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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

$$A_{shaded}=\frac{1}{3}(\frac{3^2}{1}-\frac{3^2}{2}-\frac{2^2}{1}+\frac{2^2}{2})=\frac{5}{6}$$

Answered by mr W last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

$${let}^{\prime }slookatthehatchedareabetween$$$$curvesy={ax}^{2}andy=b\sqrt{x}.$$$$bothcurvesarequadraticparabolas.$$$$y_P={ax}_P^2=b\sqrt{x_P}$$$$\Rightarrow x_P=a^{-\frac{2}{3}}{b}^{\frac{2}{3}}$$$$\Rightarrow y_P=a^{-\frac{1}{3}}{b}^{\frac{4}{3}}$$$$A_1=A_2=\frac{x_P{y}_P}{3}(\to seeQ88758)$$$$A_{hatch}=x_P{y}_P-A_1-A_2=x_P{y}_P-2\times \frac{x_P{y}_P}{3}$$$$=\frac{x_P{y}_P}{3}=\frac{b^2}{3a}$$

Commented by mr W last updated on 24/Oct/23

Commented by mr W last updated on 24/Oct/23

$$if{A}_{2\wedge 3}denotestheareabetween$$$$parabola2andparabola3,then$$$$wehave$$$$A_{shaded}=A_{2\wedge 3}-A_{1\wedge 3}-A_{2\wedge 4}+A_{1\wedge 4}$$$$\Rightarrow A_{shaded}=\frac{1}{3}(\frac{3^2}{1}-\frac{3^2}{2}-\frac{2^2}{1}+\frac{2^2}{2})=\frac{5}{6}$$

Commented by cortano12 last updated on 24/Oct/23

$$\mathrm{nice}$$

Answered by cortano12 last updated on 24/Oct/23

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