VC5.68

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Given F=|r|^{2}r=(x^{2}+y^{2}+z^{2})(xi+yj+zk)=x(x^{2}+y^{2}+z^{2})i+y(x^{2}+y^{2}+z^{2})j+z(x^{2}+y^{2}+z^{2})k\,

Therefore {\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\x(x^{2}+y^{2}+z^{2})&y(x^{2}+y^{2}+z^{2})&z(x^{2}+y^{2}+z^{2})\end{vmatrix}}\,

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

=(2yz-2zy)i-(2xz-2zx)j+(2xy-2xy)k=0\,

Hence F is conservative. If \phi (x,y,z)\, be the required scalar potential,then we must have F=\nabla \phi \, so that F\cdot dr=\nabla \phi \cdot dr\,

or F\cdot dr=[{\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j+{\frac  {\partial \phi }{\partial z}}k]\cdot (dxi+dyj+dzk)\,

={\frac  {\partial \phi }{\partial x}}dx+{\frac  {\partial \phi }{\partial y}}dy+{\frac  {\partial \phi }{\partial z}}dz=d\phi \,

Therefore,d\phi =F\cdot dr=|r|^{2}r\cdot dr={\frac  {1}{2}}|r|^{2}d(r\cdot r)={\frac  {1}{2}}r^{2}dr^{2}\,

d\phi ={\frac  {1}{2}}udu\, taking u=r^{2}\,

which implies \phi ={\frac  {1}{2}}\cdot {\frac  {1}{2}}u^{2}+c={\frac  {1}{4}}|r|^{4}+c\,, c being constant.

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