Question

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) \(y^3+xy+1,\vec{i}(3y^2 z)+\vec{k}(3-x^2)\)

The question was asked during an online exam.

Question is taken from Using Properties of Divergence and Curl in chapter Vector Differential Calculus of Engineering Mathematics

Answer 1

Right option is (b) \(y^3+2xy+1,\vec{i}(3y^2 z)+\vec{k}(3-x^2)\)

The best 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

\(\bigtriangledown.\vec{r}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\x^2 y&3x+y&(y^3 z)\end{bmatrix}\)

\(\bigtriangledown.\vec{r}=\vec{i}\left (\frac{\partial(y^3 z)}{\partial y}-\frac{\partial(3x+y)}{\partial z}\right )-\vec{j}\left (\frac{\partial(y^3 z)}{\partial x}-\frac{\partial(x^2 y)}{\partial z}\right )+\vec{k}\left (\frac{\partial(3x+y)}{\partial x}-\frac{\partial(x^2 y)}{\partial y}\right )\)

\(\bigtriangledown.\vec{r}=\vec{i}(3y^2 z)+\vec{k}(3-x^2)\).