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Ex 5
Théorème de Green ∬ S ( δ P δ y − δ G δ x ) d x d y = ∮ C P ( x , y ) d x + G ( x , y ) d y ⏟ forme diff e ˊ rentielle δ P δ y − δ G δ y = 1 ⇒ ∬ D d x d y ⏟ Aide(D) = 1 2 ∫ − y d x + x d y = 1 2 ∫ x d y − y d x \iint_S \pembrace{\frac{\delta P}{\delta y}-\frac{\delta G}{\delta x}}dxdy=\oint_C \underbrace{P(x,y)dx+G(x,y)dy}_{\text{forme différentielle}}\n\n
\begin{split}
\frac{\delta P}{\delta y} - \frac{\delta G}{\delta y}=1 \Rightarrow \underbrace{\iint_D dxdy}_{\text{Aide(D)}}&=\frac12\int-y\ dx+x\ dy\n
&=\frac12\int x\ dy-y\ dx
\end{split}
∬ S ( δy δ P − δ x δ G ) d x d y = ∮ C forme diff e ˊ rentielle P ( x , y ) d x + G ( x , y ) d y δy δ P − δy δ G = 1 ⇒ Aide(D) ∬ D d x d y = 2 1 ∫ − y d x + x d y = 2 1 ∫ x d y − y d x
Exercice d'application
<schema 4> ( x a ) 2 + ( y b ) 2 = 1 A = ∬ D d x d y = ∬ ∣ d e t J ∣ d r d θ { x = a r c o s θ y = b r s i n θ ∣ δ x δ r δ x δ θ δ y δ r δ y δ θ ∣ = ∣ a c o s θ − a r s i n θ b s i n θ − b r c o s θ ∣ d e t ∣ J c ∣ = a b r A = ∫ 0 1 ∫ 0 2 π a b r d θ = 2 r a b ∫ 0 1 r d r = 2 π a b [ r 2 2 ] 0 1 = π a b Autre m e ˊ thode : A = 1 2 ∫ C x d y − y d x { x = a c o s θ y = b s i n θ ⇒ { d x = − a s i n θ d θ d y = b c o s θ d θ A = 1 2 ∫ 0 2 π a c o s θ ∗ b c o s θ d θ − b s i n θ ( − a s i n θ d θ ) = 1 2 ∫ 0 2 π a b c o s 2 θ + a b s i n 2 θ d θ = 1 2 ∫ 0 2 π a b d θ = π a b \text{<schema 4>}\n
\pembrace{\frac{x}{a}}^2+\pembrace{\frac{y}{b}}^2=1\n
A = \iint_D\ dxdy\n=\iint|det\ J|drd\theta\n
\embrace{&x=arcos\theta\\&y=brsin\theta}{\{}{.}\n
\embrace{&\frac{\delta x}{\delta r}\frac{ \delta x}{\delta \theta}\\&\frac{\delta y}{\delta r}\frac{\delta y}{\delta \theta}}{|}{|}=\embrace{&acos\theta\ \ -arsin\theta\\&bsin\theta\ \ \phantom{-}brcos\theta}{|}{|}\n
det|J_c|=abr\n\n\n
A = \int_0^1\int_0^{2\pi}abr\ d\theta\n
= 2rab\int_0^1rdr\
=2\pi ab\cembrace{\frac{r^2}{2}}_0^1\n
=\pi ab\n\n\n
\text{Autre méthode :}\n
A = \frac12 \int_C x\ dy-y\ dx\n
\embrace{&x=acos\theta\\&y=bsin\theta}{\{}{.}\Rightarrow \embrace{&dx=-asin\theta d\theta\\&dy=bcos\theta d\theta}{\{}{.}\n
A = \frac12 \int_0^{2\pi}acos\theta*bcos\theta d\theta - bsin\theta\pembrace{-asin\theta d\theta}\n
=\frac12\int_0^{2\pi}abcos^2\theta+ab\ sin^2\theta\ d\theta\n
=\frac12 \int_0^{2\pi}ab\ d\theta = \pi ab
<schema 4> ( a x ) 2 + ( b y ) 2 = 1 A = ∬ D d x d y = ∬ ∣ d e t J ∣ d r d θ { x = a rcos θ y = b rs in θ δr δ x δ θ δ x δr δy δ θ δy = a cos θ − a rs in θ b s in θ − b rcos θ d e t ∣ J c ∣ = ab r A = ∫ 0 1 ∫ 0 2 π ab r d θ = 2 r ab ∫ 0 1 r d r = 2 πab [ 2 r 2 ] 0 1 = πab Autre m e ˊ thode : A = 2 1 ∫ C x d y − y d x { x = a cos θ y = b s in θ ⇒ { d x = − a s in θ d θ d y = b cos θ d θ A = 2 1 ∫ 0 2 π a cos θ ∗ b cos θ d θ − b s in θ ( − a s in θ d θ ) = 2 1 ∫ 0 2 π ab co s 2 θ + ab s i n 2 θ d θ = 2 1 ∫ 0 2 π ab d θ = πab
b) Soit d le domaine limité par les droites θ = θ1 et θ = θ2 et par la courbe d'équation polaire r = f(θ).
montrer que:
aire(D) = ∬ D d x d y = 1 2 ∫ θ 1 θ 2 [ f ( θ ) ] 2 d θ <schema 5> aire(D) = 1 2 ∫ x d y − y d x { x = f ( θ ) c o s θ y = f θ ) s i n θ ⇒ { d x = − f ′ ( x ) s i n θ d θ d y = − f ′ ( θ ) c o s θ d θ A = 1 2 ∫ Γ 2 + x d y − y d x { x = a c o s θ y = b s i n θ ⇒ { d x = − a s i n θ d θ d y = b c o s θ d θ A = 1 2 ∫ f ( θ ) c o s θ f ′ ( θ ) c o s θ d θ − f ( θ ) s i n θ f ′ ( θ ) s i n θ d θ = 1 2 ∫ f ( θ ) f ′ ( θ ) ( c o s 2 θ − s i n 2 θ ) d θ abandon du prof \text{aire(D)} = \iint_D dxdy = \frac12 \int_{\theta_1}^{\theta_2}\cembrace{f(\theta)}^2d\theta\n
\text{<schema 5>}\n
\text{aire(D)}=\frac12 \int x\ dy-y\ dx\n
\embrace{&x=f(\theta)cos\theta\\&y=f\theta)sin\theta}{\{}{.}\Rightarrow \embrace{&dx=-f'(x)sin\theta d\theta\\&dy=\phantom{-}f'(\theta)cos\theta d\theta}{\{}{.}\n
A=\frac12\int_{\Gamma_2^+}xdy-ydx\n
\embrace{&x=acos\theta\\&y=bsin\theta}{\{}{.}\Rightarrow \embrace{&dx=-asin\theta d\theta\\&dy=bcos\theta d\theta}{\{}{.}\n
A=\frac12\int f(\theta)cos\theta f'(\theta)cos\theta d\theta -f(\theta)sin\theta f'(\theta)sin\theta d\theta\n
=\frac12\int f(\theta)f'(\theta)(cos^2\theta-sin^2\theta)d\theta\n
\text{abandon du prof}
aire(D) = ∬ D d x d y = 2 1 ∫ θ 1 θ 2 [ f ( θ ) ] 2 d θ <schema 5> aire(D) = 2 1 ∫ x d y − y d x { x = f ( θ ) cos θ y = f θ ) s in θ ⇒ { d x = − f ′ ( x ) s in θ d θ d y = − f ′ ( θ ) cos θ d θ A = 2 1 ∫ Γ 2 + x d y − y d x { x = a cos θ y = b s in θ ⇒ { d x = − a s in θ d θ d y = b cos θ d θ A = 2 1 ∫ f ( θ ) cos θ f ′ ( θ ) cos θ d θ − f ( θ ) s in θ f ′ ( θ ) s in θ d θ = 2 1 ∫ f ( θ ) f ′ ( θ ) ( co s 2 θ − s i n 2 θ ) d θ abandon du prof