$$$$

Thm de Green-Riemann

$$\begin{split} &\int_{\delta D}Pdx+Qdy=\iint_D\pembrace{\frac{\delta Q}{\delta x}-\frac{\delta P}{\delta y}}dxdy\n &\text{on veut:}\\ &\embrace{ &\frac{\delta Q}{\delta x}-\frac{\delta P}{\delta y}\n &\text{On pose:}\n &P(x,y)=\frac{-y}{\phantom{-}2}\quad\frac{\delta P}{\delta y} = \frac{-1}{2}\n &Q(x,y)=\phantom{-}\frac{x}{2}\quad\frac{\delta Q}{\delta x}=\frac{1}{2} }{|}{.} \end{split}$$

$$\begin{split} &\text{Aide de D = }\iint_D dxdy = \int_{\delta D^+}=\frac{-y}{2}dx+\frac{x}{2}dy\n &\boxed{ \iint_D dxdy=\frac12\int_{\delta D^+}xdy-ydx\n }\n \end{split}$$

Ex5 b) : Calculer l'aide de D

$$\begin{split} &\text{<schéma cornet>}\n &\text{Aide(D)}=\frac12\int_{\theta_1}^{\theta_2}f^2(\theta)d\theta\n &\Gamma_1^+ \ \ \embrace{ &x=rcos\theta_1\\ &y=rsin\theta_1\\ &r\in\cembrace{0,f(\theta_1)} }{\{}{.}\quad \Gamma_2^+ \ \ \embrace{ &x=f(\theta)cos\theta\\ &y=f(\theta)sin\theta\\ &r\in\cembrace{\theta_1,\theta_2} }{\{}{.}\quad \Gamma_3^+ \ \ \embrace{ &x=rcos\theta_2\\ &y=rsin\theta_2\\ &r\in\cembrace{f(\theta_2),0} }{\{}{.}\n &\iint_D dxdy = \int_{\Gamma_1^+}+\int_{\Gamma_2^+}+\int_{\Gamma_3^+} = I_1+I_2+I_3\n &I_1=I_3=0\n &I_2=\frac12 \int_{\theta_1}^{\theta_2}f(\theta)cos\theta\cembrace{f'(\theta)sin\theta+f(\theta)cos\theta}d\theta\\ &\quad\quad-f(\theta)sin\theta\cembrace{f'(\theta)cos\theta+f(\theta)(-sin\theta)}d\theta\n &=\frac12\int_{\theta_1}^{\theta_2}f^2(\theta)(cos^2\theta+sin^2\theta)d\theta\n &\text{Aide(D)}=\frac12\int_{\theta_1}^{\theta_2}f^2(\theta)d\theta \end{split}$$

Ex6

$$\begin{split} &\omega=\underbrace{2\pembrace{x+y^2}}_{P(x,y)}dx+\underbrace{\pembrace{4xy+3y^2}}_{Q(x,y)}dy\n &\embrace{&\frac{\delta P}{\delta y}=4y\\&\frac{\delta Q}{\delta x}=4y}{\{}{.}\quad\Rightarrow\text{la forme différentielle est fermée}\n &\text{<schema triangle>}\n &\int_{\gamma^+}=\int_{AB}\omega+\int_{BC}\omega+\int_{CA}\omega\n &AB:\\ &\embrace{ &\embrace{&x\in\cembrace{0,1}\\&y=0}{\{}{.}\quad \embrace{&x=t\\&y=0}{\{}{.}\quad \embrace{&dx=dt\\&dy=0}{\{}{.}\n &\int_0^12tdt= 2\cembrace{\frac{t^2}{2}}_{0}^{1}=2*\frac12=1 }{|}{.}\n &BC:\\ &\embrace{ &\embrace{&x=1-t\\&y=t}{\{}{.}\quad \embrace{&dx=-dt\\&dy=dt}{\{}{.}\\ &t\in\cembrace{0,1}\n &\int_0^12\pembrace{(1-t)+t^2}(-dt)+(4(1-t)t+3t^2)dt\\ &\int_0^1(-2(1-t+t^2)+4t-4T^2+3t^2)dt\\ &\int_0^1(-2+2t-2t^2+4t-t^2)dt\\ &\int_0^1-3t^2+6t-2=-3[\frac{t^3}{3}]_0^1+6[\frac{t^2}{2}]_0^1-2[t]_0^1\\ &\quad\quad=-1+6/2-2=0 }{|}{.}\n &CA:\\ &\embrace{ &\embrace{&y=t\\&x=0}{\{}{.}\quad\embrace{dy=dt\\ \ }{\{}{.}\\ &\ t\in\cembrace{0,1} }{|}{.}\n &I=\int_0^1 3t^2dt\\ &\phantom{I} = 3\cembrace{\frac{t^3}{3}}_0^1=0-(1)^3=-1 \end{split}$$

Ex7 a)

$$\begin{split} &I=\iint_D xydxdy\\ &D=\aembrace{(x,y)\in\R^2\ |\ x\ge0,\ y\ge0,\ x+y\le1}\n &I=\int_0^1\int_0^1xydydx\n &I=\int_0^1x\int_0^{1-x}ydydx\n &I=\int_0^1x\cembrace{\frac{y^2}2}_0^{1-x}dx\n &I=\int_0^1x\pembrace{\frac{(1-x)^2}{2}}dx=\frac12\int_0^1x(1+x^2-2x)dx\n &I=\frac12\int_0^1(x+x^3-2x^2)dx\n &I=\frac12 \cembrace{\frac{x^2}{2}+\frac{x^4}{4}-2\frac{x^2}{3}}_{0}^{1}\n &I=\frac12\cembrace{\frac12+\frac14-\frac23}\n &I=\frac12\cembrace{\frac{9-8}{12}}=\frac{1}{24}\n \end{split}$$

Ex8 b)

$$\begin{split} &J = \int_{\Gamma^+}\frac{y^3}{P(x,y)}dx+\frac{x^3}{Q(x,y)}dy\n &\embrace{&\frac{\delta}{\delta y}P(x,y)=3y^2\\&\frac{\delta}{\delta x}Q(x,y)=3x^2}{\{}{.}\quad\text{la forme différentielle n'est ¯\\ \_(ツ)\_/¯}\n &\embrace{&x=acos\theta\\&y=bsin\theta}{\{}{.}\quad\quad \embrace{&dx=-asin\theta d\theta\\&dy=bcos\theta d\theta}{\{}{.}\n &J=\int_0^{2\pi}(-b^3sin^3\theta asin\theta+a^3cos^3\theta bcos\theta)d\theta\n &J=\int_0^{2\pi}\pembrace{a^3bcos^4\theta-ab^3sin^4\theta}d\theta\n &\text{abandon du prof} \end{split}$$

de retour sur l'elipse

$$\begin{split} &\iint_D(2x^3-y)dxdy\\ &D=\aembrace{(x,y)\in\R^2\ |\ x\ge0,\ y\ge0,\ \frac{x^2}{a^2}+\frac{y^2}{b^2}\le1}\n &\text{changement de var}\\ &\quad \embrace{&x=arcos\theta\\&y=brsin\theta}{\{}{.}\n &J_{\phi}= \embrace{ &\frac{\delta x}{\delta r}\frac{\delta x}{\delta\theta}\\ &\frac{\delta y}{\delta r}\frac{\delta y}{\delta\theta} }{|}{|}= \embrace{ &acos\theta\quad-arsin\theta\\ &bsin\theta\quad \phantom{-}brcos\theta }{|}{|}\n &|Det J_{phi}| = |abr*(cos^2\theta+sin^2\theta)|=|abr|\n &J=\int_0^1\int_0^{\frac{\pi}2}(2(arcos\theta)^3-brsin\theta)*|det J_{\phi}|d\theta dr\n &\phantom{J}=\int_0^1\int_0^{\frac{\pi}2}(2a^3r^3cos¨3\theta-brsin\theta)abrd\theta dr\n &\phantom{J}=\int_0^1\int_0^{\frac{\pi}2}2a^4br^4cos^3\theta d\theta dr\n &\phantom{J}-\int_0^1\int_0^{\frac{\pi}2}ab^2r^2sin\theta d\theta dr = I_1-I_2\n &I_2=ab^2\int_0^1r^2dr\int_0^{\frac{\pi}2}sin\theta d\theta\n &I_2=ab^2 \cembrace{\frac{r^3}3}_{0}^{1} \cembrace{-cos \theta}_{0}^{\frac{\pi}2}\n &I_2=ab^2\frac13*\underbrace{\cembrace{-cos\frac{\pi}2-(-cos\theta)}}_{0-(-1)=1}\n &I_2=\frac{ab^2}{3}\n &I_1=2a^4b\int_0^1r^4dr\int_0^{\frac{\pi}2} cos^3\theta d\theta\n &I_1=2a^4b \cembrace{\frac{r^5}5}_{0}^{1}\int_0^{\frac{\pi}2}cos^3\theta d\theta\n &I_1=\frac{2a^4b}{5}\int_0^{\frac{\pi}2}cos^3\theta d\theta\n &\quad\embrace{\int_0^{\frac{\pi}2}cos^3\theta d\theta = \cembrace{\frac13sin3\theta}_{0}^{\frac{\pi}2}+ \cembrace{sin^3\theta}_{0}^{\frac{\pi}2}=\frac23}{|}{.}\n &I_1= \frac{4a^4b}{15}\n &I = I_1-I_2=\boxed{\frac{4a^4b}{15}-\frac{ab^2}{3}} \end{split}$$

Volume de l'ellipsoide

$$\begin{split} \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\le1\n &\embrace{ &V_{sphere}=\frac43\pi\ r^3\\ &V_{ellipsoide}=\frac43\pi\ abc }{|}{.}\n &\embrace{ &x=arsin\theta\ cos\phi\\ &y=brsin\theta\ sin\phi\\ &z=crcos\theta }{\{}{.}\quad\quad \embrace{ &r\in [0,1]\\ &\theta\in[0,\pi]\\ &\phi\in[0,2\pi] }{.}{.}\n &det J_{\phi}=abc\ r^2sin\theta\n &V=\iiint_Vdxdydz\n &V=abc \underbrace{\int_0^1\int_0^\pi\int_0^{2\pi}r^2sin\theta\ d\theta_ d\phi\ dr}_{\frac43\pi}\n &V= \frac43\pi\ abc \end{split}$$