Question Number 41343 by maxmathsup by imad last updated on 05/Aug/18
![calculate ∫∫_D (x^2 −y^2 )dxdy with D = [−1,1]^2](https://www.tinkutara.com/question/Q41343.png)
$${calculate}\:\:\:\:\int\int_{{D}} \:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}\:\:{with} \\ $$$${D}\:=\:\left[−\mathrm{1},\mathrm{1}\right]^{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 08/Aug/18
![I = ∫_(−1) ^1 ( ∫_(−1) ^1 (x^2 −y^2 )dx)dy but ∫_(−1) ^1 (x^2 −y^2 )dx =2 ∫_0 ^1 (x^2 −y^2 )dx = 2[(x^3 /3) −y^2 x]_0 ^1 =2 { (1/3) −y^2 } =(2/3) −2y^2 ⇒ I = ∫_(−1) ^1 ((2/3)−2y^2 )dy =(4/3) −2 ∫_(−1) ^1 y^2 dy =(4/3) −4 ∫_0 ^1 y^2 dy =(4/3) −4 [(y^3 /3)]_0 ^1 =(4/3) −(4/3) ⇒ I =0](https://www.tinkutara.com/question/Q41499.png)
$${I}\:=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left(\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dx}\right){dy}\:{but} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dx}\:=\:\mathrm{2}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−{y}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \: \\ $$$$=\mathrm{2}\:\left\{\:\frac{\mathrm{1}}{\mathrm{3}}\:−{y}^{\mathrm{2}} \right\}\:=\frac{\mathrm{2}}{\mathrm{3}}\:−\mathrm{2}{y}^{\mathrm{2}} \:\Rightarrow\:{I}\:=\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\left(\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{2}{y}^{\mathrm{2}} \right){dy} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\:−\mathrm{2}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:{y}^{\mathrm{2}} {dy}\:=\frac{\mathrm{4}}{\mathrm{3}}\:−\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}} {dy}\:=\frac{\mathrm{4}}{\mathrm{3}}\:−\mathrm{4}\:\left[\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{4}}{\mathrm{3}}\:−\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\mathrm{0} \\ $$
Answered by alex041103 last updated on 08/Aug/18
![∫∫_D (x^2 −y^2 )dxdy=∫_(−1) ^1 ∫_(−1) ^1 (x^2 −y^2 )dxdy= =∫_(−1) ^1 [(x^3 /3)−y^2 x]_(x=−1) ^(x=1) dy= =∫_(−1) ^1 ((2/3)−2y^2 )dy=[(2/3)y−((2y^3 )/3)]_(y=−1) ^(y=1) = =(4/3)−(4/3)=0 ∫∫_D (x^2 −y^2 )dxdy=0](https://www.tinkutara.com/question/Q41500.png)
$$\int\int_{{D}} \:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}=\int_{−\mathrm{1}} ^{\mathrm{1}} \int_{−\mathrm{1}} ^{\mathrm{1}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}= \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{y}^{\mathrm{2}} {x}\right]_{{x}=−\mathrm{1}} ^{{x}=\mathrm{1}} {dy}= \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{2}{y}^{\mathrm{2}} \right){dy}=\left[\frac{\mathrm{2}}{\mathrm{3}}{y}−\frac{\mathrm{2}{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{{y}=−\mathrm{1}} ^{{y}=\mathrm{1}} = \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{0} \\ $$$$\int\int_{{D}} \:\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}=\mathrm{0} \\ $$