Question Number 135872 by Engr_Jidda last updated on 16/Mar/21
$${Evaluate}\:\oint_{{c}} {ydy}\:{where}\:\:{c}\:{is}\:{a}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$
Answered by mathmax by abdo last updated on 16/Mar/21
![y =ξ(√(4−x^2 )) ⇒∫_C ydy =∫_(−2) ^2 ξ(√(4−x^2 ))ξ((−2x)/(2(√(4−x^2 ))))dx =∫_(−2) ^2 xdx =0 (ξ^2 =1) and ∫_C ydx =∫_(−2) ^2 ξ(√(4−x^2 ))dx =_(x=2sint) ξ ∫_(−(π/2)) ^(π/2) 2cost (2cost)dt =4ξ ∫_(−(π/2)) ^(π/2) ((1+cos(2t))/2)dt =2ξ ∫_(−(π/2)) ^(π/2) (1+cos(2t))dt =2πξ +ξ[sin(2t)]_(−(π/2)) ^(π/2) =2πξ](Q135883.png)
$$\mathrm{y}\:=\xi\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{C}} \mathrm{ydy}\:=\int_{−\mathrm{2}} ^{\mathrm{2}} \xi\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\xi\frac{−\mathrm{2x}}{\mathrm{2}\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$=\int_{−\mathrm{2}} ^{\mathrm{2}} \:\mathrm{xdx}\:=\mathrm{0}\:\:\left(\xi^{\mathrm{2}} =\mathrm{1}\right)\:\:\mathrm{and}\: \\ $$$$\int_{\mathrm{C}} \mathrm{ydx}\:=\int_{−\mathrm{2}} ^{\mathrm{2}} \xi\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{x}=\mathrm{2sint}} \:\xi\:\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2cost}\:\left(\mathrm{2cost}\right)\mathrm{dt} \\ $$$$=\mathrm{4}\xi\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2t}\right)}{\mathrm{2}}\mathrm{dt}\:=\mathrm{2}\xi\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2t}\right)\right)\mathrm{dt} \\ $$$$=\mathrm{2}\pi\xi\:+\xi\left[\mathrm{sin}\left(\mathrm{2t}\right)\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{2}\pi\xi \\ $$
Commented by Engr_Jidda last updated on 17/Mar/21
$${thank}\:{you}\:{so}\:{much}\:{sir}.\: \\ $$
Commented by mathmax by abdo last updated on 17/Mar/21
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$