Past exam of the mathematics course of the University of Cambridge 2012 ia Paper 3 11C b Solution Created 2026-09-24 Updated 2026-10-07
Assume the continuously differentiable irrotational vector field is defined on all of , as in this vector calculus setting. A constructive vector-field potential with the requested negative-gradient convention isSince zero curl means , differentiation under the integral givesThus . This is the Poincare lemma in its straight-line form, and works on any star-shaped set about the integration base point. Zero curl on a domain with holes would not by itself guarantee a global vector-field potential; the full-space hypothesis matters.
For the given field, the second component requires . Integration in gives . Its derivative is , and its derivative is . Comparison with the other two components forces . HenceIts negative gradient is exactly the given field. Because this vector-field potential is smooth everywhere, commutation of mixed partial derivatives gives , verifying irrotationality. All such vector-field potentials differ by a constant on the connected domain.
Past exam of the mathematics course of the University of Cambridge 2017 ia Paper 3 9B c Solution Created 2026-09-24 Updated 2026-10-05
Use the usual setting of a continuously differentiable solenoidal vector field defined on all radial segments from the origin, including the origin. More generally an open star-shaped set containing zero suffices. Differentiation under the integral sign givesApply divergence and curl of a cross product with the second field , using and :The scaling identity in part (a) now converts the last expression to a one-variable derivative:Continuity at the origin ensures the lower endpoint is zero. Hence the radial vector potential of a solenoidal vector field isThe order of the cross product matters: would produce . This construction is not valid on every punctured or multiply connected domain. For example is solenoidal away from zero, but its radial integral diverges at zero and it has nonzero flux through a surrounding sphere. The stated formula therefore implicitly needs the regular radial-domain hypothesis, rather than just local zero divergence.
Past exam of the mathematics course of the University of Cambridge 2017 ib Paper 4 4F Solution Created 2026-09-24 Updated 2026-10-05
Choose a star centre of the open star-shaped set and define the candidate antiderivative by a straight-segment contour integral:For each fixed and all sufficiently small , the filled triangle with vertices lies in . Indeed, the compact segment has a neighbourhood of some positive radius contained in the open set , and every point of this triangle lies within of that segment. The zero triangle integral consequently givesBy continuity, the last integral tends to as . HenceIn particular, this proves the existence of the antiderivative without incorrectly assuming the star-shaped set is convex.
On a general domain, the conclusion can fail. Take and . Every filled triangle contained in has zero boundary integral by Cauchy integral theorem, but the unit circle has integral . An antiderivative would make every closed contour integral zero. This is the period obstruction to a holomorphic antiderivative: the local conclusion of Morera's theorem does not remove the global obstruction.
Past exam of the mathematics course of the University of Cambridge 2022 ib Paper 1 12G a Solution Created 2026-09-24 Updated 2026-09-25
The hypothesis makes star-shaped with centre . For small with the triangle having vertices contained in , the vanishing triangular integral givesParametrizing the final segment,by continuity. Thus is holomorphic andThis is the triangle-integral construction behind Morera's theorem.
Let be a continuously differentiable solenoidal vector field on an open star-shaped set containing the origin. Thenis a vector potential for . For , differentiation gives and . Boundedness near the origin removes the lower endpoint. A field defined only on a punctured domain need not satisfy these hypotheses.