Question : Use Gauss Divergence theorem to evaluate the surface integral over the portion of z=9-x^2-y^2 with z>=0
\[\iint_{\gamma}{\left(\nabla\ \times\ \vec{F}\right).\hat{n}\ dS\ }\]where\[ \vec{F}\ =\ (y-z)\ \hat{i}-(x+z)\hat{j}\ +\ (x+y)\ \hat{k} \]γ is the portion of \[z =\ 9-x^2\ \ -y^2\] with z ≥ 0 and ˆn is the upward-pointing unit normal.
Solution :
To Evaluate the surface integral
\[\iint_{\gamma}{\left(\nabla\ \times\ \vec{F}\right).\hat{n}\ dS\ }\]For the given vector field\[ \vec{F}\ =\ (y-z)\ \hat{i}-(x+z)\hat{j}\ +\ (x+y)\ \hat{k}\]It’s curl is given as
\[\nabla\ \times\ \vec{F}=\left|\begin{matrix}\ \hat{i}&\hat{j}&\hat{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\y-z&-x-z&x+y\\\end{matrix}\right|\]\[=\hat{i}\left(1+1\right)-\hat{j}\left(1+1\right)+\hat{k}\left(-1-1\right)
\nabla\ \times\ \vec{F}=2\hat{i}-2\hat{j}-2\hat{k}\]Now we shall compute
\[\nabla.\left(\nabla\ \times\ \vec{F}\right)=\nabla.\left(2\hat{i}-2\hat{j}-2\hat{k}\right)=0\]By Gauss Divergence theorem
\[\iint_{\gamma}{\vec{F}.\hat{n}\ dS}=\iiint_{V}{\nabla.\vec{F}}dV\]\[=>\iint_{\gamma}{\left(\nabla\ \times\ \vec{F}\right).\hat{n}\ dS}=\iiint_{V}{\nabla.\left(\nabla\ \times\ \vec{F}\right)}dV=0\]Therefore by Gauss Divergence theorem the value of the given surface integral is zero.