Question

Use Stokes' Theorem to evaluate ∮C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 5xzj + exyk, C is the circle x^2 + y^2 = 9, z = 2.

          Use Stokes' Theorem to evaluate ∮C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 5xzj + exyk, C is the circle x^2 + y^2 = 9, z = 2.
        

Added by Emily H.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Use Stokes' Theorem to evaluate ∮C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 5xzj + exyk, C is the circle x^2 + y^2 = 9, z = 2.
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Transcript

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00:01 Hi, from the question given that f of x comma y comma z is equal to y z z i plus phi x z j plus e to the power of x y k and also given that x squared plus y squared is equal to nine and z z and z z is equal to now the polar coordinates are x is equal to r cos theta, y is equal to r sine theta, and x squared plus y squared is equal to r squared.
01:04 Here value of r is 3.
01:06 The normal will be n is equal to k or n is equal to 0 .0.
01:20 So curl f, that is del cross f vector is equal to modules of i, j, k, do by do x, do by do y, d, d o by do z, with y, f, f, e to the power of x, y, f, e to the power of x.
01:47 Now simplified further we obtain this is equal to i of do by do y of e power xy is x e power x y minus do by do z of phi x z z is 5 x x x minus j of do by do x of e power x x minus j of du by do x of e power x is y e power x minus do by do y z of y z is that is minus y plus k do by do x of five x z is five z do by do y of y z is that is minus z so simplified further we obtain x e to the power of x y minus five x comma y minus y minus y to the power of x y minus y to the power of x y 1 y is also there so y and 5 z minus z is 4 z so it can be written in a inner product form now curl f that is del cross f dot n vector or n at is equal to x e to the power of x y minus 5 x y minus y to the power of x y 4 z dot product with 0 .0 .3 so this can be written as 12 z now integral close to integral over the region c f vector dot d r vector is equal to double integral over the region d 12 z d a so this is equal to integral to integral 0 to 2 pi integral 0 to 3 12 z d r d theta multiplied with r and we know that in the question it is given that the value of z is 2 therefore 12 into 2 is 24 integral 0 to 2 5 d theta r square divided by 2 with limit 0 to 3 so this is equal to 20 4 times 2 pi times 9 divided by 2.
04:49 So canceling the common terms...
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