Question Number 46103 by rahul 19 last updated on 21/Oct/18
$${Find}\:{the}\:{area}\:{enclosed}\:{between}\:{curves} \\ $$$${y}^{\mathrm{2}} \left(\mathrm{2}{a}−{x}\right)={x}^{\mathrm{3}} \:{and}\:{line}\:{x}=\mathrm{2}\:{above}\:{the} \\ $$$${x}−{axis}\:? \\ $$$${Graphing}\:{calculators}\:{are}\:{not}\:{allowed}.. \\ $$
Commented by rahul 19 last updated on 21/Oct/18
$${I}'{m}\:{stuck}\:{in}\:{drawing}\:{graph}\:{of}\:{above} \\ $$$${curve}….{pls}\:{tell}\:{steps}\:{to}\:{draw}\:{it}. \\ $$
Commented by ajfour last updated on 21/Oct/18
$${any}\:{info}\:{about}\:{a}\:? \\ $$
Commented by rahul 19 last updated on 21/Oct/18
$${Sir},\:{here}\:{it}'{s}\:\:{a}\:+{ve}\:{constant}. \\ $$
Answered by ajfour last updated on 21/Oct/18
Commented by ajfour last updated on 21/Oct/18
$${y}^{\mathrm{2}} \left(\mathrm{2}{a}−{x}\right)={x}^{\mathrm{3}} \\ $$$${y}^{\mathrm{2}} \:=\:\frac{{x}^{\mathrm{3}} }{\mathrm{2}{a}−{x}}\:\:>\:\mathrm{0} \\ $$$${if}\:{x}\:>\mathrm{0}\:\:\Rightarrow\:\:\mathrm{2}{a}−{x}\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{x}\:<\:\mathrm{2}{a} \\ $$$${if}\:{x}\:<\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{2}{a}−{x}\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{x}\:>\:\mathrm{2}{a}\:\:\left({not}\:{possible}\right) \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} \left(\mathrm{2}{a}−{x}\right)+{x}^{\mathrm{3}} }{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}{y}\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{3}{a}−{x}\right)}{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\:\:{required}\:{area}=\:\int_{\mathrm{0}} ^{\:\:\mathrm{2}} \frac{{x}\sqrt{{x}}}{\:\sqrt{\mathrm{2}{a}−{x}}}{dx} \\ $$$$\:\:\:\:{let}\:\:\:{x}=\mathrm{2}{a}\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${for}\:{x}=\mathrm{2},\:\:\:\theta_{\mathrm{0}} \:=\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{a}}} \\ $$$$\:\:{Area}\:=\:\int_{\mathrm{0}} ^{\:\:\theta_{\mathrm{0}} } \:\frac{\mathrm{2}{a}\sqrt{\mathrm{2}{a}}\:\mathrm{sin}\:^{\mathrm{3}} \theta\left(\mathrm{4}{a}\mathrm{sin}\:\theta\mathrm{cos}\:\theta\right){d}\theta}{\:\sqrt{\mathrm{2}{a}}\:\mathrm{cos}\:\theta} \\ $$$$\:\:=\:\mathrm{2}{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\:\:\theta_{\mathrm{0}} } \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right)^{\mathrm{2}} {d}\theta\:\:. \\ $$
Commented by rahul 19 last updated on 21/Oct/18
Thank you sir ! ����