Question Number 36195 by prof Abdo imad last updated on 30/May/18
$${let}\:\:{C}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:{y}=\mathrm{2}{x}^{\mathrm{2}} \right\} \\ $$$${calculate}\:\int_{{C}} \:{x}^{\mathrm{2}} {ydx}\:+\left({x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} \right){dy} \\ $$
Commented by math khazana by abdo last updated on 18/Aug/18
![∫_C x^2 ydx +(x^2 −y^2 )dy =∫_C x^2 (2x^2 )dx +∫_C (x^2 −4x^4 )(4x)dx =∫_0 ^1 4x^4 dx + 4∫_0 ^1 (x^3 −4x^5 )dx =(4/5)[x^5 ]_0 ^1 + 4 [ (x^4 /4) −(2/3) x^6 ]_0 ^1 =(4/5) +4{(1/4) −(2/3)} =(4/5) +1 −(8/3) = (9/5) −(8/3) =((27−40)/(15)) =−((13)/(15)) .](https://www.tinkutara.com/question/Q42104.png)
$$\int_{{C}} \:\:\:{x}^{\mathrm{2}} {ydx}\:\:+\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dy} \\ $$$$=\int_{{C}} {x}^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} \right){dx}\:\:\:+\int_{{C}} \left({x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{4}} \right)\left(\mathrm{4}{x}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{4}{x}^{\mathrm{4}} \:{dx}\:\:+\:\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left({x}^{\mathrm{3}} \:−\mathrm{4}{x}^{\mathrm{5}} \right){dx} \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}\left[{x}^{\mathrm{5}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\:\mathrm{4}\:\left[\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\:−\frac{\mathrm{2}}{\mathrm{3}}\:{x}^{\mathrm{6}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{4}}{\mathrm{5}}\:+\mathrm{4}\left\{\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{2}}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}\:+\mathrm{1}\:−\frac{\mathrm{8}}{\mathrm{3}}\:=\:\frac{\mathrm{9}}{\mathrm{5}}\:−\frac{\mathrm{8}}{\mathrm{3}}\:=\frac{\mathrm{27}−\mathrm{40}}{\mathrm{15}}\:=−\frac{\mathrm{13}}{\mathrm{15}}\:. \\ $$