Question Number 198731 by cortano12 last updated on 23/Oct/23
Commented by cortano12 last updated on 24/Oct/23
$$\mathrm{what}\:\mathrm{this}\:\mathrm{formula}\:? \\ $$
Commented by mr W last updated on 24/Oct/23
$${A}_{{shaded}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}}−\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{5}}{\mathrm{6}} \\ $$
Answered by mr W last updated on 24/Oct/23
Commented by mr W last updated on 24/Oct/23
$${let}'{s}\:{look}\:{at}\:{the}\:{hatched}\:{area}\:{between} \\ $$$${curves}\:{y}={ax}^{\mathrm{2}} \:{and}\:{y}={b}\sqrt{{x}}. \\ $$$${both}\:{curves}\:{are}\:{quadratic}\:{parabolas}. \\ $$$${y}_{{P}} ={ax}_{{P}} ^{\mathrm{2}} ={b}\sqrt{{x}_{{P}} } \\ $$$$\Rightarrow{x}_{{P}} ={a}^{−\frac{\mathrm{2}}{\mathrm{3}}} {b}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\Rightarrow{y}_{{P}} ={a}^{−\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$${A}_{\mathrm{1}} ={A}_{\mathrm{2}} =\frac{{x}_{{P}} {y}_{{P}} }{\mathrm{3}}\:\:\:\:\:\:\:\:\left(\rightarrow\:{see}\:{Q}\mathrm{88758}\right) \\ $$$${A}_{{hatch}} ={x}_{{P}} {y}_{{P}} −{A}_{\mathrm{1}} −{A}_{\mathrm{2}} ={x}_{{P}} {y}_{{P}} −\mathrm{2}×\frac{{x}_{{P}} {y}_{{P}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}_{{P}} {y}_{{P}} }{\mathrm{3}}=\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}} \\ $$
Commented by mr W last updated on 24/Oct/23
Commented by mr W last updated on 24/Oct/23
$${if}\:{A}_{\mathrm{2}\wedge\mathrm{3}} \:{denotes}\:{the}\:{area}\:{between} \\ $$$${parabola}\:\mathrm{2}\:{and}\:{parabola}\:\mathrm{3},\:{then} \\ $$$${we}\:{have} \\ $$$${A}_{{shaded}} ={A}_{\mathrm{2}\wedge\mathrm{3}} −{A}_{\mathrm{1}\wedge\mathrm{3}} −{A}_{\mathrm{2}\wedge\mathrm{4}} +{A}_{\mathrm{1}\wedge\mathrm{4}} \\ $$$$\Rightarrow{A}_{{shaded}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{1}}−\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{5}}{\mathrm{6}} \\ $$
Commented by cortano12 last updated on 24/Oct/23
$$\mathrm{nice}\: \\ $$
Answered by cortano12 last updated on 24/Oct/23
