Given a vector  and  ˆn as the unit normal vector to the surface of the hemisphere (x² + y² + z² = 1;z ≥0), the value of integral  evaluated on the curved surface of the hemisphere S is

Correct Answer: $\frac{\pi }{2}$

1. Understanding the Problem and Stokes' Theorem
We are given a vector field:
$$\overrightarrow{u}=\frac{1}{3}\left(-y^3\hat{i}+x^3\hat{j}+z^3\hat{k}\right)$$
We want to evaluate the surface integral:
$${\int }_S\left(\nabla \times \overrightarrow{u}\right)·\hat{n}dS$$
where $S$ is the curved surface of the hemisphere defined by:
$$x^2+y^2+z^2=1,z\ge 0$$
and $\hat{n}$ is the unit outward normal vector to the curved surface $S$.
By applying Stokes' Theorem (as shown in the formula in the second attached image):
$${\int }_S\left(\nabla \times \overrightarrow{u}\right)·\hat{n}dS={\oint }_C\overrightarrow{u}·d\overrightarrow{r}$$
where $C$ is the boundary curve enclosing the hemisphere. The boundary of this hemisphere (depicted in the first attached image) is a circle lying in the $xy$-plane where $z=0$.

2. Simplifying the Line Integral
The boundary curve $C$ is the unit circle in the $xy$-plane:
$$x^2+y^2=1,z=0$$
Since $z=0$ along the curve $C$, we have $dz=0$.
Substituting $\overrightarrow{u}$ and $d\overrightarrow{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}$ into the line integral:
$${\oint }_C\overrightarrow{u}·d\overrightarrow{r}={\oint }_C\frac{1}{3}\left(-y^{3}dx+x^{3}dy+z^{3}dz\right)$$
Setting $z=0$ and $dz=0$ (as visible in the third attached image):
$${\oint }_C\overrightarrow{u}·d\overrightarrow{r}={\oint }_C\frac{1}{3}\left(-y^{3}dx+x^{3}dy\right)$$

3. Applying Green's Theorem
We can convert the line integral over the closed curve $C$ to a double integral over the region $D$ (the region enclosed by the circle $x^2+y^2\le 1$ in the $xy$-plane) using Green's Theorem:
$${\oint }_C\left(Pdx+Qdy\right)={\&iint;}_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$$
Here, our components are:
$$P=-\frac{1}{3}{y}^3\text{and}Q=\frac{1}{3}{x}^3$$
Taking the partial derivatives:
$$\frac{\partial P}{\partial y}=-y^2\text{and}\frac{\partial Q}{\partial x}=x^2$$
Substituting these into the formula:
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=x^2-\left(-y^2\right)=x^2+y^2$$
Therefore, the double integral over the disk $D$ becomes:
$${\&iint;}_D\left(x^2+y^2\right)dxdy$$

4. Evaluating in Polar Coordinates
To evaluate this integral, we convert to polar coordinates:
$$x=rcos\theta ,y=rsin\theta ,dxdy=rdrd\theta$$
For the unit disk $D$, the limits of integration are:
$$r\in \left[0,1\right]\text{and}\theta \in \left[0,2\pi \right]$$
Substituting these into the integral (as shown in the fourth attached image):
$${\int }_{\theta =0}^{2\pi }{\int }_{r=0}^1{r}^2·rdrd\theta ={\int }_{\theta =0}^{2\pi }\left({\int }_{r=0}^1{r}^{3}dr\right)d\theta$$
First, integrate with respect to $r$:
$${\int }_0^1{r}^{3}dr={\left[\frac{r^4}{}\right]}_0^1$$
Next, integrate with respect to $\theta$:
$${\int }_0^{2\pi }\frac{1}{4}d\theta =\frac{1}{4}{\left[\theta \right]}_0^{2\pi }=\frac{2\pi }{4}=\frac{\pi }{2}$$