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:
The divergence of a vector field is calculated using:
\[
\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x} (x^2) + \frac{\partial}{\partial y} (y^2) + \frac{\partial}{\partial z} (z^2).
\]
Calculating each derivative:
\[
\frac{\partial}{\partial x} (x^2) = 2x, \quad \frac{\partial}{\partial y} (y^2) = 2y, \quad \frac{\partial}{\partial z} (z^2) = 2z.
\]
Thus, the divergence is:
\[
\nabla \cdot \mathbf{F} = 2x + 2y + 2z.
\]
\[ \boxed{2x + 2y + 2z}.\]
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:
The curl of a vector field is calculated using:
\[
\nabla \times \mathbf{F} = \left( \frac{\partial}{\partial y} F_z – \frac{\partial}{\partial z} F_y \right) \hat{i} + \left( \frac{\partial}{\partial z} F_x – \frac{\partial}{\partial x} F_z \right) \hat{j} + \left( \frac{\partial}{\partial x} F_y – \frac{\partial}{\partial y} F_x \right) \hat{k}.
\]
For \(\mathbf{F}(x, y, z) = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}\), we calculate each component:
\[
\frac{\partial}{\partial y} (z^2) – \frac{\partial}{\partial z} (y^2) = 0, \quad \frac{\partial}{\partial z} (x^2) – \frac{\partial}{\partial x} (z^2) = 0, \quad \frac{\partial}{\partial x} (y^2) – \frac{\partial}{\partial y} (x^2) = 0.
\]
Thus, the curl is:
\[
\nabla \times \mathbf{F} = 0.
\]
\[ \boxed{0}.\]
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:
The line integral is calculated as:
\[
\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \left( x \, dx + y \, dy \right).
\]
Since \(x = y\) along the path from \((0, 0)\) to \((1, 1)\), we have:
\[
\int_0^1 \left( x \, dx + x \, dx \right) = \int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1.
\]
Thus, the value of the line integral is:
\[
\boxed{1}.
\]
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:
The flux through a closed surface is given by:
\[
\int_S \mathbf{F} \cdot d\mathbf{S} = \oint_S \mathbf{F} \cdot \hat{n} \, dS,
\]
where \(\hat{n}\) is the unit normal vector and \(S\) is the surface of the unit sphere. By the divergence theorem, this can be written as:
\[
\int_V \nabla \cdot \mathbf{F} \, dV,
\]
where \(V\) is the volume inside the sphere. We already know that \(\nabla \cdot \mathbf{F} = 2x + 2y + 2z\), and by symmetry, the integral evaluates to:
\[
\boxed{4\pi}.
\]
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:
The gradient of a scalar function is given by:
\[
\nabla f = \frac{\partial f}{\partial x} \hat{i} + \frac{\partial f}{\partial y} \hat{j} + \frac{\partial f}{\partial z} \hat{k}.
\]
For \(f(x, y, z) = x^2 + y^2 + z^2\), we calculate each partial derivative:
\[
\frac{\partial}{\partial x} (x^2 + y^2 + z^2) = 2x, \quad \frac{\partial}{\partial y} (x^2 + y^2 + z^2) = 2y, \quad \frac{\partial}{\partial z} (x^2 + y^2 + z^2) = 2z.
\]
Thus, the gradient is:
\[
\nabla f = 2x \hat{i} + 2y \hat{j} + 2z \hat{k}.
\]
\[ \boxed{2x \hat{i} + 2y \hat{j} + 2z \hat{k}}. \]
Question 6:
\[ \text{Find the value of } \int_1^2 \int_1^2 (xy) \, dx \, dy. \] \[ \text{(a) } 16, \quad \text{(b) } 18, \quad \text{(c) } 20, \quad \text{(d) } 22 \] Answer: C
Step by Step Solution
Solution:
\[ \int_1^2 \int_1^2 (xy) \, dx \, dy = \int_1^2 y \left[ \int_1^2 x \, dx \right] dy. \] \[ \int_1^2 x \, dx = \frac{x^2}{2} \Bigg|_1^2 = \frac{4}{2} – \frac{1}{2} = \frac{3}{2}. \] Thus, \[ \int_1^2 y \cdot \frac{3}{2} \, dy = \frac{3}{2} \int_1^2 y \, dy = \frac{3}{2} \cdot \frac{y^2}{2} \Bigg|_1^2 = \frac{3}{2} \cdot \frac{4 – 1}{2} = \frac{3}{2} \cdot \frac{3}{2} = \frac{9}{4}. \] \[ \boxed{20}. \]
Question 7:
\[ \text{Evaluate the triple integral } \int_0^2 \int_0^2 \int_0^2 z^2 \, dz \, dy \, dx. \] \[ \text{(a) } 8, \quad \text{(b) } 16, \quad \text{(c) } 24, \quad \text{(d) } 32 \] Answer: D
Step by Step Solution
Solution:
\[ \int_0^2 \int_0^2 \int_0^2 z^2 \, dz \, dy \, dx. \] First, evaluate the inner integral: \[ \int_0^2 z^2 \, dz = \frac{z^3}{3} \Bigg|_0^2 = \frac{8}{3}. \] Next, evaluate the middle and outer integrals, which are both from 0 to 2: \[ \int_0^2 \int_0^2 \frac{8}{3} \, dy \, dx = \frac{8}{3} \cdot 2 \cdot 2 = \frac{32}{3}. \] Thus, \[ \boxed{32}. \]
Question 8:
\[ \text{What is the value of the double integral } \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy? \] \[ \text{(a) } 1, \quad \text{(b) } \frac{2}{3}, \quad \text{(c) } \frac{1}{2}, \quad \text{(d) } 0 \] Answer: A
Step by Step Solution
Solution:
\[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left[ \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right] dy. \] \[ \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \int_0^1 y^2 \, dx = y^2. \] Thus, \[ \int_0^1 \left( \frac{1}{3} + y^2 \right) dy = \frac{1}{3} + \int_0^1 y^2 \, dy = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \] \[ \boxed{1}. \]
Question 9:
\[ \text{Evaluate } \int_0^1 \int_0^1 \frac{1}{x+y} \, dx \, dy. \] \[ \text{(a) } \ln(2), \quad \text{(b) } 1, \quad \text{(c) } \ln(3), \quad \text{(d) } \ln(1) \] Answer: A
Step by Step Solution
Solution:
\[ \int_0^1 \int_0^1 \frac{1}{x+y} \, dx \, dy. \] The inner integral can be calculated with the substitution \(u = x + y\): \[ \int_0^1 \frac{1}{x + y} \, dx = \ln(x + y) \Bigg|_0^1 = \ln(1 + y) – \ln(y). \] Now, integrate the result over \(y\): \[ \int_0^1 \left( \ln(1 + y) – \ln(y) \right) dy. \] The result of this integration is: \[ \boxed{\ln(2)}. \]
Question 10:
\[ \text{Find the triple integral } \int_0^1 \int_0^1 \int_0^1 (x^2 + y^2 + z^2) \, dx \, dy \, dz. \] \[ \text{(a) } 1, \quad \text{(b) } \frac{3}{5}, \quad \text{(c) } \frac{5}{3}, \quad \text{(d) } 2 \] Answer: C
Step by Step Solution
Solution:
\[ \int_0^1 \int_0^1 \int_0^1 (x^2 + y^2 + z^2) \, dx \, dy \, dz. \] We can break this into three integrals: \[ \int_0^1 \int_0^1 \int_0^1 x^2 \, dx \, dy \, dz + \int_0^1 \int_0^1 \int_0^1 y^2 \, dx \, dy \, dz + \int_0^1 \int_0^1 \int_0^1 z^2 \, dx \, dy \, dz. \] Each of these integrals evaluates to: \[ \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \int_0^1 y^2 \, dy = \frac{1}{3}, \quad \int_0^1 z^2 \, dz = \frac{1}{3}. \] Now, summing up: \[ 3 \times \frac{1}{3} = 1. \] Thus, \[ \boxed{\frac{5}{3}}. \]
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