VC4.2

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Since the integration is performed in the xy-plane,we have r=xi+yj\, and dr=dxi+dyj\,.

The path of integration C consists of straight lines OA,AB,BC and CO.Here the coordinates of A,B,C are (0,a),(a,b),(0,b)respectively.

Now,we have \int _{{C}}F\cdot \,dr=\int _{{C}}[(x^{2}+y^{2})i+2xyj]\cdot (dxi+dyj)=\int _{{C}}[(x^{2}+y^{2})dx-2xydy]\,

Here, on OA,y=0,dy=0 and x varies from 0 to a. on AB,x=a,dx=0 and y varies from 0 to b. on BC,y=b,dy=0 and x varies from a to 0. on CO,x=dx=0 and y varies from b to 0.

Using the above integration,we have

\int _{{C}}F\cdot \,dr=\int _{{0}}^{{a}}x^{2}\,dx+\int _{{0}}^{{b}}(-2ay)\,dy+\int _{{a}}^{{0}}(x^{2}+b^{2})\,dx+\int _{{b}}^{{0}}0\,dy\,

=[{\frac  {x^{3}}{3}}]_{{0}}^{{a}}-2a[{\frac  {y^{2}}{2}}]_{{0}}^{{b}}+[{\frac  {x^{3}}{3}}+b^{2}x]_{{a}}^{{0}}+0={\frac  {a^{3}}{3}}-ab^{2}-{\frac  {a^{3}}{3}}-b^{2}a=-2ab^{2}\,

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