University:
University of California, Berkeley
Course:
MATH 56 Linear Algebra
Academic year:

2020
Views:

55

Pages:

1
Author:

SpaceCadet101

Stokes’ Theorem Stokes’ Theorem: Let S be an oriented piecewise-smooth surface that is bounded by a simple, → − closed, piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in R3 that contains S. Then, R → RR − → − → − − F · d→ r = curl F · d S C S → − → − → − → − → − − Exercise 1. Evaluate the integral S (∇ X F ) · → n dS, where F = −xz i + yz j + xyez k → − 2 2 and S is the cap of the paraboloid z = 5 − x − y above the plane z = 3. Assume n points in the upward direction on S. (Briggs Sec 14.7 Ex 3) RR Exercise 2. Let the surface S be the elliptical paraboloid z = x2 + 4y 2 lying beneath the plane − z = 1. We define the orientation of S by taking the inner normal vector → n to the surface, which → − → − is the normal vector having a positive k -component. Find the flux of ∇ X F across S in the → − → − → − → − − direction → n for the vector field F = y i − xz j + xz 2 k . (Hass Sec 16.7 Ex 10) Class Exercise 1. Use Stokes’ Theorem to evaluate RR → − → − curl F · d S . (#2,4,6) S → − → − → − → − (a) F (x, y, z) = 2y cos z i + ex sin z j + xey k , S is the hemisphere x2 + y 2 + z 2 = 9, z ≥ 0, oriented upward p → − → − → − → − (b) F (x, y, z) = tan−1 (x2 yz 2 ) i + x2 y j + x2 z 2 k , S is the cone x = y 2 + z 2 , 0 ≤ x ≤ 2, oriented in the direction of the positive x-axis → − → − → − → − (c) F (x, y, z) = exy i + exz j + x2 z k , S is the half of the ellipsoid 4x2 + y 2 + 4z 2 = 4 that lies to the right of the xz-plane, oriented in the direction of the positive y-axis H → → − − → − → − → − − Exercise 3. Evaluate the line integral F · d→ r , where F = z i − z j + (x2 − y 2 ) k and C consists of the three line segments that bound the plane z = 8 − 4x − 2y in the first octant. (Briggs Sec 14.7 Ex 2) R → → − − → − → − → − − Exercise 4. Evaluate C F · d→ r , where F (x, y, z) = −y 2 i + x j + z 2 k and C is the curve of intersection of the plane y + z = 2 and the cylinder x2 + y 2 = 1. (Stew Sec 16.8 Ex 1) R → − − Class Exercise 2. Use Stokes’ Theorem to evaluate C F · d→ r . In each case C is oriented counterclockwise as viewed from above. (#8,10) √ → − → − → − → − (a) F (x, y, z) = i + (x + yz) j + (xy − z) k , C is the boundary of the part of the plane 3x + 2y + z = 1 in the first octant → − → − → − → − (b) F (x, y, z) = xy i + 2z j + 3y k , C is the curve of intersection of the plane x + z = 5 and the cylinder x2 + y 2 = 9 Exercise 5. Let S be the part of the paraboloid z = 9 − x2 − y 2 with z ≥ 0, and let C be the → − → − → − → − trace of S on the xy-plane. Verify Stokes’ Theorem for the vector field F = 3z i + 4x j + 2y k . (Swok Sec 18.7 Ex 1) → − Class Exercise 3. Verify that Stokes’ Theorem is true for the given vector field F and surface → − → − → − → − S. F (x, y, z) = −2yz i + y j + 3x k , S is the part of the paraboloid z = 5 − x2 − y 2 that lies above the plane z = 1, oriented upward. (#14) Class Exercise 4. Let C be a simple closed smooth curve that lies in the plane x + y + z = 1. Show that the line integral R z dx − 2x dy + 3y dz C depends only on the area of the region enclosed by C and not on the shape of C or its location in the plane. (#16) Class Exercise 5. Evaluate R (y + sin x ) dx + (z 2 + cos y) dy + x3 dz C − where C is the curve → r (t) = < sin t, cos t, sin 2t >, 0 ≤ t ≤ 2π. (#18)