Man berechne das Kurvenintegral der skalarwertigen Funktion f{\displaystyle f} längs der Kurve c{\displaystyle c}: f(x,y)=xyx2+y2{\displaystyle f(x,y)={\frac {xy}{x^{2}+y^{2}}}}, c(t)=(cost,sint){\displaystyle c(t)=(\cos t,\sin t)}, 0≤t≤π/2{\displaystyle 0\leq t\leq \pi /2}.
Kurvenintegral einer Funktion f:Rn→R{\displaystyle f:\mathbb {R} ^{n}\to \mathbb {R} } entlang einer Kurve k:[a,b]→Rn{\displaystyle k:[a,b]\to \mathbb {R} ^{n}}:
Satz:
∫cf(x,y)ds=∫abf(c(t))⋅||c′(t)||dt={\displaystyle \int \limits _{c}f(x,y)ds=\int _{a}^{b}f(c(t))\cdot \left||c'(t)\right||dt=}
∫0π2costsintcos2t+sin2t⏟1⋅cos2t+sin2t⏟1dt={\displaystyle \int _{0}^{\tfrac {\pi }{2}}{\frac {\cos t\sin t}{\underbrace {\cos ^{2}t+\sin ^{2}t} _{1}}}\cdot \underbrace {\sqrt {\cos ^{2}t+\sin ^{2}t}} _{1}\;dt=}
∫0π2costsintdt=∫0π2sin2t2dt=12∫0π2sin2tdt={\displaystyle \int _{0}^{\tfrac {\pi }{2}}\cos t\sin t\;dt=\int _{0}^{\tfrac {\pi }{2}}{\frac {\sin 2t}{2}}dt={\tfrac {1}{2}}\int _{0}^{\tfrac {\pi }{2}}\sin 2t\;dt=}
12⋅−cos2t⋅12|0π2=14⋅−cost|0π=14(−cosπ+cos0⏟1+1)=12{\displaystyle \left.{\tfrac {1}{2}}\cdot -\cos 2t\cdot {\tfrac {1}{2}}\right|_{0}^{\tfrac {\pi }{2}}=\left.{\tfrac {1}{4}}\cdot -\cos t\right|_{0}^{\pi }={\tfrac {1}{4}}(\underbrace {-\cos \pi +\cos 0} _{1+1})={\frac {1}{2}}}
Ähnliche Beispiele:
entlang einer Kurve ![{\displaystyle k:[a,b]\to \mathbb {R} ^{n}}](/index.php?title=Spezial:MathShowImage&hash=3b95ac2b11e74d5aaf5188beb4538860&mode=mathml)
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