VC5.25

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Here the curve C is given by x^{2}+y^{2}+z^{2}-2ax-2ay=0,x+y=2a\, or (x-a)^{2}+(y-a)^{2}+z^{2}=2a^{2},x+y=2a\, which is the curve of intersection of the sphere (x-a)^{2}+(y-a)^{2}+z^{2}=({\sqrt  {2a}})^{2}\, and the plane x+y=2a\,

Since the center of the sphere (a,a,0)lies on the plane x+y=2a\,

C is a circle lying on the plane and radius a{\sqrt  {2}}\,. Let S1 be taken as the region bounded by this circle C.

Now,we have ydx+zdy+xdz=(yi+zj+xk)\cdot (dxi+dyj+dzk)=ydx+zdy+xdz=F\cdot dr\, --(1)

Where,let F=yi+zj+xk\, --(2)

Now,we have {\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\y&z&x\end{vmatrix}}=-i-j-k\, --(3)

Equation of the surface S1 is \phi (x,y,z)=x+y-2a=0\,

Therefore,\nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}]\phi =i+j\, --(4)

Therefore,n=the unit normal vector to surface S1={\frac  {\nabla \phi }{|\nabla \phi |}}={\frac  {i+j}{{\sqrt  {1+1}}}}={\frac  {i+j}{{\sqrt  {2}}}}\,

Now,\int _{C}(ydx+zdy+xdz)=\int _{F}F\cdot dr\,,using (1)

=\int _{{S_{1}}}{\mathrm  {curl}}F\cdot ndS\, by Stokes'theorm

=\int _{{S_{1}}}{-(i+j+k)}\cdot [{\frac  {i+j}{{\sqrt  {2}}}}]dS\,,by (3) and (4)

=-{\sqrt  {2}}\int _{{S_{1}}}dS=-{\sqrt  {2}}(areaofthecircleC)=-{\sqrt  {2}}\cdot \pi (a{\sqrt  {2}})^{2}=-2{\sqrt  {2}}\pi a^{2}\,

Here,the sense of description of the circle C,namely,starting from the point (2a,0,0) and going below the xy-plane,ensures that the positive direction of n is i+j.

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