\(\dot{\gamma} (t) = \left( \begin{matrix} -\sin t \\ \cos t \\ 2 \end{matrix} \right)\)

\(|| \dot{\gamma} (t) || = \sqrt{ \sin^2 t + \cos^2 t +2^2} = \sqrt{5}\)

\(\int_0^{2 \pi} f(\gamma (t)) || \dot{\gamma} (t) || \mathrm{d}t = \int_0^{2 \pi} (\cos t + \sin t + 2t) \sqrt{5} \qquad \mathrm{d}t =\)

\(\sqrt{5} \ (\sin t - \cos t + t^2 \qquad |_{0}^{2 \pi} ) = \sqrt{5} \cdot 4 \pi^2\)

Das Wegintegral über ein Vektorfeld ist \(\int_a^b V(x(t))\cdot x'(t) \mathrm{d}t\)

\(\vec V= \begin{pmatrix} x+y \\ yz \\ z-x\end{pmatrix}\)   \(\vec x = \begin{pmatrix} t+1 \\ t \\ 1-t \end{pmatrix}\)

\(\vec V(x(t))= \begin{pmatrix} 2t+1 \\ t-t^2 \\ -2t\end{pmatrix}\)   \(\vec x' = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\)

\(\int_0^1 \ 2t+1+t-t^2+2t \ \mathrm{d}t = \int_0^1 \ 5t-t^2+1 \ \mathrm{d}t =\)

\(\frac 52 t^2-\frac 13 t^3+t |_0^1= \frac{19}{6}\)

Das Wegintegral über ein Vektorfeld ist

\(\int_a^b V(x(t))\cdot x'(t) \ \mathrm{d}t\)

\(\vec V= \begin{pmatrix} 2xy \\ x+y^2\end{pmatrix}\)  \(\vec x = \begin{pmatrix} t+1 \\ t \end{pmatrix}\)

\(\vec V(x(t))= \begin{pmatrix} 2t(t+1) \\ t^2+t+1 \end{pmatrix}\)   \(\vec x' = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)

Skalarprodukt \(V(x(t))\cdot x'(t)= 2t(t+1)+t^2+t+1=3t^2+3t+1\)

\(\int_1^2 \ 3t^2+3t+1 \ \mathrm{d}t = t^3 + \frac{3t^2}{2}+t \ |_1^2 = \frac{25}{2}\)

\(\gamma '= \begin{pmatrix} 1 \\ 2 \end{pmatrix}\)

\(|\gamma '| = \sqrt{1^2+2^2}=\sqrt{5}\)   

\(f(\gamma(t))= t+4t^2\)

\(\int_0^2 \ (t+4t^2 ) \sqrt{5} \ \mathrm{d}t = \sqrt 5( \frac{4t^3}{3}+\frac{t^2}{2}) \ |_0^2=\sqrt 5 \cdot \frac{38}{3}\)

\(\vec{A} (x(t), \; y(t)) = \left( \begin{array}{c} 2x(t) + 2y(t) \\\ x(t) \\\ \end{array}\right) = \left( \begin{array}{c} 2 + t^2 \\\ 1 \\\ \end{array}\right)\)

Nun wird der Ortsvektor abgeleitet:

\(\frac{d\vec{r}}{dt} = \left( \begin{array}{c} 0 \\\ 2t \\\ \end{array}\right)\)

Nun kann das Vektorfeld entlang dieses Weges integriert werden:

\(\bigint_C \vec{A}(\vec{r}) \cdot d\vec{r} = \\ = \bigint_{t_1}^{t_2} \Big( \vec{A}(\vec{r}) \cdot \frac{d\vec{r}}{dt} \Big) dt = \)

\(= \bigint_{0}^{1} \Big( \left( \begin{array}{c} 2 + t^2 \\\ 1 \\\ \end{array}\right) \cdot \left( \begin{array}{c} 0 \\\ 2t \\\ \end{array}\right) \Big) dt = \\ = \bigint_{0}^{1} ( 2t ) dt = \\ \)

\(= 2 \frac{t^2}{2} \; \mid _0^1 = \\ =1\)

\(\vec{A} (x(t), \; y(t)) = \left( \begin{array}{c} 2x(t)y(t) \\\ x(t)^2 \\\ \end{array}\right) = \left( \begin{array}{c} 2 t^3 \\\ t^2 \\\ \end{array}\right)\)

Als nächstes wird der Ortsvektor abgeleitet:

\(\frac{d\vec{r}}{dt} = \left( \begin{array}{c} 1 \\\ 2t \\\ \end{array}\right)\)

Nun kann das Vektorfeld entlang dieses Weges integriert werden:

\(\bigint_C \vec{A}(\vec{r}) \cdot d\vec{r} = \\ = \bigint_{t_1}^{t_2} \Big( \vec{A}(\vec{r}) \cdot \frac{d\vec{r}}{dt} \Big) dt = \)

\(= \bigint_{0}^{1} \Big( \left( \begin{array}{c} 2t^3 \\\ t^2 \\\ \end{array}\right) \cdot \left( \begin{array}{c} 1 \\\ 2t \\\ \end{array}\right) \Big) dt = \\ \)

\(= \bigint_{0}^{1} ( 2t^3 + 2t^3 ) dt = \\ = \bigint_{0}^{1} ( 4t^3 ) dt = \\ = 4 \frac{t^4}{4} \; \mid _0^1 = \\ =1\)

\(\vec{A} (x(t), \; y(t)) = \left( \begin{array}{c} 2y(t) \\\ 4x(t)^2 - y(t) \\\ \end{array}\right) = \left( \begin{array}{c} 2t^2 \\\ 4 \cdot (2t)^2 -t^2\\\ \end{array}\right) = \)

\(= \left( \begin{array}{c} 2t^2 \\\ 15t^2 \\\ \end{array}\right)\)

Nun wird der Ortsvektor abgeleitet:

\(\frac{d\vec{r}}{dt} = \left( \begin{array}{c} 2 \\\ 2t \\\ \end{array}\right)\)

Nun kann das Vektorfeld entlang dieses Weges integriert werden:

\(\bigint_C \vec{A}(\vec{r}) \cdot d\vec{r} = \\ = \bigint_{t_1}^{t_2} \Big( \vec{A}(\vec{r}) \cdot \frac{d\vec{r}}{dt} \Big) dt = \)

\(= \bigint_{1}^{5} \Big( \left( \begin{array}{c} 2t^2 \\\ 15t^2 \\\ \end{array}\right) \left( \begin{array}{c} 2 \\\ 2t \\\ \end{array}\right) \Big) dt = \\ = \bigint_{1}^{5} ( 4t^2 + 30t^3) dt = \)

\(= \big( 4 \cdot \frac{t^3}{3} + 30 \cdot \frac{t^4}{4} \big) \; \mid _1^5 = \\ = \big( \frac{4}{3} 5^3 + \frac{30}{4}5^4 \big) - \big( \frac{4}{3} + \frac{30}{4} \big) = \\ \)

\(= \frac{500}{3} + \frac{9375}{2} - \frac{53}{6} = \\ = \frac{14536}{3}\)

\(x^{2}+y^{2}=1\)  beschreibt einen Kreis mit Radius \(r=1\)

Dieser Kreis kann folgendermaßen parametrisiert werden:

\(\vec{r}(t)=\left(\begin{array}{c} {r \cos t} \\ {r \sin t} \end{array}\right) = \left(\begin{array}{c} {\cos t} \\ { \sin t} \end{array}\right)\)

Außerdem benötigt man noch:

\(\frac{ d\vec{r} }{dt} = \left(\begin{array}{c} {- \sin t} \\ { \cos t} \end{array}\right)\)

Damit erhält man:

\(\vec{u}(t)= \left(\begin{array}{c} {2 y-4 x y} \\ {4 y^2 -2x} \end{array}\right) = \left(\begin{array}{c} {2 \sin (t)-4 \cos (t) \sin (t)} \\ {4 \sin^2 (t) - 2\cos (t) } \end{array}\right)\)

\(\bigoint_C \vec{u}(t) d\vec{s} = \bigoint_C \vec{u}(t) \cdot \frac{ d\vec{r} }{dt} dt = \)

\(= \bigoint_C \left(\begin{array}{c} {2 \sin (t)-4 \cos (t) \sin (t)} \\ {4 \sin^2 (t) - 2\cos (t) } \end{array}\right) \cdot \)\( \left(\begin{array}{c} {- \sin t} \\ { \cos t} \end{array}\right) dt\)

Damit der gesamte Kreis durchlaufen wird, muss von 0 bis \(2\pi\) integriert werden:

\(\int_0^{2\pi} ( -2 \sin^2 (t) + 4\cos (t) \sin^2 (t) + 4\sin^2(t)\cos(t) - 2\cos^2 (t) ) dt =\)

\(= \int_0^{2\pi} \big( -2 ( \sin^2 (t) + \cos^2 (t) ) + 4\cos (t) \sin^2 (t) + 4\cos (t) \sin^2 (t) \big) dt = \\ = \int_0^{2\pi} \big( -2 + 4\cos (t) \sin^2 (t) + 4\cos (t) \sin^2 (t) \big) dt = \)

\(= \int_0^{2\pi} \big( -2 + 8\cos (t) \sin^2 (t)\big) dt = \\ = -2 \int_0^{2\pi} dt + 8 \int_0^{2\pi} \cos (t) \sin^2 (t)dt = \\ = -2 \cdot 2\pi + 8 \int_0^{2\pi} \cos (t) \sin^2 (t)dt\)

Um dieses Integral zu lösen wird substituiert:

\(u=\sin (t) \rightarrow \frac{du}{dt} = \cos (t) \rightarrow dt = \frac{du}{\cos (t)}\)

\(8 \int_0^{2\pi} \cos (t) \sin^2 (t)dt = 8 \int_0^{2\pi} \cos (t) u^2 \frac{du}{\cos (t)} = \\ 8 \int_0^{2\pi} u^2 du = 8 \frac{u^3}{3} \; \mid_0^{2\pi} = \)

\(= \frac{8}{3} (\sin^3 (2\pi) - \sin^3 (0) ) = \frac{8}{3} (0 -0 ) =0\)

Das Endergebnis lautet somit:

\(\bigoint_C \vec{u}(t) d\vec{s} = -4\pi\)