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Question-191652




Question Number 191652 by 073 last updated on 28/Apr/23
Commented by 073 last updated on 28/Apr/23
solution????
$$\mathrm{solution}???? \\ $$
Answered by aleks041103 last updated on 28/Apr/23
R is a rectangle⇒∫∫_(R) ≡∫_0 ^2 dx∫_1 ^3 dy  ⇒I=∫_1 ^( 3) ∫_0 ^( 2) (2−x^2 −y^2 )dxdy=  =∫_1 ^( 3) (2x−(x^3 /3)−y^2 x)_0 ^2 dy=  =∫_1 ^( 3) (4−(8/3)−2y^2 )dy=  =∫_1 ^( 3) ((4/3)−2y^2 )dy=((4/3)y−(2/3)y^3 )_1 ^3 =  =(8/3)−(2/3) (27−1)=  =(2/3)(4−26)=−((44)/3)
$${R}\:{is}\:{a}\:{rectangle}\Rightarrow\underset{{R}} {\int\int}\equiv\int_{\mathrm{0}} ^{\mathrm{2}} {dx}\int_{\mathrm{1}} ^{\mathrm{3}} {dy} \\ $$$$\Rightarrow{I}=\int_{\mathrm{1}} ^{\:\mathrm{3}} \int_{\mathrm{0}} ^{\:\mathrm{2}} \left(\mathrm{2}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}= \\ $$$$=\int_{\mathrm{1}} ^{\:\mathrm{3}} \left(\mathrm{2}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{y}^{\mathrm{2}} {x}\right)_{\mathrm{0}} ^{\mathrm{2}} {dy}= \\ $$$$=\int_{\mathrm{1}} ^{\:\mathrm{3}} \left(\mathrm{4}−\frac{\mathrm{8}}{\mathrm{3}}−\mathrm{2}{y}^{\mathrm{2}} \right){dy}= \\ $$$$=\int_{\mathrm{1}} ^{\:\mathrm{3}} \left(\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{2}{y}^{\mathrm{2}} \right){dy}=\left(\frac{\mathrm{4}}{\mathrm{3}}{y}−\frac{\mathrm{2}}{\mathrm{3}}{y}^{\mathrm{3}} \right)_{\mathrm{1}} ^{\mathrm{3}} = \\ $$$$=\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}\:\left(\mathrm{27}−\mathrm{1}\right)= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{4}−\mathrm{26}\right)=−\frac{\mathrm{44}}{\mathrm{3}} \\ $$

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