Find the curl of f 3yz 3x xyz and g cosh y sinh z cosh x a...

Find the curl of F = [3y+z, −3x, xyz] and G = [cosh y, sinh z, cosh x]. (a) Use Stokes’ Theorem to find the surface integral ∬S1 [xz, 1−yz, −6] ●ndA over the hemisphere S1 = {[x, y, z]∶ x 2 + y 2 + z 2 = 25, z <= 0} where n points outside/downwards. (b) Use Stokes’ Theorem to find ∮C H(\gamma ) ● d\gamma of H = [4z+e x , 4x+e y , 4y+e z ] over the surface boundary C of S2 repr. by r(u, v) = [u 2 , 2u+2v, v2 ] def. on R = {[u, v]∶ 0<=u<=2, 0<=v<=2−u}. (c) Use Stokes’ Theorem to find the surface integral ∬S3 [cosh z, sinh x, sinh y] ●ndA over the ellipsoid S3 = {[x, y, z]∶ x 2+3y 2+2z 2 = 4} where n points outside. You can use Matlab to calculate partial derivatives, cross products and curls; show where you use it.

Transcript

00:01 In part a, we're given a vector field f and the curve c, and we're asked the stasis theorem to evaluate the line integral over c of f.

00:13 F is the vector field x squared y i plus one third x -g to j plus x y k, and c is the curved intersection of the hyperbolic paraboloid z equals y squared minus x squared and the cylinder x squared plus y squared equals y squared equals one oriented counterclockwise as viewed from above.

00:44 So we have first of all that s is the part of the surface, which is our hyperbolic paraboloid, c equals y squared minus x squared, that lies above the unit disk, which we'll call d.

01:15 Now the curl of the vector field f that we're given this is x i minus y j plus x squared minus x squared k which reduces to x i minus y j and using previous equation function g of x and y so our surface is a function of y and x so our surface is a function of y and x and x and and we'll take function p to be the first component of the curl of f, which is x, q will be the second component of the curl of f, which is negative y, and r will be 0, the third component.

02:56 Therefore, we have that the line integral over c of f, dystokesis theorem is equal to the surface integral over s of the curl of f, and because of surface is a function of x and y, this is the double integral over the domain of integration d, sorry unit disk, of the opposite of p, so negative x times the partial derivative of g with respect to x, which is negative 2x, minus q, which is negative y, times the partial derivative of g with respect to x, which is negative 2x, minus q, which is negative y, times the partial derivative of g with respect to y, which is 2y, plus r, which is 0, da.

04:03 And simplifying, this is 2 times the double integral over d of x squared plus y squared, da.

04:19 And converting to polar coordinates, this is two times the integral from 0 to pi, integral from 0 to 1, the radius of the disk of r squared, times r is r cubed, d r d theta.

04:42 And taking antiderivatives, you get 2 times 2 pi times 1 fourth, and this simplifies to pi.

05:05 Then in part b, we're asked to graph first the hyperbolic paraboloid and the cylinder, so that we can see the curve c and the surface s using part a.

05:15 To do this, i'm going to use desmosis parametric surface graph...

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