VC4.21

From Exampleproblems

Here the integration is performed in the xy-plane.

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

$F\cdot dr=(e^x\sin y i+e^x\cos y j)\cdot(dxi+dyj)=e^x\sin y dx+e^x\cos y dy\,$

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, $\int_{C_1}F\cdot dr=\int_{C_1} (e^x\sin y dx+e^x\cos y dy)=\int_{x=0}^{1} dx=0\,$ --(1)

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

$\int_{C_2} F\cdot dr=\int_{C_2} (e^x\sin y dx+e^x\cos y dy)=\int_{y=0}^{\frac{\pi}{2}} e\cos y dy=e[\sin y]_{0}^{\frac{\pi}{2}}=e\,$ --(2)

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

Therefore, $\int_{C_3}F\cdot dr=\int_{C_3}(e^x\sin y dx+e^x\cos y dy)=\int_{x=0}^{1}e^x dx=[e^x]_1^0=1-e^x\,$ --(3)

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

$\int_{C_4}F\cdot dr=\int_{C_4}(e^x\sin y dx+e^x\cos y dy)=\int_{y=\frac{\pi}{2}}^{0}=[\sin y]_{\frac{\pi}{2}}^{0}=-1\,$ --(4)

Using (1),(2),(3) and (4), $\oint_C F\cdot dr=0+e+1-e-1=0\,$

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