 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/13/f/solution

For

a = 0

, differentiate the proposed scalar potential directly:

\nabla (- \frac{1}{2} e^{- (x^{2} + y^{2} + z^{2})}) = e^{- r^{2}} (x, y, z) .

Thus

Φ (r) = - \frac{1}{2} e^{- r^{2}} + C when a = 0.

For

a \neq = 0

, the curl is the nonzero vector

(0, 0, 2 a)

. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of

R^{3}

, so no excluded singular set complicates this conclusion.

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