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Question Number 98020 by bobhans last updated on 11/Jun/20
What is the area of the region bounded  by x^2 +y^2  ≤ 9 ; x+y ≤ 3 and y ≤ x
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded} \\ $$$$\mathrm{by}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\leqslant\:\mathrm{9}\:;\:\mathrm{x}+\mathrm{y}\:\leqslant\:\mathrm{3}\:\mathrm{and}\:\mathrm{y}\:\leqslant\:\mathrm{x}\: \\ $$
Answered by bemath last updated on 11/Jun/20
Commented by bobhans last updated on 11/Jun/20
very..==afdolll
$$\mathrm{very}..==\mathrm{afdolll} \\ $$
Commented by Rio Michael last updated on 11/Jun/20
yeah! i got it now. you are correct   A = ((27π + 18)/8) square units
$$\mathrm{yeah}!\:\mathrm{i}\:\mathrm{got}\:\mathrm{it}\:\mathrm{now}.\:\mathrm{you}\:\mathrm{are}\:\mathrm{correct} \\ $$$$\:{A}\:=\:\frac{\mathrm{27}\pi\:+\:\mathrm{18}}{\mathrm{8}}\:\mathrm{square}\:\mathrm{units} \\ $$
Commented by bobhans last updated on 11/Jun/20

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