Question

Given a vector $\overrightarrow{u}$ = $\frac{1}{3}$$(-y^3\hat{i}+x^3\hat{j}+z^3\hat{k})$ and $\hat{n}$ as the unit normal vector to the surface of the hemisphere ($x^2+y^2+z^2=1;z>0$), the value of integral $\int (▽\times \overrightarrow{u}).\hat{n}dS$ evaluated on the curved surface of the hemisphere S is

• A. -$\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is C $\frac{\pi }{2}$

$▽\times u=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{-y^3}{3}& \frac{x^3}{3}& \frac{z^3}{3}\end{array}\right]$
= ($x^2+y^2)\hat{k}$
$\hat{n}$ = $\hat{k}$
$(▽\times u).\hat{n}=(▽\times u).\hat{k}=x^2+y^2$
$\int (x^2+y^2)\hat{k}.\hat{n}dS=\int (x^2+y^2)dS$
=$\int \int (x^2+y^2)dxdy$
=${\int }_0^{2\pi }{\int }_0^1{r}^2\left(rdrd\theta \right)=\frac{\pi }{2}$