Vector Calculus Questions and Answers

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 \( 2\pi \)

Answer: C

The vector \(\vec{c}\) is orthogonal to \(\vec{a} + \vec{b}\).

A cross product is perpendicular to both of its factors: \(\vec{c}\cdot\vec{a} = 0\) and \(\vec{c}\cdot\vec{b} = 0\). Adding these gives \(\vec{c}\cdot(\vec{a} + \vec{b}) = 0\), so \(\vec{c}\) is orthogonal to the sum, and indeed to every vector in the plane of \(\vec{a}\) and \(\vec{b}\).

\(\vec{c}\) is not zero, because a cross product vanishes only when the vectors are parallel (or one of them is zero), and these are not. The two 'alone' options are wrong because \(\vec{c}\) is orthogonal to both vectors, not to one of them only.

Answer: B

The field is curl-free but not divergence-free.

For \(\vec{F} = x\hat{\imath} + y\hat{\jmath} + z\hat{k}\):

\[ \begin{aligned} \nabla\cdot\vec{F} &= \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} \\ &\quad + \frac{\partial z}{\partial z} \\ &= 1 + 1 + 1 = 3 \end{aligned} \]

so it is not divergence-free; it is the field pointing straight outward from the origin, a uniform source. Each component of the curl, such as \(\partial F_{z}/\partial y - \partial F_{y}/\partial z = 0 - 0\), vanishes because each component depends only on its own coordinate, so \(\nabla\times\vec{F} = 0\). Indeed \(\vec{F}\) is the gradient of \((x^{2} + y^{2} + z^{2})/2\), and every gradient field is curl-free.

That rules out 'divergence-free and curl-free', 'divergence-free but not curl-free' and 'neither'.

Answer: A

The direction is that of the gradient, \(\nabla\phi\).

The rate of change of \(\phi\) in the direction of a unit vector \(\hat{n}\) is the directional derivative \(\nabla\phi \cdot \hat{n} = |\nabla\phi|\cos\alpha\), where \(\alpha\) is the angle between \(\hat{n}\) and the gradient. It is largest when \(\alpha = 0\), that is, along \(\nabla\phi\), and its largest value is \(|\nabla\phi|\).

\(\nabla \times (\phi\vec{r})\) equals \(\nabla\phi \times \vec{r}\), which is perpendicular to the gradient, so along it the rate of change is zero. \(\phi\vec{r}\) points along the position vector, which has nothing to do with how \(\phi\) varies. \((\nabla\phi \cdot d\vec{r})\vec{r}\) is also a multiple of \(\vec{r}\), scaled by the change in \(\phi\), and again points radially rather than along the gradient.