VC4.3

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Since the integration is performed in xy-plane, we have r=xi+yj,dr=dxi+dyj\, The path of integration C consists of the straight line OA and AB. Now, we have

\int _{{C}}F\cdot \,dr=\int _{{C}}[(2x+y)i+(3y-x)j]\cdot (dxi+dyj)\,

=\int _{{C}}[(2x+y)dx+(3y-x)dy]\, --(1)

The equation of the straight line AB is y-0={\frac  {2-0}{3-2}}(x-2)\, i.e y=2x-4\,

Therefore,on OA,y=0,dy=0 and x varies from 0 to 2, and on AB,y=2x-4,dy=2dx\, and x varies from 2 to 3. Then from the equation (1),

\int _{{C}}F\cdot \,dr=\int _{{0}}^{{2}}2x\,dx+\int _{{2}}^{{3}}[(2x+2x-4)dx+(6x-12-x)\cdot \,2dx]\,

=[x^{2}]_{{0}}^{{2}}+\int _{{2}}^{{3}}(14x-28)dx=4+14\int _{{2}}^{{3}}(x-2)dx=4+14[{\frac  {(x-2)^{{2}}}{2}}]_{{2}}^{{3}}=4+7=11\,

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