Help-me-Use-double-integral-to-find-the-area-of-the-region-bounded-by-the-following-curves-given-in-the-plane-shown-below-y-2-4x-and-x-2-4y-

Question Number 179181 by neinhaltsieger369 last updated on 26/Oct/22

Help - me!

Use double integral to find the area of the

region bounded by the following curves

given in the plane shown below :

y^{2} = 4 x and x^{2} = 4 y

Answered by Acem last updated on 26/Oct/22

L e t C_{1} : t h e c u r v e x = \frac{1}{4} y^{2}, C_{2} : y = \frac{1}{4} x^{2}

C o m m o n p o i n t s :

B y c o m p e n s a t i o n x i n C_{2}

y = \frac{1}{4} \frac{1}{16} y^{4} \Rightarrow y (y^{3} - 64) = 0 \Rightarrow y = 0 o r y = 4

p_{1} (0, 0), p_{2} (4, 4); x : 0 \to 4

W i t h i n t h e r e g i o n b o u n d e d C_{1} i s a b o v e t h e C_{2}

b e c a u s e a n d f o r e x a m p l e f o r x = 1 \in [0, 4]

y_{C_{1}} = 2 > \frac{1}{4} = y_{C_{2}} I t p r e f e r t o g r a p h t h e m

S = \int_{0}^{4} (2 x^{\frac{1}{2}} - \frac{1}{4} x^{2}) d x = \frac{4}{3} \sqrt{x^{3}} - \frac{1}{12} x^{3} ∣_{0}^{4}

S = \frac{32}{3} - \frac{16}{3} - 0 \Rightarrow S = \frac{16}{3} {u n i t}^{2}

Commented by neinhaltsieger369 last updated on 26/Oct/22

Thank you!

Commented by Acem last updated on 26/Oct/22

C h e c k {m a n x s o l}^{'} s o l u t i o n

Commented by Acem last updated on 26/Oct/22

{Y o u}^{'} r e v e r y m u c h w e l c o m e!

Commented by manxsol last updated on 27/Oct/22

G r a c i a s S r . A c e m . u n f o r o ? m u y i n t e r e s a n t e y d e g r a n n i v e l

Answered by manxsol last updated on 26/Oct/22

Commented by Acem last updated on 26/Oct/22

V e r y w e l l, y o u r s o l u t i o n i s e x a c t l y w h a t i s n e e d e d

T h a n k y o u

Commented by neinhaltsieger369 last updated on 26/Oct/22

Thank you!

Answered by CElcedricjunior last updated on 26/Oct/22

On demande de calculer l^{'} aire du caves

delimit \overset{´}{e} pas

{\begin{cases} y^{2} = 4 x \\ x^{2} = 4 y \end{cases} => {\begin{cases} x = \frac{y^{2}}{4} \\ x = \pm 2 \sqrt{y} \end{cases}

x = x

=> \frac{y^{2}}{4} = \pm 2 \sqrt{y}

pour \frac{y^{2}}{4} = - 2 \sqrt{y} impossible

\frac{y^{2}}{4} = 2 \sqrt{y} => y^{4} = 64 y

=> y (y^{3} - 64) = 0 => y (y^{3} - 4^{3}) = 0

=> y (y - 4) (y^{2} + 4 y + 16) = 0

=> y = 0 ou y = 4 or y^{2} + 4 y + 16 > 0 n^{'} adnet pas de solution dans R

Donc

A = \int_{0}^{4} \int_{\frac{y^{2}}{4}}^{2 \sqrt{y}} dxdy

A = \int_{0}^{4} (2 \sqrt{y} - \frac{y^{2}}{4}) dy

A = {[\frac{4}{3} \sqrt{y^{3}} - \frac{y^{3}}{12}]}_{0}^{4} = \frac{32}{3} - \frac{64}{12} = \frac{32}{3} - \frac{16}{3}

A = \frac{16}{3} um

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