφdφdθ = 4πc, 0 S which doesn’t depend on R. 0 2. Stewart 16.8.10 [4 pts] Evaluate C F· dr, where F = xy,2z,3yand C is the curve of intersection of the plane x+z =5 and the cylinder x2 +y2 = 9. Solution: The curve C is the boundary of an elliptical region across the middle of the cylinder. Call this region S. To match the counterclockwise orientation of C, we give S the upwards orientation. By Stokes’ Theorem, the line integral we are asked to compute is equal to the surface integral S curlF · ndS. The parametrization of S is r(r, t) = rcost,rsint,5 − rcost, 0 ≤ r ≤ 3, 0 ≤ t ≤ 2π, because the x-and y-coordinates of points on this ellipse must lie inside the disk x2 +y2 ≤ 1 in the xy-plane, as the ellipse is inside the cylinder x2 + y2 = 9. The z-component of our parametrization was found by using the equation of the plane, z = 5−x (the region S lies on that plane). Using the right hand rule, we find that the desired normal vector will be rr×rt, which we calculate to be r,0,r. Finally, we compute the curl of F: curlF = 1,0,x. Thus our integral is F· ndS = S = = =9π 1, 0,r cost· r,0,rdtdr 2π 0 0 2π 3 0 ( 9 (r +r2cost)drdt 2 +9cost)dt 1