Hallo Leute!
Ich habe weitere Kurvenintegral-Aufgaben gelöst. Könntet ihr wieder einen Blick werfen und mir kurz eine Rückmeldung geben, ob die Berechnungen so stimmen.
Aufgabe:
Seien
\( \begin{array}{l} \gamma:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}^{3} \quad f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} \\ t \mapsto\left(\begin{array}{c} 2 \sin (t) \\ -\cos (t) \\ t^{2} \end{array}\right) \quad \text { und } \quad\left(\begin{array}{l} x \\ y \\ z \end{array}\right) \mapsto\left(\begin{array}{c} \frac{x}{4} \\ -4 z \\ \sqrt{z} \cdot x \end{array}\right) \\ \end{array} \)
zwei Abbildungen. Bestimme das Kurvenintegral von \( f \) längs \( \gamma \).
\( \begin{array}{l}\text { 5) } \gamma:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}^{3} \quad f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} \\ t \longmapsto\left(\begin{array}{c}2 \sin (t) \\ -\cos (t) \\ t^{2}\end{array}\right) \quad\left(\begin{array}{c}x \\ y \\ z\end{array}\right) \mapsto\left(\begin{array}{c}x \\ -4 z \\ \mid z \cdot x\end{array}\right) \\ \dot{\gamma}=\left(\begin{array}{c}2 \cos (t) \\ \sin (t) \\ 2 t\end{array}\right) ; \quad f(\gamma(t))=\left(\begin{array}{l}\frac{2 \cdot \sin (t)}{4} \\ -4 t^{2} \\ \left.\sqrt{t^{2}} \cdot 2 \sin (t) n\right)\end{array}\right) \\ \langle f(\gamma(t)), \dot{\gamma}(t)\rangle=\frac{4 \sin (t) \cos (t)}{4}-4 t^{2} \sin (t)+ \\ 4 t^{2} \sin (t) \\ =\sin (t) \cos (t) \\ \int \limits_{0}^{\pi / 2} \sin (t) \cos (t) d t=\int \limits_{0}^{1} u \cdot d u= \\ u=\sin (t) \quad \frac{\pi}{2} \longmapsto u_{2}=1 \\ d u=\cos (t) d t \quad 0 \longmapsto u_{1}=0 \\ {\left[\frac{y^{2}}{2}\right]_{0}^{1}=\left[\frac{1}{2}-\frac{0}{2}\right]=\frac{1}{2}} \\\end{array} \)
Seien
\( \gamma:[0,3] \rightarrow \mathbb{R}^{3}, t \mapsto\left(\begin{array}{c} 2 t \\ t^{2}-1 \\ -t \end{array}\right)_{z}^{\alpha} \text { und } \quad f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3},\left(\begin{array}{l} x \\ y \\ z \end{array}\right) \mapsto\left(\begin{array}{c} y+1 \\ 4 \\ x z \end{array}\right) . \)
Bestimmen Sie das Kurvenintegral von \( f \) längs \( \gamma \). Art
\( \begin{array}{l}\text { 2) } \gamma:[0,3] \rightarrow \mathbb{R}^{3}, t \mapsto\left(\begin{array}{c}2 t \\ t^{2}-1 \\ -t\end{array}\right) \\ f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3},\left(\begin{array}{l}x \\ y \\ z\end{array}\right) \mapsto\left(\begin{array}{l}y+1 \\ 4 \\ x z\end{array}\right) \\ \dot{\gamma}=\left(\begin{array}{c}2 \\ 2 t \\ -1\end{array}\right) \quad f(\gamma(t))=\left(\begin{array}{c}t^{2}-1+1 \\ 4 \\ 2 t \cdot(-t)\end{array}\right)= \\ =\left(\begin{array}{c}t^{2} \\ 4 \\ -2 t^{2}\end{array}\right) \\ \langle f(\gamma(t)), \dot{\gamma}(t)\rangle=\left\langle\left(\begin{array}{c}t^{2} \\ 4 \\ -2 t^{2}\end{array}\right),\left(\begin{array}{c}2 \\ 2 t \\ -1\end{array}\right)\right\rangle=\left(\begin{array}{l}2 \\ 2\end{array}\right) \\ =2 t^{2}+8 t+2 t^{2} \\ \int \limits_{0}^{3}\left(4 t^{2}+8 t\right) d t=\left[\frac{4 t^{3}}{3}+\frac{8 t^{2}}{2}\right]_{0}^{3}= \\ {\left[\frac{108}{3}+\frac{72}{2}\right]=\frac{216}{6}+\frac{216}{6}=\frac{432}{6}} \\ =72 \\\end{array} \)
