Question Number 179181 by neinhaltsieger369 last updated on 26/Oct/22
$$\:\mathrm{Help}-\mathrm{me}! \\ $$$$\: \\ $$$$\:\mathrm{Use}\:\mathrm{double}\:\mathrm{integral}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{following}\:\mathrm{curves}\: \\ $$$$\:\mathrm{given}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{shown}\:\mathrm{below}: \\ $$$$\: \\ $$$$\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{4x}\:\mathrm{and}\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:=\:\mathrm{4y} \\ $$
Answered by Acem last updated on 26/Oct/22
![Let C_1 : the curve x= (1/4) y^2 , C_2 : y= (1/4) x^2 Common points: By compensation x in C_2 y= (1/4) (1/(16)) y^4 ⇒ y(y^3 −64)= 0 ⇒ y= 0 or y= 4 p_1 (0, 0) , p_2 (4,4) ; x: 0 →4 Within the region bounded C_1 is above the C_2 because and for example for x=1∈ [0, 4] y_C_1 = 2 > (1/4)= y_C_2 It prefer to graph them S= ∫_0 ^( 4) (2x^(1/2) − (1/4) x^2 )dx= (4/3) (√x^3 )− (1/(12)) x^3 ∣_0 ^4 S= ((32)/3) −((16)/3) − 0 ⇒ S= ((16)/3) unit^2](https://www.tinkutara.com/question/Q179183.png)
$$\:{Let}\:{C}_{\mathrm{1}} :\:{the}\:{curve}\:{x}=\:\frac{\mathrm{1}}{\mathrm{4}}\:{y}^{\mathrm{2}} \:,\:{C}_{\mathrm{2}} :\:{y}=\:\frac{\mathrm{1}}{\mathrm{4}}\:{x}^{\mathrm{2}} \\ $$$$\:{Common}\:{points}: \\ $$$$\:{By}\:{compensation}\:{x}\:{in}\:{C}_{\mathrm{2}} \\ $$$$\:{y}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\mathrm{1}}{\mathrm{16}}\:{y}^{\mathrm{4}} \:\Rightarrow\:{y}\left({y}^{\mathrm{3}} −\mathrm{64}\right)=\:\mathrm{0}\:\Rightarrow\:{y}=\:\mathrm{0}\:{or}\:{y}=\:\mathrm{4} \\ $$$$\:{p}_{\mathrm{1}} \left(\mathrm{0},\:\mathrm{0}\right)\:,\:{p}_{\mathrm{2}} \left(\mathrm{4},\mathrm{4}\right)\:;\:{x}:\:\mathrm{0}\:\rightarrow\mathrm{4} \\ $$$$\:{Within}\:{the}\:{region}\:{bounded}\:{C}_{\mathrm{1}} \:{is}\:{above}\:{the}\:{C}_{\mathrm{2}} \\ $$$$\:{because}\:{and}\:{for}\:{example}\:{for}\:{x}=\mathrm{1}\in\:\left[\mathrm{0},\:\mathrm{4}\right] \\ $$$$\:{y}_{{C}_{\mathrm{1}} } =\:\mathrm{2}\:>\:\frac{\mathrm{1}}{\mathrm{4}}=\:{y}_{{C}_{\mathrm{2}} } \:\boldsymbol{{It}}\:\boldsymbol{{prefer}}\:\boldsymbol{{to}}\:\boldsymbol{{graph}}\:\boldsymbol{{them}} \\ $$$$\:{S}=\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \:\left(\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\:\frac{\mathrm{1}}{\mathrm{4}}\:{x}^{\mathrm{2}} \right){dx}=\:\frac{\mathrm{4}}{\mathrm{3}}\:\sqrt{{x}^{\mathrm{3}} }−\:\frac{\mathrm{1}}{\mathrm{12}}\:{x}^{\mathrm{3}} \:\mid_{\mathrm{0}} ^{\mathrm{4}} \\ $$$$\:{S}=\:\frac{\mathrm{32}}{\mathrm{3}}\:−\frac{\mathrm{16}}{\mathrm{3}}\:−\:\mathrm{0}\:\Rightarrow\:{S}=\:\frac{\mathrm{16}}{\mathrm{3}}\:{unit}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by neinhaltsieger369 last updated on 26/Oct/22
$$\:\mathrm{Thank}\:\:\mathrm{you}! \\ $$
Commented by Acem last updated on 26/Oct/22
$${Check}\:{manxsol}'{solution} \\ $$
Commented by Acem last updated on 26/Oct/22
$${You}'{re}\:{very}\:{much}\:{welcome}! \\ $$
Commented by manxsol last updated on 27/Oct/22
$${Gracias}\:{Sr}.{Acem}.{un}\:{foro}?{muy}\:{interesante}\:{y}\:{de}\:{gran}\:{nivel} \\ $$
Answered by manxsol last updated on 26/Oct/22
Commented by Acem last updated on 26/Oct/22
$${Very}\:{well},\:{your}\:{solution}\:{is}\:{exactly}\:{what}\:{is}\:{needed} \\ $$$$\:{Thank}\:{you} \\ $$
Commented by neinhaltsieger369 last updated on 26/Oct/22
$$\:\mathrm{Thank}\:\:\mathrm{you}! \\ $$
Answered by CElcedricjunior last updated on 26/Oct/22
![On demande de calculer l′aire du caves delimite^ pas { ((y^2 =4x)),((x^2 =4y)) :}=> { ((x=(y^2 /4))),((x=±2(√y))) :} x=x =>(y^2 /4)=±2(√y) pour (y^2 /4)=−2(√y) impossible (y^2 /4)=2(√y)=>y^4 =64y =>y(y^3 −64)=0=>y(y^3 −4^3 )=0 =>y(y−4)(y^2 +4y+16)=0 =>y=0 ou y=4 or y^2 +4y+16>0 n′adnet pas de solution dans R Donc A=∫_0 ^4 ∫_(y^2 /4) ^(2(√y)) dxdy A=∫_0 ^4 (2(√y)−(y^2 /4))dy A=[(4/3)(√y^3 )−(y^3 /(12))]_0 ^4 =((32)/3)−((64)/(12))=((32)/3)−((16)/3) A=((16)/3) um ....................le celebre cedric junior.........](https://www.tinkutara.com/question/Q179210.png)
$$\boldsymbol{\mathrm{On}}\:\boldsymbol{\mathrm{demande}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{calculer}}\:\boldsymbol{\mathrm{l}}'\boldsymbol{\mathrm{aire}}\:\boldsymbol{\mathrm{du}}\:\boldsymbol{\mathrm{caves}} \\ $$$$\boldsymbol{\mathrm{delimit}}\acute {\boldsymbol{\mathrm{e}}}\:\boldsymbol{\mathrm{pas}}\: \\ $$$$\begin{cases}{\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{4}\boldsymbol{\mathrm{x}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\mathrm{4}\boldsymbol{\mathrm{y}}}\end{cases}=>\begin{cases}{\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{4}}}\\{\boldsymbol{\mathrm{x}}=\pm\mathrm{2}\sqrt{\boldsymbol{{y}}}}\end{cases} \\ $$$$\boldsymbol{{x}}=\boldsymbol{{x}} \\ $$$$=>\frac{\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{4}}=\pm\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}} \\ $$$$\boldsymbol{\mathrm{pour}}\:\frac{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{4}}=−\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}}\:\boldsymbol{\mathrm{impossible}} \\ $$$$\frac{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}}=>\boldsymbol{\mathrm{y}}^{\mathrm{4}} =\mathrm{64}\boldsymbol{\mathrm{y}} \\ $$$$=>\boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{3}} −\mathrm{64}\right)=\mathrm{0}=>\boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$=>\boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{y}}−\mathrm{4}\right)\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{y}}+\mathrm{16}\right)=\mathrm{0} \\ $$$$=>\boldsymbol{\mathrm{y}}=\mathrm{0}\:\boldsymbol{\mathrm{ou}}\:\boldsymbol{\mathrm{y}}=\mathrm{4}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{y}}+\mathrm{16}>\mathrm{0}\:\boldsymbol{\mathrm{n}}'\boldsymbol{\mathrm{adnet}}\:\boldsymbol{\mathrm{pas}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{dans}}\:\mathbb{R} \\ $$$$\boldsymbol{\mathrm{Donc}} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{4}} \int_{\frac{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{4}}} ^{\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}}} \boldsymbol{\mathrm{dxdy}} \\ $$$$\boldsymbol{\mathrm{A}}=\int_{\mathrm{0}} ^{\mathrm{4}} \left(\mathrm{2}\sqrt{\boldsymbol{\mathrm{y}}}−\frac{\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{4}}\right)\boldsymbol{\mathrm{dy}} \\ $$$$\boldsymbol{\mathrm{A}}=\left[\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\boldsymbol{\mathrm{y}}^{\mathrm{3}} }−\frac{\boldsymbol{\mathrm{y}}^{\mathrm{3}} }{\mathrm{12}}\right]_{\mathrm{0}} ^{\mathrm{4}} =\frac{\mathrm{32}}{\mathrm{3}}−\frac{\mathrm{64}}{\mathrm{12}}=\frac{\mathrm{32}}{\mathrm{3}}−\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{A}}=\frac{\mathrm{16}}{\mathrm{3}}\:\:\:\boldsymbol{\mathrm{um}} \\ $$$$\:\: \\ $$$$\:………………..{le}\:{celebre}\:{cedric}\:{junior}……… \\ $$