From Example Problems
Jump to: navigation, search

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\,


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\,

Main Page