Vector Differential Calculus Questions and Answers – Using Properties of Divergence and Curl

This set of Engineering Mathematics Problems focuses on “Using Properties of Divergence and Curl”.

1. Find the divergence of this given vector \(\vec{F}=x^3 y\vec{i}+3xy^2 z\vec{j}+3zx\vec{k}\).
a) 3x2 y+6xyz+x
b) 2x2 y+6xyz+3x
c) 3x2 y+3xyz+3x
d) 3x2 y+6xyz+3x
View Answer

Answer: d
Explanation: We know that divergence of a vector is given by
\(\bigtriangledown.\vec{F}=\frac{\partial(x^3 y)}{\partial x}+\frac{\partial(3xy^2 z)}{\partial y}+\frac{\partial(3xz)}{\partial z}\)
\(\bigtriangledown.\vec{F}=3x^2 y+6xyz+3x\).

2. Find the divergence of this given vector \(\vec{r}=12x^6 y^6 \vec{i}+3x^3 y^3 z\vec{j}+3x^2 yz^2 \vec{k}\).
a) \(12x^5 y^6+2x^3 yz+6x^2 yz\)
b) \(72x^5 y^6+2x^3 yz+3x^2 yz\)
c) \(72x^5 y^6+2x^3 yz+6x^2 yz\)
d) \(6x^5 y^6+2x^3 yz+6x^2 yz\)
View Answer

Answer: c
Explanation: We know that divergence of a vector is given by
\(\bigtriangledown.\vec{r}=\frac{\partial(12x^6 y^6)}{\partial x}+\frac{\partial(3x^3 y^3 z)}{\partial y}+\frac{\partial(3x^2 yz^2)}{\partial z}\)
\(\bigtriangledown.\vec{r}=12×6x^5 y^6+x^3×2y×z+3x^2×y×2z\)
\(\bigtriangledown.\vec{r}=72x^5 y^6+2x^3 yz+6x^2 yz\).

3. Find the curl for \(\vec{r}=x^2 yz\vec{i}+(3x+2y)z\vec{j}+21z^2 x\vec{k}\).
a) \(\vec{i}(3x+2y)-\vec{j}(11z^2-x^2 y)+\vec{k}(3z-x^2 z)\)
b) \(\vec{i}(x+2y)-\vec{j}(21z^2-x^2 y)+\vec{k}(3z-x^2 z)\)
c) \(-\vec{i}(3x+2y)-\vec{j}(21z^2-x^2 y)+\vec{k}(3z-x^2 z)\)
d) \(\vec{i}(3x+2y)-\vec{j}(21z^2-x^2 y)+\vec{k}(3z-x^2 z)\)
View Answer

Answer: c
Explanation: We know that the curl for any vector quantity is given by
\(\bigtriangledown.\vec{r}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2 yz&(3x+2y)z&21z^2 x\end{bmatrix}\)
\(\bigtriangledown.\vec{r}=\vec{i}\left (\frac{\partial(21z^2 x)}{\partial y}-\frac{\partial((3x+2y)z)}{\partial z}\right)-\vec{j}\left (\frac{\partial(21z^2 x)}{\partial x}-\frac{\partial(x^2 yz)}{\partial z}\right)\)
\(+\vec{k}\left (\frac{\partial((3x+2y)z)}{\partial x}-\frac{\partial(x^2 yz)}{\partial y}\right)\)
\(\bigtriangledown.\vec{r}=-\vec{i}(3x+2y)-\vec{j}(21z^2-x^2 y)+\vec{k}(3z-x^2 z)\).
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4. Find the curl for \((\vec{r})=y^2 z^3 \vec{i}+x^2 z^2 \vec{j}+(x-2y)\vec{k}\).
a) \(-2\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\)
b) \(-2\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{k}(xz^2-yz^3)\)
c) \(-2\vec{i}(1+x^2 z)-\vec{j}(1-32z^2)+\vec{2k}(xz^2-yz^3)\)
d) \(\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\)
View Answer

Answer: a
Explanation: We know that the curl for any vector quantity is given by
\(\bigtriangledown.\vec{r}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\y^2 z^3&x^2 z^2&(x-2y)\end{bmatrix}\)
\(\bigtriangledown.\vec{r}=\vec{i}\left (\frac{\partial(x-2y)}{\partial y}-\frac{\partial(x^2 z^2)}{/\partial z}\right )-\vec{j}\left (\frac{\partial(x-2y)}{\partial x}-\frac{\partial(y^2 z^3)}{\partial z}\right )+\vec{k}\left(\frac{\partial(x^2 z^2)}{\partial x}-\frac{\partial(y^2 z^3)}{\partial y}\right )\) \(\bigtriangledown.\vec{r}=-2\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\).

5. What is the divergence and curl of the vector \(\vec{F}=x^2 y\vec{i}+(3x+y) \vec{j}+y^3 z\vec{k}\).
a) \(y^3+2xy+1,\vec{i}(3y^2 z)+\vec{j}(3-x^2)\)
b) \(y^3+2xy+1,\vec{i}(3y^2 z)+\vec{k}(3-x^2)\)
c) \(3y^3+2xy+1,\vec{i}(3y^2 z)+\vec{k}(3-x^2)\)
d) \(y^3+xy+1,\vec{i}(3y^2 z)+\vec{k}(3-x^2)\)
View Answer

Answer: b
Explanation: We know that divergence of a vector is given by
\(\bigtriangledown.\vec{F}=\frac{\partial(x^2 y)}{\partial x}+\frac{\partial(3x+y)}{\partial y}+\frac{\partial(y^3 z)}{\partial z}\)
\(\bigtriangledown.\vec{F}=y^3+2xy+1\)
We know that the curl for any vector quantity is given by
\(\bigtriangledown.\vec{r}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2 y&3x+y&(y^3 z)\end{bmatrix}\)
\(\bigtriangledown.\vec{r}=\vec{i}\left (\frac{\partial(y^3 z)}{\partial y}-\frac{\partial(3x+y)}{\partial z}\right )-\vec{j}\left (\frac{\partial(y^3 z)}{\partial x}-\frac{\partial(x^2 y)}{\partial z}\right )+\vec{k}\left (\frac{\partial(3x+y)}{\partial x}-\frac{\partial(x^2 y)}{\partial y}\right )\)
\(\bigtriangledown.\vec{r}=\vec{i}(3y^2 z)+\vec{k}(3-x^2)\).
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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