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Exercice Intégrale curviligne
Calculons la valeur des int e ˊ grales de la forme : ∫ C + x y d x + ( x + y ) d y Avec C + prenant diff e ˊ rentes valeurs \text{Calculons la valeur des intégrales de la forme : }\int_{C^+} xydx+(x+y)dy\\
\text{Avec}\ C^+\ \text{prenant différentes valeurs}
Calculons la valeur des int e ˊ grales de la forme : ∫ C + x y d x + ( x + y ) d y Avec C + prenant diff e ˊ rentes valeurs
1)
a)
C + = { ( x , y ) ∈ R 2 ∣ x 2 + y 2 = 1 } { x = c o s θ d x = s i n θ d θ y = s i n θ d y = c o s θ d θ I = ∫ 0 2 π c o s θ s i n θ ( − s i n θ d θ ) + ( c o s θ + s i n θ ) c o s θ d θ = ∫ 0 2 π ( − c o s θ s i n 2 θ + c o s 2 θ + c o s θ s i n θ ) d θ = − ∫ 0 2 π c o s θ s i n 2 θ d θ + ∫ 0 2 π c o s 2 θ d θ + ∫ 0 2 π c o s θ sin θ d θ = − [ 1 3 s i n θ ] 0 2 π + [ 1 2 θ − 1 2 s i n 2 θ ] 0 2 π + [ 1 2 sin 2 θ ] 0 2 π = 0 + [ 1 2 θ ] 0 2 π + 0 = π
C^+ = \aembrace{(x,y)\in\R^2\ |\ x^2+y^2=1}\\
\embrace{x=cos\theta\quad\quad &dx = sin\theta d\theta\\ y=sin\theta\quad\quad &dy=cos\theta d\theta}{\{}{.}\n
\begin{split}
I&=\int^{2\pi}_0cos\theta sin\theta\pembrace{-sin\theta d\theta}+\pembrace{cos\theta +sin\theta}cos\theta d\theta\n
&=\int^{2\pi}_0\pembrace{-cos\theta sin^2\theta + cos^2\theta +cos\theta sin\theta}d\theta\n
&=-\int^{2\pi}_0cos\theta sin^2\theta d\theta + \int^{2\pi}_0cos^2\theta d\theta +\int^{2\pi}_0cos\theta \sin\theta d\theta\n
&=-\cembrace{\frac13sin\theta}_{0}^{2\pi}+\cembrace{\frac12\theta-\frac12sin2\theta}_{0}^{2\pi}+\cembrace{\frac12\sin^2\theta}_{0}^{2\pi}\n
&=0 + \cembrace{\frac12\theta}_{0}^{2\pi}+0=\pi
\end{split}
C + = { ( x , y ) ∈ R 2 ∣ x 2 + y 2 = 1 } { x = cos θ y = s in θ d x = s in θ d θ d y = cos θ d θ I = ∫ 0 2 π cos θ s in θ ( − s in θ d θ ) + ( cos θ + s in θ ) cos θ d θ = ∫ 0 2 π ( − cos θ s i n 2 θ + co s 2 θ + cos θ s in θ ) d θ = − ∫ 0 2 π cos θ s i n 2 θ d θ + ∫ 0 2 π co s 2 θ d θ + ∫ 0 2 π cos θ sin θ d θ = − [ 3 1 s in θ ] 0 2 π + [ 2 1 θ − 2 1 s in 2 θ ] 0 2 π + [ 2 1 sin 2 θ ] 0 2 π = 0 + [ 2 1 θ ] 0 2 π + 0 = π
b) intégration sur un arc de parabole
∫ C + x y d x + ( x + y ) d y = π Arc de parabole y = x 2 A = ( − 1 ; 1 ) , B = ( 2 ; 4 ) Trouver une param e ˊ trisation de C + { x ( t ) = t t ∈ [ − 1 ; 2 ] y ( t ) = t 2 ( d x = d t d y = 2 t d t I = ∫ − 1 2 t ∗ t 2 + ( t + t 2 ) 2 t d t = ∫ − 1 2 ( t 3 + 2 t 2 + 2 t 3 ) d t ∫ − 1 2 ( 3 t 3 + 2 t 2 ) d t = 3 [ t 4 4 ] − 1 2 + [ t 3 3 ] − 1 2 = 3 ( 2 4 4 − 1 4 ) + 2 ( Z 3 3 − ( − 1 ) 3 3 ) = 3 ( 4 − 1 2 ) + 2 ( 8 3 + 1 3 ) = 3 ∗ 15 4 + 18 3 = 45 4 + 24 4 I = 69 4 \int_{C^+}xydx+(x+y)dy=\pi\n
\text{Arc de parabole}\quad y=x^2\quad A=(-1;1),\ B=(2;4)\n
\text{Trouver une paramétrisation de}\ C^+\n
\embrace{&x(t)=t\quad\quad\quad t\in\cembrace{-1;2}\\&y(t)=t^2}{\{}{.}\n
\embrace{&dx=dt\\&dy=2tdt}{(}{.}\n
\begin{split}
I &=\int^2_{-1}t*t^2+\pembrace{t+t^2}2t\ dt\n
&=\int^2_{-1}\pembrace{t^3+2t^2+2t^3}dt\
\int^2_{-1}\pembrace{3t^3+2t^2}dt\n
&=3 \cembrace{\frac{t^4}4}_{-1}^{2}+ \cembrace{\frac{t^3}3}_{-1}^{2}\n
&= 3\pembrace{\frac{2^4}4-\frac14}+2\pembrace{\frac{Z^3}3-\frac{(-1)^3}3}\n
&=3\pembrace{4-\frac12}+2\pembrace{\frac83+\frac13}\n
&=3*\frac{15}4+\frac{18}3 = \frac{45}4+\frac{24}{4}\n
I&=\frac{69}4
\end{split}
∫ C + x y d x + ( x + y ) d y = π Arc de parabole y = x 2 A = ( − 1 ; 1 ) , B = ( 2 ; 4 ) Trouver une param e ˊ trisation de C + { x ( t ) = t t ∈ [ − 1 ; 2 ] y ( t ) = t 2 ( d x = d t d y = 2 t d t I I = ∫ − 1 2 t ∗ t 2 + ( t + t 2 ) 2 t d t = ∫ − 1 2 ( t 3 + 2 t 2 + 2 t 3 ) d t ∫ − 1 2 ( 3 t 3 + 2 t 2 ) d t = 3 [ 4 t 4 ] − 1 2 + [ 3 t 3 ] − 1 2 = 3 ( 4 2 4 − 4 1 ) + 2 ( 3 Z 3 − 3 ( − 1 ) 3 ) = 3 ( 4 − 2 1 ) + 2 ( 3 8 + 3 1 ) = 3 ∗ 4 15 + 3 18 = 4 45 + 4 24 = 4 69
3)
∫ C ( y + z ) d x + ( z + x ) d y + d z x 2 + y 2 \int_C \frac{\pembrace{y+z}dx+\pembrace{z+x}dy+dz}{x^2+y^2}\n
∫ C x 2 + y 2 ( y + z ) d x + ( z + x ) d y + d z
1. le long de la droite joingannt A(1,1,1) à B(2,2,2)
{ x = t y = t z = t ⇒ { d x = d t d y = d t d z = d t I = ∫ 1 2 2 t d t + 2 t d t + d t 2 t 2 = ∫ 1 2 4 t + 1 2 t 2 d t = ∫ 1 2 2 d t t + ∫ 1 2 1 2 t 2 d t = 2 [ ln t ] 1 2 + 1 2 [ − 1 t ] 1 2 = 2 ln 2 + 1 2 ( − 1 2 + 1 ) I = 2 ln 2 + 1 4 \embrace{x=t\\y=t\\z=t}{\{}{.}\quad\Rightarrow\quad\embrace{dx=dt\\dy=dt\\dz=dt}{\{}{.}\n
\begin{split}
I&=\int_1^2\frac{2tdt+2tdt+dt}{2t^2}\n
&=\int_1^2\frac{4t+1}{2t^2}dt\n
&=\int_1^2\frac{2dt}{t}+\int_1^2\frac1{2t^2}dt\n
&=2 \cembrace{\ln{t}}_{1}^{2}+\frac12 \cembrace{-\frac1t}_{1}^{2}\n
&=2\ln{2}+\frac12\pembrace{-\frac12+1}\n
I&=2\ln{2}+\frac14
\end{split}
⎩ ⎨ ⎧ x = t y = t z = t ⇒ ⎩ ⎨ ⎧ d x = d t d y = d t d z = d t I I = ∫ 1 2 2 t 2 2 t d t + 2 t d t + d t = ∫ 1 2 2 t 2 4 t + 1 d t = ∫ 1 2 t 2 d t + ∫ 1 2 2 t 2 1 d t = 2 [ ln t ] 1 2 + 2 1 [ − t 1 ] 1 2 = 2 ln 2 + 2 1 ( − 2 1 + 1 ) = 2 ln 2 + 4 1
2. Selon une hélice de rayon 1
{ x = c o s θ y = s i n θ z = θ θ ∈ [ 0 , 2 π ] ⇒ { d x = − s i n θ d θ d y = c o s θ d θ d z = d θ I = ∫ 0 2 π ( s i n θ + θ ) ( − s i n θ d θ ) + ( θ + c o s θ ) c o s θ d θ + d θ 1 = ∫ 0 2 π ( − s i n 2 θ − θ s i n θ + θ c o s θ + θ 2 θ + 1 ) d θ = ∫ 0 2 π ( c o s 2 θ + 1 − θ ( s i n θ − c o s θ ) ) d θ = ∫ 0 2 π d θ + ∫ 0 2 π c o s 2 θ d θ − ∫ 0 2 π θ ( s i n θ − c o s θ ) d θ = 2 π + 0 − ∫ 0 2 π θ ( s i n θ − c o s θ ) d θ ∣ ( u = θ v ′ = s i n θ − c o s θ u ′ = 1 v = − c o s θ − s i n θ ∫ u v ′ = [ u v ] − ∫ u ′ v I = 2 π + [ [ θ ( − c o s θ − s i n θ ) ] 0 2 π + ∫ 0 2 π ( c o s θ + s i n θ ) d θ ] = 2 π + 2 π I = 4 π \embrace{&x=cos\theta\\&y=sin\theta\\&z=\theta}{\{}{.}\quad\theta\in\cembrace{0,2\pi}\Rightarrow\quad \embrace{&dx=-sin\theta d\theta\\&dy=cos\theta d\theta\\&dz=d\theta}{\{}{.}\n
\begin{split}
I&=\int_0^{2\pi}\frac{(sin\theta+\theta)(-sin\theta d\theta)+(\theta + cos\theta)cos\theta d\theta +d\theta}{1}\n
&=\int_0^{2\pi}\pembrace{-sin^2\theta-\theta sin\theta+\theta cos\theta+\theta^2\theta+1}d\theta\n
&=\int_0^{2\pi}\pembrace{cos2\theta+1-\theta\pembrace{sin\theta-cos\theta}}d\theta\n
&=\int_0^{2\pi}d\theta+\int_0^{2\pi}cos2\theta d\theta - \int_0^{2\pi}\theta\pembrace{sin\theta-cos\theta}d\theta\n
&=2\pi + 0 -\int_0^{2\pi}\theta\pembrace{sin\theta-cos\theta}d\theta
\end{split}\n\n
\embrace{
&\embrace{&u=\theta\\&v'=sin\theta-cos\theta}{(}{.}\quad\embrace{&u'=1\\&v=-cos\theta-sin\theta}{.}{.}\n
&\int uv'=\cembrace{uv}-\int u'v
}{|}{.}\n\n
\begin{split}
I&=2\pi+\cembrace{\cembrace{\theta(-cos\theta-sin\theta)}_0^{2\pi} +\int_0^{2\pi}\pembrace{cos\theta+sin\theta}d\theta}\n
&=2\pi + 2\pi\n
I&=4\pi
\end{split}
⎩ ⎨ ⎧ x = cos θ y = s in θ z = θ θ ∈ [ 0 , 2 π ] ⇒ ⎩ ⎨ ⎧ d x = − s in θ d θ d y = cos θ d θ d z = d θ I = ∫ 0 2 π 1 ( s in θ + θ ) ( − s in θ d θ ) + ( θ + cos θ ) cos θ d θ + d θ = ∫ 0 2 π ( − s i n 2 θ − θ s in θ + θ cos θ + θ 2 θ + 1 ) d θ = ∫ 0 2 π ( cos 2 θ + 1 − θ ( s in θ − cos θ ) ) d θ = ∫ 0 2 π d θ + ∫ 0 2 π cos 2 θ d θ − ∫ 0 2 π θ ( s in θ − cos θ ) d θ = 2 π + 0 − ∫ 0 2 π θ ( s in θ − cos θ ) d θ ( u = θ v ′ = s in θ − cos θ u ′ = 1 v = − cos θ − s in θ ∫ u v ′ = [ uv ] − ∫ u ′ v I I = 2 π + [ [ θ ( − cos θ − s in θ ) ] 0 2 π + ∫ 0 2 π ( cos θ + s in θ ) d θ ] = 2 π + 2 π = 4 π