Question

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) \(\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\)

I got this question in semester exam.

The doubt is from Using Properties of Divergence and Curl in chapter Vector Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (a) \(-2\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\)

The best I can explain: We know that the curl for any vector quantity is given by

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

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

\(\bigtriangledown.\vec{r}=-2\vec{i}(1+x^2 z)-\vec{j}(1-3y^2 z^2)+\vec{2k}(xz^2-yz^3)\).