Kurvenintegral - 2 Aufgaben < mehrere Veränderl. < reell < Analysis < Hochschule < Mathe < Vorhilfe
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hi
ich habe folgende Aufgaben gelöst und möchte gerne wissen, ob ich alles richtig gemacht habe und das Ergebnis stimmt.
3a) [mm] $\textbf{F} [/mm] = [mm] \textbf{i}x [/mm] + [mm] \textbf{j}y^2 [/mm] + [mm] \textbf{k}z; \quad \textbf{C} [/mm] : [mm] \textbf{r} [/mm] = [mm] \textbf{i}t+\textbf{j}t^2+\textbf{k}t,~0 \le [/mm] t [mm] \le [/mm] 1$
[mm] $\frac{dx}{dt} [/mm] = 1; [mm] ~\frac{dy}{dt} [/mm] = [mm] 2t;~\frac{dz}{dt} [/mm] = 1 [mm] \quad \rightarrow d\textbf{r} [/mm] = [mm] (\textbf{i}+2t\textbf{j}+\textbf{k}) [/mm] dt$
[mm] $\int_C \textbf{F} \cdot d\textbf{r} [/mm] &= [mm] \int_0^1 (\textbf{i}t [/mm] + [mm] \textbf{j}t^4+\textbf{k}t) \cdot (\textbf{i}+2t\textbf{j}+\textbf{k}) [/mm] dt = [mm] \int_0^1 (t+2t^5+t) [/mm] dt [mm] =\int_0^1 (2t+2t^5) [/mm] dt = [mm] \Big[t^2 [/mm] + [mm] \frac{1}{3}t^6 \Big]_0^1 [/mm] = 1 + [mm] \frac{1}{3} [/mm] = [mm] \frac{4}{3}
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3b) [mm] $\textbf{F} [/mm] = [mm] y\textbf{i} [/mm] + [mm] xz^3\textbf{j}-y^3z\textbf{k}; \quad \textbf{C} [/mm] : [mm] \textbf{r} [/mm] = [mm] x^2+y^2=4; [/mm] ~z=-3$
$x = [mm] 2\cos\theta; [/mm] = [mm] 2\sin\theta; [/mm] z = -3 [mm] \quad{0\le \theta \le 2\pi}$
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[mm] $\textbf{r} [/mm] = [mm] 2\cos\theta \textbf{i} [/mm] + [mm] 2\sin\theta\textbf{j} [/mm] - [mm] 3\textbf{k}$\\
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[mm] $d\textbf{r} [/mm] = [mm] (-2\sin\theta\textbf{i}+2\cos\theta\textbf{j} [/mm] + [mm] 0\textbf{k})d\theta$
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[mm] \oint_c \textbf{F}\cdot d\textbf{r} [/mm] = [mm] \int_0^{2\pi} (2\sin\theta\textbf{i}-54\cos\theta\textbf{j} [/mm] + [mm] 24\sin^3\theta\textbf{k}) \cdot (-2\sin\theta\textbf{i}+2\cos\theta\textbf{j} [/mm] + [mm] 0\textbf{k})d\theta \\$
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$= [mm] \int_0^{2\pi} (-4\sin^2\theta-108\cos^2\theta)d\theta =-4\int_0^{2\pi} (\sin^2\theta+27\cos^2\theta)d\theta= [/mm] -4 [mm] \cdot (\pi+27\pi) [/mm] = -4 [mm] \cdot 28\pi [/mm] = [mm] -112\pi$
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Vielen Dank!
Gruß GB
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