Vector Calculus Questions and Answers

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.

Answer: A

The line integral equals \(\frac{8}{3}\).

By Green's theorem with \(M = 2y^{2}+2xy+4y\) and \(N = x^{2}+4xy+8x\):

\[ \begin{aligned} \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} &= (2x+4y+8)-(4y+2x+4) \\ &= 4 \end{aligned} \]

So the integral is 4 times the enclosed area. The curves \(y = 2x^{2}\) and \(y = 2\sqrt{x}\) meet at x = 0 and x = 1, with the square-root curve on top:

\[ \begin{aligned} A &= \int_{0}^{1}(2\sqrt{x}-2x^{2})\,dx = \frac{4}{3}-\frac{2}{3} = \frac{2}{3} \\ \oint &= 4 \times \frac{2}{3} = \frac{8}{3} \end{aligned} \]

2/3 is the area alone, without the factor 4. 4/3 and 1/3 come from using only one of the curves or halving the integrand.

Answer: D

The line integral equals \(25\pi\).

By Green's theorem, with \(M = x^{5}-3y\) and \(N = -2x+e^{y^{2}}\):

\[ \begin{aligned} \oint_{C}(M\,dx+N\,dy) &= \iint_{D}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA \\ &= \iint_{D}(-2+3)\,dA = \text{area of } D \end{aligned} \]

The circle has radius 5 (its centre at (1, 0) does not matter), so the area is \(\pi \times 5^{2} = 25\pi\).

The awkward terms \(x^{5}\) and \(e^{y^{2}}\) drop out because each depends only on the variable it is integrated against. Taking the integrand as -2 - 3 or 3 alone gives multiples of \(25\pi\) not offered; 10, 15 and 20 times \(\pi\) do not match the area.

Answer: D

The divergence is \(2z \cos z^{2}\).

Take the partial derivative of each component with respect to its own variable and add them.

\[ \begin{aligned} \frac{\partial V_{x}}{\partial x} &= -\cos xy + xy\sin xy \\ \frac{\partial V_{y}}{\partial y} &= \cos xy - xy\sin xy \\ \frac{\partial V_{z}}{\partial z} &= 2z\cos z^{2} \end{aligned} \]

The first two cancel exactly, leaving \(\nabla \cdot \mathbf{V} = 2z \cos z^{2}\). The \(x^{2} + y^{2}\) inside the k-component does not depend on \(z\), so it drops out.

The options with \(\sin xy\) or \(\cos z\) keep terms that cancel or differentiate wrongly, and \(2x \cos x^{2}\) differentiates the wrong component.