 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/11b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

B_{1} = \nabla (e^{x^{2}} cos (yz)),

so its integral around

C

is zero and it is conservative with potential

e^{x^{2}} cos (yz)

Along

C

d r = (- sin s, cos s, 0) d s

and

B_{2} = (- sin s, cos s, 0),

\oint_{C} B_{2} \cdot d r = \int_{0}^{2 π} 1 d s = 2 π .

Thus

B_{2}

is not conservative and cannot be a gradient on the stated punctured region.

Finally,

B_{3} = \nabla lo g (x^{2} + y^{2} + z^{2}) .

Its integral around

C

is zero, and it is conservative with the displayed potential. Therefore the three integrals are respectively

0, 2 π, 0,

and exactly

B_{1}, B_{3}

are conservative gradient fields.

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