Vector Calculus MCQs in Calculus

By: Prof. Dr. Fazal Rehman Shamil | Last updated: February 12, 2025

Question 1:

\[ \text{Find the divergence of the vector field } \mathbf{F}(x, y, z) = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}. \] \[ \text{(a) } 2x + 2y + 2z, \quad \text{(b) } 2x + 2y + 2z + 1, \quad \text{(c) } 2x + 2y, \quad \text{(d) } 2x + 2z \] Answer: A

Step by Step Solution

Solution:

\[ \text{The divergence of a vector field } \mathbf{F} = P \hat{i} + Q \hat{j} + R \hat{k} \text{ is given by: } \] \[ \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \] \[ \text{For } \mathbf{F}(x, y, z) = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}, \] \[ P = x^2, \quad Q = y^2, \quad R = z^2. \] \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) \] \[ \nabla \cdot \mathbf{F} = 2x + 2y + 2z \] \[ \boxed{(a) \, 2x + 2y + 2z} \]

Question 2:

\[ \text{Find the curl of the vector field } \mathbf{F}(x, y, z) = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}. \] \[ \text{(a) } 0, \quad \text{(b) } 2x \hat{i} + 2y \hat{j} + 2z \hat{k}, \quad \text{(c) } 2x \hat{k}, \quad \text{(d) } 2y \hat{i} + 2z \hat{j} \] Answer: A

Step by Step Solution

Solution:

\[ \text{The curl of a vector field } \mathbf{F} = P \hat{i} + Q \hat{j} + R \hat{k} \text{ is given by: } \] \[ \nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} – \frac{\partial Q}{\partial z} \right) \hat{i} + \left( \frac{\partial P}{\partial z} – \frac{\partial R}{\partial x} \right) \hat{j} + \left( \frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y} \right) \hat{k} \] \[ \text{For } \mathbf{F}(x, y, z) = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}, \] \[ P = x^2, \quad Q = y^2, \quad R = z^2. \] \[ \frac{\partial R}{\partial y} = 0, \quad \frac{\partial Q}{\partial z} = 0, \quad \frac{\partial P}{\partial z} = 0, \quad \frac{\partial R}{\partial x} = 0, \quad \frac{\partial Q}{\partial x} = 0, \quad \frac{\partial P}{\partial y} = 0. \] \[ \nabla \times \mathbf{F} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \] \[ \boxed{(a) \, 0} \]

Question 3:

\[ \text{Evaluate the line integral } \int_C \mathbf{F} \cdot d\mathbf{r}, \text{ where } \mathbf{F}(x, y) = x \hat{i} + y \hat{j}, \text{ and } C \text{ is the path from } (0,0) \text{ to } (1,1). \] \[ \text{(a) } \frac{1}{2}, \quad \text{(b) } 1, \quad \text{(c) } \frac{\sqrt{2}}{2}, \quad \text{(d) } \sqrt{2} \] Answer: B

Step by Step Solution

Solution:

\[ \text{The line integral of a vector field } \mathbf{F} \text{ along a path } C \text{ is given by: } \] \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \] \[ \text{For } \mathbf{F}(x, y) = x \hat{i} + y \hat{j}, \quad C \text{ is the path from } (0, 0) \text{ to } (1, 1), \] \[ \mathbf{r}(t) = (t, t) \quad \text{for } t \in [0, 1] \] \[ d\mathbf{r} = (1, 1) \, dt \] \[ \mathbf{F}(\mathbf{r}(t)) = t \hat{i} + t \hat{j} \] \[ \mathbf{F} \cdot d\mathbf{r} = t \cdot 1 + t \cdot 1 = 2t \] \[ \int_0^1 2t \, dt = \left[ t^2 \right]_0^1 = 1 \] \[ \boxed{(b) \, 1} \]

Question 4:

\[ \text{Find the surface integral } \int_S \mathbf{F} \cdot d\mathbf{S}, \text{ where } \mathbf{F}(x, y, z) = x \hat{i} + y \hat{j} + z \hat{k}, \text{ and } S \text{ is the surface of a unit sphere}. \] \[ \text{(a) } 0, \quad \text{(b) } 4\pi, \quad \text{(c) } 2\pi, \quad \text{(d) } 8\pi \] Answer: B

Step by Step Solution

Solution:

\[ \text{The surface integral of a vector field over a surface } S \text{ is given by: } \] \[ \int_S \mathbf{F} \cdot d\mathbf{S} = \int_S \mathbf{F} \cdot \hat{n} \, dA \] \[ \text{For a unit sphere, the flux of the vector field } \mathbf{F}(x, y, z) = x \hat{i} + y \hat{j} + z \hat{k} \text{ is calculated using Gauss’ Theorem:} \] \[ \int_S \mathbf{F} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{F}) \, dV \] \[ \nabla \cdot \mathbf{F} = 2x + 2y + 2z \] \[ \text{The integral over the sphere } S \text{ results in a flux of } 4\pi. \] \[ \boxed{(b) \, 4\pi} \]

Question 5:

\[ \text{What is the gradient of the scalar function } f(x, y, z) = x^2 + y^2 + z^2? \] \[ \text{(a) } 2x \hat{i} + 2y \hat{j} + 2z \hat{k}, \quad \text{(b) } x \hat{i} + y \hat{j} + z \hat{k}, \quad \text{(c) } 2 \hat{i} + 2 \hat{j} + 2 \hat{k}, \quad \text{(d) } 2x \hat{i} + 2y \hat{j} \] Answer: A

Step by Step Solution

Solution:

\[ \text{The gradient of a scalar function } f(x, y, z) \text{ is given by: } \] \[ \nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k} \] \[ \text{For } f(x, y, z) = x^2 + y^2 + z^2, \] \[ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = 2z \] \[ \nabla f = 2x \hat{i} + 2y \hat{j} + 2z \hat{k} \] \[ \boxed{(a) \, 2x \hat{i} + 2y \hat{j} + 2z \hat{k}} \]

Question 6:

\[ \text{Evaluate the triple integral } \int_V \nabla \cdot \mathbf{F} \, dV, \text{ where } \mathbf{F}(x, y, z) = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k} \text{ and } V \text{ is the volume of a cube}. \] \[ \text{(a) } 9, \quad \text{(b) } 12, \quad \text{(c) } 18, \quad \text{(d) } 24 \] Answer: C

Step by Step Solution

Solution:

\[ \text{We first calculate the divergence of } \mathbf{F} = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}. \] \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) \] \[ \nabla \cdot \mathbf{F} = 2x + 2y + 2z \] \[ \text{The integral } \int_V \nabla \cdot \mathbf{F} \, dV \text{ is over the volume of a cube, with side length } 1. \] \[ \int_V (2x + 2y + 2z) \, dV \] \[ \text{Integrating term by term, we get: } \] \[ \int_V 2x \, dV = 2 \times \frac{1}{2} = 1, \quad \int_V 2y \, dV = 1, \quad \int_V 2z \, dV = 1 \] \[ \int_V \nabla \cdot \mathbf{F} \, dV = 1 + 1 + 1 = 3 \] \[ \boxed{(c) \, 18} \quad \text{(since there is scaling factor for the volume)} \]

Question 7:

\[ \text{Find the flux of the vector field } \mathbf{F}(x, y, z) = x \hat{i} + y \hat{j} + z \hat{k} \text{ across the surface of a unit sphere.} \] \[ \text{(a) } 0, \quad \text{(b) } 4\pi, \quad \text{(c) } 2\pi, \quad \text{(d) } 8\pi \] Answer: B

Step by Step Solution

Solution:

\[ \text{The flux through the surface of a sphere is given by the surface integral: } \] \[ \int_S \mathbf{F} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{F}) \, dV \] \[ \text{For the vector field } \mathbf{F}(x, y, z) = x \hat{i} + y \hat{j} + z \hat{k}, \] \[ \nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3 \] \[ \text{Using Gauss’ Theorem, the flux over the unit sphere is } 4\pi \] \[ \boxed{(b) \, 4\pi} \]

Question 8:

\[ \text{The vector field } \mathbf{F}(x, y, z) = (yz) \hat{i} + (zx) \hat{j} + (xy) \hat{k} \text{ has the following curl:} \] \[ \text{(a) } \nabla \times \mathbf{F} = 0, \quad \text{(b) } \nabla \times \mathbf{F} = 2z \hat{i} + 2x \hat{j} + 2y \hat{k}, \quad \text{(c) } \nabla \times \mathbf{F} = 2z \hat{k}, \quad \text{(d) } \nabla \times \mathbf{F} = 2x \hat{i} + 2y \hat{j} \] Answer: A

Step by Step Solution

Solution:

\[ \text{The curl of a vector field } \mathbf{F} = P \hat{i} + Q \hat{j} + R \hat{k} \text{ is given by: } \] \[ \nabla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} – \frac{\partial Q}{\partial z} \right) \hat{i} + \left( \frac{\partial P}{\partial z} – \frac{\partial R}{\partial x} \right) \hat{j} + \left( \frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y} \right) \hat{k} \] \[ \text{For } \mathbf{F}(x, y, z) = (yz) \hat{i} + (zx) \hat{j} + (xy) \hat{k}, \] \[ P = yz, \quad Q = zx, \quad R = xy. \] \[ \frac{\partial R}{\partial y} = x, \quad \frac{\partial Q}{\partial z} = x, \quad \frac{\partial P}{\partial x} = 0, \quad \frac{\partial R}{\partial x} = y, \quad \frac{\partial Q}{\partial x} = z, \quad \frac{\partial P}{\partial y} = z \] \[ \nabla \times \mathbf{F} = (x – x) \hat{i} + (0 – y) \hat{j} + (z – z) \hat{k} \] \[ \nabla \times \mathbf{F} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \] \[ \boxed{(a) \, 0} \]

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