VC4.21

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Here the integration is performed in the xy-plane.

Therefore,r=xi+yj,dr=dxi+dyj\,

F\cdot dr=(e^{x}\sin yi+e^{x}\cos yj)\cdot (dxi+dyj)=e^{x}\sin ydx+e^{x}\cos ydy\,

Now,by defintion,the circulation of F round C = \oint _{C}F\cdot dr=\int _{{C_{1}}}F\cdot dr+\int _{{C_{2}}}F\cdot dr+\int _{{C_{3}}}F\cdot dr+\int _{{C_{4}}}F\cdot dr\,

where C_{1},C_{2},C_{3},C_{4}\, denote straight lines OP,PQ,QR and RO respectively.

Now,on C1,y=0,dy=0 and x varies from 0 to 1,therefore,\int _{{C_{1}}}F\cdot dr=\int _{{C_{1}}}(e^{x}\sin ydx+e^{x}\cos ydy)=\int _{{x=0}}^{{1}}dx=0\, --(1)

Next,on C2,x=1,dx=0 and y varies from 0 to {\frac  {\pi }{2}}\,,therefore,

\int _{{C_{2}}}F\cdot dr=\int _{{C_{2}}}(e^{x}\sin ydx+e^{x}\cos ydy)=\int _{{y=0}}^{{{\frac  {\pi }{2}}}}e\cos ydy=e[\sin y]_{{0}}^{{{\frac  {\pi }{2}}}}=e\, --(2)

Again on C3,y={\frac  {\pi }{2}}\,,dy=0 and x varies from 1 to 0.

Therefore,\int _{{C_{3}}}F\cdot dr=\int _{{C_{3}}}(e^{x}\sin ydx+e^{x}\cos ydy)=\int _{{x=0}}^{{1}}e^{x}dx=[e^{x}]_{1}^{0}=1-e^{x}\, --(3)

Finally,on C4,x=0,dx=0 and y varies from {\frac  {\pi }{2}}\, to 0.

\int _{{C_{4}}}F\cdot dr=\int _{{C_{4}}}(e^{x}\sin ydx+e^{x}\cos ydy)=\int _{{y={\frac  {\pi }{2}}}}^{{0}}=[\sin y]_{{{\frac  {\pi }{2}}}}^{{0}}=-1\, --(4)

Using (1),(2),(3) and (4),\oint _{C}F\cdot dr=0+e+1-e-1=0\,

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