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Question Number 78283 by msup trace by abdo last updated on 15/Jan/20
calculate ∫∫_D  (x^2 +2y)dxdy  with D={(x,y)∈R^2 / x^2 ≥y and y≥x^2 }
$${calculate}\:\int\int_{{D}} \:\left({x}^{\mathrm{2}} +\mathrm{2}{y}\right){dxdy} \\ $$$${with}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:{x}^{\mathrm{2}} \geqslant{y}\:{and}\:{y}\geqslant{x}^{\mathrm{2}} \right\} \\ $$
Commented by mr W last updated on 15/Jan/20
how can it be y≤x^2  and y≥x^2  ?
$${how}\:{can}\:{it}\:{be}\:{y}\leqslant{x}^{\mathrm{2}} \:{and}\:{y}\geqslant{x}^{\mathrm{2}} \:? \\ $$
Commented by msup trace by abdo last updated on 15/Jan/20
error of typo  D={(x,y)∈R^2 /x^2 ≥y  and x≥y^2 }
$${error}\:{of}\:{typo} \\ $$$${D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /{x}^{\mathrm{2}} \geqslant{y}\:\:{and}\:{x}\geqslant{y}^{\mathrm{2}} \right\} \\ $$
Commented by mr W last updated on 15/Jan/20
but i think it makes sense only when  x^2 ≤y and x≥y^2
$${but}\:{i}\:{think}\:{it}\:{makes}\:{sense}\:{only}\:{when} \\ $$$${x}^{\mathrm{2}} \leqslant{y}\:{and}\:{x}\geqslant{y}^{\mathrm{2}} \\ $$

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