Question Number 215390 by MathematicalUser2357 last updated on 05/Jan/25
![∮_γ x^2 dx [γ: x^2 +y^2 =1]](Q215390.png)
$$\oint_{\gamma} {x}^{\mathrm{2}} {dx}\:\left[\gamma:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\right] \\ $$
Answered by MrGaster last updated on 05/Jan/25
![solve:∫_0 ^(2π) (cos t)^2 (−sin t)dt[∵x=cos t,y=sin t,dx=−sin t dt] − ∫_(0 ) ^(2π) cos^2 t sin t dt =−[((cos^3 t)/3)]_0 ^(2π) =−(((cos^3 (2π))/3)−((cos^3 (0))/3)) =−((1/3)−(1/3)) =0](Q215391.png)
$$\mathrm{solve}:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{cos}\:{t}\right)^{\mathrm{2}} \left(−\mathrm{sin}\:{t}\right){dt}\left[\because{x}=\mathrm{cos}\:{t},{y}=\mathrm{sin}\:{t},{dx}=−\mathrm{sin}\:{t}\:{dt}\right] \\ $$$$\:\:\:−\:\:\int_{\mathrm{0}\:} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{2}} {t}\:\mathrm{sin}\:{t}\:{dt} \\ $$$$=−\left[\frac{\mathrm{cos}^{\mathrm{3}} {t}}{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$=−\left(\frac{\mathrm{cos}^{\mathrm{3}} \left(\mathrm{2}\pi\right)}{\mathrm{3}}−\frac{\mathrm{cos}^{\mathrm{3}} \left(\mathrm{0}\right)}{\mathrm{3}}\right) \\ $$$$=−\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\mathrm{0} \\ $$