Vector Calculus Questions and Answers

Answer: D

The work done is −15.

The resultant force and the displacement from A to B are

\[ \begin{aligned} \mathbf{F} &= (2 - 1)\mathbf{i} + (-5 + 2)\mathbf{j} + (6 - 1)\mathbf{k} \\ &= \mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \\ \overrightarrow{AB} &= 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} \end{aligned} \]

Work is their dot product:

\[ W = (1)(2) + (-3)(4) + (5)(-1) = 2 - 12 - 5 = -15 \]

The negative sign means the force opposes the displacement on balance. 2 and −5 are single terms of the sum, and 12 drops the signs.

Answer: D

The moment is \(-112\hat{i} - 16\hat{j} + 50\hat{k}\).

The moment of a force about a point A is \(\vec{r} \times \vec{F}\), where \(\vec{r}\) runs from A to any point on the force's line of action. The force passes through the origin, so \(\vec{r} = O - A = (-3, -4, -8)\).

\[ \begin{aligned} M_{x} &= (-4)(8) - (-8)(-10) = -112 \\ M_{y} &= (-8)(5) - (-3)(8) = -16 \\ M_{z} &= (-3)(-10) - (-4)(5) = 50 \end{aligned} \]

Using \(\vec{r}\) from the origin to A, (3, 4, 8), reverses every sign and gives \(112\hat{i} + 16\hat{j} - 50\hat{k}\), the moment about the origin of a force through A. The other two options rearrange the same three numbers into the wrong components.

Answer: A

The divergence of a curl is 0.

For a twice continuously differentiable vector field \(\mathbf{F} = (F_{1}, F_{2}, F_{3})\),

\[ \begin{aligned} \nabla \cdot (\nabla \times \mathbf{F}) &= \partial_{x}(\partial_{y}F_{3} - \partial_{z}F_{2}) \\ &\quad + \partial_{y}(\partial_{z}F_{1} - \partial_{x}F_{3}) \\ &\quad + \partial_{z}(\partial_{x}F_{2} - \partial_{y}F_{1}) \end{aligned} \]

The mixed partial derivatives cancel in pairs, since \(\partial_{x}\partial_{y}F_{3} = \partial_{y}\partial_{x}F_{3}\) and so on (equality of mixed partials, which is why the field must be twice continuously differentiable). The result is identically zero, so 1, 2 and infinity are all wrong. A curl field is always solenoidal.

Answer: D

The outward flux is \(4\pi abc\).

By the divergence theorem, the flux through the closed ellipsoid equals the integral of \(\nabla\cdot\vec{u}\) over the solid it bounds. Here

\[ \begin{aligned} \nabla\cdot\vec{u} &= 1 + 1 + 1 = 3 \\ \text{flux} &= 3\cdot\tfrac{4}{3}\pi abc = 4\pi abc \end{aligned} \]

using the volume \(\frac{4}{3}\pi abc\) of the ellipsoid. The condition \(a \gt b \gt c\) plays no part. \(\pi abc\), \(2\pi abc\) and \(3\pi abc\) come from using the wrong volume or dropping the factor 3 from the divergence; the volume alone would be \(\frac{4}{3}\pi abc\), which is not offered.

Answer: B

The work done is \(-\frac{\pi}{2}\).

Parametrise the upper semicircle by x = cos t, y = sin t, t from 0 to \(\pi\). On it \(x^{2} + y^{2} = 1\), so

\[ \begin{aligned} F \cdot dr &= (x+y)\,dx - (x^{2}+y^{2})\,dy \\ &= -(\cos t + \sin t)\sin t\,dt - \cos t\,dt \end{aligned} \]

\[ \begin{aligned} W &= \int_{0}^{\pi} (-\sin t\cos t - \sin^{2} t - \cos t)\,dt \\ &= 0 - \frac{\pi}{2} - 0 = -\frac{\pi}{2} \end{aligned} \]

The terms \(\sin t\cos t\) and \(\cos t\) integrate to 0 over [0, \(\pi\)], and \(\sin^{2} t\) integrates to \(\pi/2\). \(\pi/2\) has the sign reversed (the path traversed the other way), and \(\pm\pi\) would come from integrating \(\sin^{2} t\) as if it were 1.

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.