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Question Number 191652 by 073 last updated on 28/Apr/23

Commented by 073 last updated on 28/Apr/23

$$\mathrm{solution}????$$

Answered by aleks041103 last updated on 28/Apr/23

$$Risarectangle\Rightarrow \underset{R}{\int \int }\equiv {\int }_0^{2}dx{\int }_1^{3}dy$$$$\Rightarrow I={\int }_1^3{\int }_0^2(2-x^2-y^2)dxdy=$$$$={\int }_1^3{(2x-\frac{x^3}{3}-y^{2}x)}_0^{2}dy=$$$$={\int }_1^3(4-\frac{8}{3}-2y^2)dy=$$$$={\int }_1^3(\frac{4}{3}-2y^2)dy={(\frac{4}{3}y-\frac{2}{3}{y}^3)}_1^3=$$$$=\frac{8}{3}-\frac{2}{3}(27-1)=$$$$=\frac{2}{3}(4-26)=-\frac{44}{3}$$

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