Vector Calculus Questions and Answers

Answer: C

The integral equals 3V.

By the divergence (Gauss) theorem, the flux of a vector field through a closed surface equals the volume integral of its divergence:

\[ \begin{aligned} \oint\!\!\oint_{S} \vec{r} \cdot \vec{n}\, dS &= \iiint_{V} \nabla \cdot \vec{r}\, dV \\ \nabla \cdot \vec{r} &= 1 + 1 + 1 = 3 \\ \text{flux} &= 3V \end{aligned} \]

Check with a sphere of radius a: \(\vec{r} \cdot \vec{n} = a\) everywhere on the surface, so the flux is \(a \times 4\pi a^{2} = 3 \times \frac{4}{3}\pi a^{3}\). V and 2V count only one or two of the three coordinate directions, and 4V has no basis.

Answer: A

The surface integral equals \(4\pi\).

The gradient of \(\varphi = \frac{1}{2}(x^{2} + y^{2} + z^{2})\) is the position vector \(\nabla\varphi = (x, y, z) = \vec{r}\). On the unit sphere the outward normal is \(\hat{\mathbf{n}} = \vec{r}\) itself, so \(\hat{\mathbf{n}} \cdot \nabla\varphi = |\vec{r}|^{2} = 1\) everywhere, and the integral is just the area of the sphere, \(4\pi\).

The divergence theorem confirms it: \(\nabla^{2}\varphi = 3\), and \(3 \times \frac{4}{3}\pi = 4\pi\).

\(4\pi/3\) is the volume of the sphere without the factor 3, \(3\pi\) has no basis, and 0 would require \(\varphi\) to be harmonic.

Answer: C

The divergence is \(2e^{x}\cos y\).

The field has components \(u_{x} = e^{x}\cos y\) and \(u_{y} = e^{x}\sin y\). The divergence is the sum of each component differentiated along its own direction:

\[ \begin{aligned} \nabla\cdot\vec{u} &= \frac{\partial}{\partial x}(e^{x}\cos y) + \frac{\partial}{\partial y}(e^{x}\sin y) \\ &= e^{x}\cos y + e^{x}\cos y \\ &= 2e^{x}\cos y \end{aligned} \]

\(e^{x}\cos y + e^{x}\sin y\) forgets to differentiate the sine with respect to y, and \(2e^{x}\sin y\) mixes up which component is differentiated. The divergence is not zero, since both terms are positive where cos y is.

Answer: A

The surface integral is 14.

By the divergence (Gauss) theorem, the outward flux through a closed surface equals the volume integral of the divergence inside it:

\[ \begin{aligned} \nabla \cdot \vec{F} &= \frac{\partial (3x)}{\partial x} + \frac{\partial (5y)}{\partial y} + \frac{\partial (6z)}{\partial z} \\ &= 3 + 5 + 6 = 14 \end{aligned} \]

The divergence is a constant, so the flux is 14 times the volume of the cube, which is 1. The position of the centroid at (1, 2, 3) does not matter, because the divergence is the same everywhere.

27, 28 and 31 cannot arise: with a constant divergence of 14 and a volume of 1, the flux is 14 wherever the cube sits; 28, for instance, is double the true flux.

Answer: A

The unit outward normal is \(\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}\).

Take \(\phi = x^{2}+y^{2}+z^{2}\). Its gradient is normal to the surface:

\[ \begin{aligned} \nabla\phi &= 2x\hat{i}+2y\hat{j}+2z\hat{k} \\ &= \sqrt{2}\hat{i}+\sqrt{2}\hat{j} \text{ at the point} \\ |\nabla\phi| &= 2 \end{aligned} \]

Dividing by 2 gives the unit normal. On a sphere centred at the origin, the outward normal is simply the position vector of the point, which is the same answer.

The minus sign in the second option makes the vector tangent to the sphere, \(\hat{k}\) is the normal at the north pole, and the last option has a z-component although z = 0 at this point.

Answer: C

The surface integral equals \(\dfrac{4\pi}{3}R^{3}\).

The integral \(\oint z\,dx\,dy\) over a closed surface is the flux of the vector field \(\mathbf{F} = (0, 0, z)\). By the divergence theorem it equals the volume integral of \(\nabla \cdot \mathbf{F} = \partial z/\partial z = 1\):

\[ \begin{aligned} \oint_{S} z\,dx\,dy &= \iiint_{V} 1\,dV \\ &= \frac{4}{3}\pi R^{3} \end{aligned} \]

which is the volume of the sphere. Zero would be the answer only if the divergence vanished; \(4\pi R^{3}\) is three times the volume (the flux of the full position vector \(x\,\hat{i} + y\,\hat{j} + z\,\hat{k}\)), and \(\pi R^{3}\) matches no step.

Answer: D

The divergence of the curl of any vector field is zero.

Writing it out, \(\nabla\cdot(\nabla\times\vec{F})\) is a sum of mixed second partial derivatives that cancel in pairs, such as \(\partial^{2}F_{z}/\partial x\partial y - \partial^{2}F_{z}/\partial y\partial x\), for any field with continuous second derivatives.

It is a scalar identity, not related to the magnitude or direction of the field or of its curl, so the other three options are wrong.

Answer: A

The gradient is always perpendicular to the surface of constant \(\varphi\).

Along any curve lying in a level surface \(\varphi = c\), the rate of change \(\nabla\varphi\cdot\vec{t}\) is zero, where \(\vec{t}\) is the tangent. So \(\nabla\varphi\) is orthogonal to every tangent direction of the surface, that is, normal to it.

It cannot be parallel to the surface for the same reason. Its magnitude is the maximum rate of change of \(\varphi\), not the minimum, and it vanishes only at stationary points, not always.

Answer: C

The direction is along \(3\hat{i} - 10\hat{j}\).

A function increases fastest along its gradient:

\[ \begin{aligned} \nabla f &= (4x^{3} - 5y^{2})\hat{i} - 10xy\,\hat{j} \\ \nabla f(2,2) &= (32 - 20)\hat{i} - 40\hat{j} \\ &= 12\hat{i} - 40\hat{j} = 4(3\hat{i} - 10\hat{j}) \end{aligned} \]

A direction is fixed only up to a positive multiple, so \(3\hat{i} - 10\hat{j}\) is the same direction. \(-12\hat{i} + 40\hat{j}\) points the opposite way, the direction of fastest decrease, and the other two options have a wrong sign on one component.

Answer: B

The surface integral equals 4π.

By the divergence theorem the flux through the closed sphere equals the volume integral of the divergence:

\[ \begin{aligned} \nabla \cdot \vec{F} &= \frac{\partial e^{x}}{\partial x} + \frac{\partial (3y)}{\partial y} + \frac{\partial (-ze^{x})}{\partial z} \\ &= e^{x} + 3 - e^{x} = 3 \\ \iint_{S} \vec{F} \cdot \hat{n}\,dA &= 3 \times \tfrac{4}{3}\pi(1)^{3} = 4\pi \end{aligned} \]

4π/3 is the volume of the unit sphere without the factor 3, and 12π multiplies by 3 twice or uses the surface area 4π. The answer is not 0, because the divergence is a non-zero constant.