VC5.74

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{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\x^{2}+xy^{2}&y^{2}+x^{2}y&0\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(0)-{\frac  {\partial }{\partial z}}(y^{2}+x^{2}y)]i-[{\frac  {\partial }{\partial x}}(0)-{\frac  {\partial }{\partial z}}(x^{2}+xy^{2})]j+[{\frac  {\partial }{\partial x}}(y^{2}+x^{2}y)-{\frac  {\partial }{\partial y}}(x^{2}+xy^{2})]k\,

=(0-0)i-(0-0)j+(2xy-2xy)k=0\,

Therefore,the vector field given by F is irrotational.

Let \phi (x,y,z)\, be the required scalar potential such that

F=\nabla \phi \, or (x^{2}+xy^{2})i+(y^{2}+x^{2}y)j={\frac  {\partial \phi }{\partial x}}i+{\frac  {\partial \phi }{\partial y}}j\,

Therefore {\frac  {\partial \phi }{\partial x}}=x^{2}+xy^{2}\, whence \phi ={\frac  {x^{3}}{3}}+{\frac  {x^{2}y^{2}}{2}}+f_{1}(y)\, --(1)

and {\frac  {\partial \phi }{\partial y}}=y^{2}+x^{2}y\, whence \phi ={\frac  {y^{3}}{3}}+{\frac  {x^{2}y^{2}}{2}}+f_{2}(x)\, --(2)

(1) and (2) each represent phi.These agree if we choose f_{1}(y)={\frac  {y^{3}}{3}}\,

and f_{2}(x)={\frac  {x^{3}}{3}}\,

Hence the required scalar potential is given by \phi ={\frac  {x^{3}+y^{3}}{3}}+{\frac  {x^{2}y^{2}}{2}}+c\,, c being constant.

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