# VC4.23

Here,$\nabla \times F={\begin{vmatrix}i&j&k\\{\frac {\partial }{\partial x}}&{\frac {\partial }{\partial y}}&{\frac {\partial }{\partial z}}\\y&(x-2xz)&-xy\end{vmatrix}}\,$

=$[{\frac {\partial (-xy)}{\partial y}}-{\frac {\partial (x-2xz)}{\partial z}}]i-[{\frac {\partial (-xy)}{\partial z}}-{\frac {\partial y}{\partial z}}]j+[{\frac {\partial (x-2xz)}{\partial x}}-{\frac {\partial y}{\partial z}}]k\,$

=$[-x-(-2x)]i-[-y-0]j+[1-2z-1]k=xi+yj-2zk\,$

Therefore,given surface integral=$\iint _{S}(\nabla \times F)\cdot ndS=\iint (\nabla \times F)\cdot n{\frac {dxdy}{|n\cdot k|}}\,$ --(1)

If the given surface can be written as $\phi (x,y,z)=x^{2}+y^{2}+z^{2}=a^{2}--(2)\,$,then the normal vector to S is $\nabla \phi =[i{\frac {\partial }{\partial x}}+j{\frac {\partial }{\partial y}}j+{\frac {\partial }{\partial z}}k](x^{2}+y^{2}+z^{2})=2xi+2yj+2zk\,$

Let n denote the unit vector,then we have $n={\frac {\nabla \phi }{|\nabla \phi |}}={\frac {2(xi+yj+zk)}{{\sqrt {(4x^{2}+4y^{2}+4z^{2})}}}}={\frac {xi+yj+zk}{a}}\,$,by using(2)

Therefore,$(\nabla \times F)\cdot n=(xi+yj-2zk)\cdot ({\frac {xi+yj+zk}{a}})={\frac {x^{2}+y^{2}-2z^{2}}{a}}\,$ --(3)

Also,$n\cdot k={\frac {z}{a}}\,$ --(4)

With (3) and (4),(1) gives $\iint _{S}(\nabla \times f)\cdot ndS\,$

=$\iint _{R}{\frac {x^{2}+y^{2}-2z^{2}}{a}}\cdot ({\frac {dxdy}{{\frac {z}{a}}}})\,$

=$\iint _{R}{\frac {x^{2}+y^{2}-2(a^{2}-x^{2}-y^{2})}{{\sqrt {(a^{2}-x^{2}-y^{2})}}}}dxdy=\iint _{R}{\frac {3(x^{2}+y^{2})-2a^{2}}{{\sqrt {(a^{2}-x^{2}-y^{2})}}}}dxdy\,$ --(5)

To transform the double integral on RHS of (5) into polar coordinates,we write $x=r\cos \theta ,y=r\sin \theta \,$ so that $dxdy=rd\theta dr,x^{2}+y^{2}=r^{2}\,$

Again for the circular region r varies from 0 to a and theta varies from 0 to $2\pi \,$,then (5) can be written as

=$\int _{{\theta =0}}^{{2\pi }}\int _{{r=0}}^{{a}}{\frac {3r^{2}-2a^{2}}{{\sqrt {(a^{2}-r^{2})}}}}rd\theta dr\,$

=$\int _{{r=0}}^{{a}}{\frac {r(3r^{2}-2a^{2})}{{\sqrt {(a^{2}-r^{2})}}}}[\theta ]_{{0}}^{{2\pi }}]dr\,$ =$2\pi \int _{{0}}^{{a}}{\frac {r(3r^{2}-2a^{2})}{{\sqrt {(a^{2}-r^{2})}}}}dr\,$ =$2\pi \int _{{0}}^{{{\frac {\pi }{2}}}}{\frac {a\sin \phi (3a^{2}\sin ^{2}\phi -2a^{2})}{a\cos \phi }}a\cos \phi d\phi \,$ [on putting $r=a\sin \phi ,dr=a\cos \phi d\phi \,$] =$2\pi a^{3}[3\int _{{0}}^{{{\frac {\pi }{2}}}}\sin ^{3}\phi d\phi -2\int _{{0}}^{{{\frac {\pi }{2}}}}\sin \phi d\phi ]\,$ =$2\pi a^{3}[3\int _{{0}}^{{{\frac {\pi }{2}}}}{\frac {3\sin \phi -\sin 3\phi }{4}}d\phi -2(-\cos \phi )_{{0}}^{{{\frac {\pi }{2}}}}]\,$ [By using the formula of $\sin 3\phi \,$]

=$2\pi a^{3}[{\frac {3}{4}}(-3\cos \phi +{\frac {1}{3}}\cos 3\phi )_{{0}}^{{{\frac {\pi }{2}}}}-2]\,$ =$2\pi a^{3}[{\frac {3}{4}}(3-{\frac {1}{3}})-2]=2\pi a^{3}[2-2]=0\,$