Vector Calculus Questions and Answers

Answer: D

With \( \mathbf{F} = \nabla\phi \) the integrand becomes an exact differential, \( \nabla\phi \cdot d\mathbf{r} = d\phi \), so

\[ \int_{A}^{B} \mathbf{F} \cdot d\mathbf{r} \;=\; \phi(B) - \phi(A). \]

It vanishes only when the two endpoints share a potential value, which for a closed path is automatic but for an open one is not

Answer: C

Only the component crossing the surface counts, so each element contributes \( \mathbf{F} \cdot \mathbf{n}\,dS \) and

\[ \Phi \;=\; \iint_{S} \mathbf{F} \cdot \mathbf{n}\,dS. \]

A field lying wholly in the surface carries no flux through it, which the dot product records and an integral of the magnitude would not

Answer: B

On the sphere the field is radial with magnitude \( a \), and the outward normal is radial too, so \( \mathbf{F} \cdot \mathbf{n} = a \) everywhere:

\[ \Phi \;=\; a \times 4\pi a^{2} \;=\; 4\pi a^{3}. \]

The divergence theorem confirms it, since \( \nabla \cdot \mathbf{F} = 3 \) and \( 3 \times \tfrac{4}{3}\pi a^{3} = 4\pi a^{3} \)

Answer: A

The theorem states

\[ \oint_{C}\left(M\,dx + N\,dy\right) \;=\; \iint_{R}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dA. \]

The other pairings describe Gauss's divergence theorem and Stokes's theorem, both of which are three-dimensional generalisations of this plane result

Answer: D

Take \( M = -y \) and \( N = x \), so that \( \partial N/\partial x - \partial M/\partial y = 1 + 1 = 2 \):

\[ \oint_{C}\left(x\,dy - y\,dx\right) \;=\; \iint_{R} 2\,dA \;=\; 2A. \]

Hence the half. This is how a planimeter measures an area from its boundary alone, without ever sampling the interior

Answer: C

The theorem reads

\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \;=\; \iint_{S} (\nabla \times \mathbf{F}) \cdot \mathbf{n}\,dS. \]

That the surface may be any one spanning the curve follows because the curl is solenoidal, so its flux depends only on the boundary. It also explains at once why an irrotational field has zero circulation

Answer: B

The theorem reads

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

It is the statement that what leaves a region is what the sources inside it produce, and it is what turns the integral form of a conservation law into its differential form. A closed surface has no boundary curve, so no line integral arises

Answer: A

The divergence is constant:

\[ \nabla \cdot \mathbf{F} \;=\; 1 + 1 + 1 \;=\; 3, \]

so the divergence theorem gives \( \iiint 3\,dV = 3V \) whatever the shape of the surface. For a sphere of radius \( a \) this is \( 3 \times \tfrac{4}{3}\pi a^{3} = 4\pi a^{3} \), matching the direct calculation

Answer: D

By Stokes's theorem the circulation equals the flux of the curl through a spanning surface, and an irrotational field has zero curl:

\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \;=\; \iint_{S} \mathbf{0} \cdot \mathbf{n}\,dS \;=\; 0. \]

Simple connectedness matters: on a region with a hole the spanning surface may not exist, and the circulation can then be non-zero

Answer: C

Green's theorem is the plane case, trading a boundary line integral for an integral over the enclosed area. Stokes lifts it into space, the area integral becoming a flux of the curl through a spanning surface. Gauss relates a closed surface to the volume it encloses, and has no boundary curve at all