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

The correct answer is π/2.

1. Identifying the Given Information:
From the provided images, we have:
The vector field:
$\overrightarrow{u}=\frac{1}{3}\left(-y^3\hat{i}+x^3\hat{j}+z^3\hat{k}\right)$
The surface integral to evaluate over the curved surface $S$ of the hemisphere:
$I={\iint }_S\left(\nabla \times \overrightarrow{u}\right)·\hat{n}dS$
where $S$ is the hemisphere $x^2+y^2+z^2=1$ with $z\ge 0$, and $\hat{n}$ is the unit outward normal vector to the surface.

2. Applying Stokes' Theorem:
According to Stokes' Theorem, the surface integral of the curl of a vector field over an open surface $S$ is equal to the line integral of the vector field along the boundary curve $C$ of the surface:
${\iint }_S\left(\nabla ��\overrightarrow{u}\right)·\hat{n}dS={oint}_C\overrightarrow{u}·d\overrightarrow{r}$
where $C$ is the boundary of the hemisphere in the $xy$-plane.

3. Defining the Boundary Curve $C$:
The boundary curve of the hemisphere $x^2+y^2+z^2=1$ with $z\ge 0$ lies in the plane $z=0$.
Thus, the boundary $C$ is the unit circle in the $xy$-plane:
$x^2+y^2=1\text{,}z=0$

Since $z=0$ on the boundary curve $C$, we have $dz=0$. The position vector and its differential are:
$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\Rightarrow d\overrightarrow{r}=dx\hat{i}+dy\hat{j}+dz\hat{k}$
Substituting $z=0$ and $dz=0$ into the dot product $\overrightarrow{u}·d\overrightarrow{r}$:
$\overrightarrow{u}·d\overrightarrow{r}=\frac{1}{3}\left(-y^{3}dx+x^{3}dy\right)$

4. Evaluating the Line Integral using Green's Theorem:
The line integral around the closed curve $C$ in the $xy$-plane is:
${oint}_C\overrightarrow{u}·d\overrightarrow{r}=\frac{1}{3}{oint}_C\left(-y^{3}dx+x^{3}dy\right)$
By Green's Theorem in a plane:
${oint}_C\left(Pdx+Qdy\right)={\iint }_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$
Here, $P=-y^3$ and $Q=x^3$.
Calculating the partial derivatives:
$\frac{\partial Q}{\partial x}=3x^2$
$\frac{\partial P}{\partial y}=-3y^2$
Thus, the integrand becomes:
$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=3x^2-\left(-3y^2\right)=3\left(x^2+y^2\right)$

Substituting this back into the double integral over the disk $R$ bounded by $x^2+y^2=1$:
${oint}_C\overrightarrow{u}·d\overrightarrow{r}=\frac{1}{3}{\iint }_{R}3\left(x^2+y^2\right)dxdy={\iint }_R\left(x^2+y^2\right)dxdy$

5. Computing the Double Integral in Polar Coordinates:
Let $x=rcos\theta$ and $y=rsin\theta$, where:
$x^2+y^2=r^2$
$dxdy=rdrd\theta$
The limits of integration for the unit disk $R$ are $r$ from $0$ to $1$ and $\theta$ from $0$ to $2\pi$:
${\iint }_R\left(x^2+y^2\right)dxdy={\int }_0^{2\pi }{\int }_0^1{r}^2·rdrd\theta$
$={\int }_0^{2\pi }d\theta {\int }_0^1{r}^{3}dr$
$={\left[\theta \right]}_0^{2\pi }{\left[\frac{r^4}{4}\right]}_0^1$
$=2\pi ·\frac{1}{4}=\frac{\pi }{2}$

Therefore, the value of the evaluated integral is π/2.