Vector Calculus Questions and Answers

Answer: B

A divergence-free field is the curl of a vector potential, since every curl is solenoidal. A curl-free field is the gradient of a scalar, since every gradient is irrotational. A field with both properties has \( \nabla \cdot \nabla\phi = 0 \), so its scalar potential is harmonic

Answer: A

The components are identified first, then the cross-derivative test decides whether a potential exists at all - there is no point integrating if it fails. One component is then integrated, leaving a function of the other variable, which the second component fixes. Integrating before testing is what wastes the effort

Answer: D

The first two are standard. The third confuses the two conditions: the radial field \( x\mathbf{i} + y\mathbf{j} + z\mathbf{k} \) is irrotational yet has divergence three. The two properties are independent, and a field satisfying both is the special harmonic case

Answer: C

All three hold. The freedom of surface follows from the curl being solenoidal. Closedness is essential to the divergence theorem, since an open surface bounds no volume. And Green's theorem is Stokes's theorem applied to a plane field, the curl then having only a component perpendicular to the plane

Answer: B

The divergence contracts the operator with the field and returns a scalar rate of outflow per unit volume. The curl and the gradient both return vectors, and the curl of a gradient is a vector too, albeit the zero vector. Keeping track of which operations change the rank is what stops ill-formed identities being written

Answer: A

Conservativeness constrains the curl, not the divergence: the field \( x\mathbf{i} + y\mathbf{j} \) has potential \( (x^{2}+y^{2})/2 \) and divergence two. The other three are equivalent restatements of conservativeness on a simply connected region

Answer: D

Solenoidal means \( \nabla \cdot \mathbf{F} = 0 \), so the divergence theorem gives

\[ \oiint_{S} \mathbf{F} \cdot \mathbf{n}\,dS \;=\; \iiint_{V} 0\,dV \;=\; 0 \]

for every closed surface in the region, whatever its shape or position. This is what \"source-free\" means in integral terms

Answer: C

On a simply connected region a vanishing curl does give a potential, but an annulus has a hole, so a loop encircling it bounds no surface lying wholly inside. The field \( (-y, x)/(x^{2}+y^{2}) \) is the standard case: its curl is zero yet its circulation round the hole is \( 2\pi \)