Question: There is exactly one number a such that the vector field F = 4y i+ (ax + 4yz)j + 2y k is conservative. (A)


There is exactly one number a such that the vector field F= 4y i+ (ax + 4yz)j + 2y k is conservative. (A)

There is exactly one number a such that the vector field F = 4y i+ (ax + 4yz)j + 2y k is conservative. (A) Find that number, and (B) substitute it and find a potential for the corresponding vector field. Let C be the intersection of the cylinder x + y = 1 with the plane z = y, with counterclockwise orientation, as viewed from above. (A) Compute the line integral of F along C, expressing your answer as a function of a. (B) Show that the line integral is zero exactly when the vector field F is conservative. Use the same F as in Question 1.

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