\(\frac{\part x}{\part \vartheta} \times \frac{\part x}{\part \varphi} = \)\( \left( \begin{matrix} \cos(\vartheta) \cos(\varphi) \\ \cos(\vartheta) \sin(\varphi) \\ - \sin(\vartheta) \end{matrix} \right) \times \)\( \left( \begin{matrix} - \sin(\vartheta) \sin(\varphi) \\ \sin(\vartheta) \cos(\varphi) \\ 0 \end{matrix} \right)=\)\( \left( \begin{matrix} \sin^2(\vartheta) \cos(\varphi) \\ \sin^2(\vartheta) \sin(\varphi) \\ \sin(\vartheta) \cos(\vartheta) \end{matrix} \right)\)

\(|| \frac{\part x}{\part \vartheta} \times \frac{\part x}{\part \varphi} || = \sin(\vartheta) \)

\(\iint_F f dF = \int_0^{2\pi} ( \int_0^{\pi} \sin(\vartheta) \ \mathrm{d} \vartheta)\mathrm{d}\varphi = 4\pi \)

\(\iint_F f dF = \int_0^{2\pi} \int_0^{3} f(x(\varphi, \vartheta))|| \frac{\part x}{\part \vartheta} \times \frac{\part x}{\part \varphi} || \ \mathrm{d} \vartheta\mathrm{d}\varphi \)

\(\frac{\part x}{\part \varphi} \times \frac{\part x}{\part \vartheta} = \)\( \left( \begin{matrix} - r \sin(\varphi) \\ r \cos(\varphi) \\ 0 \end{matrix} \right) \times \)\( \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)=\)\( \left( \begin{matrix} r \cos(\varphi) \\ r \sin(\varphi) \\ 0\end{matrix} \right)\)

\(|| \frac{\part x}{\part \vartheta} \times \frac{\part x}{\part \varphi} || = \sqrt{r^2 \sin ^2(\varphi)+r^2 \cos ^2(\varphi)} =r \)

\(f(x_1,x_2,x_3)=x_1+x_3= r \cos(\varphi)+ \vartheta\)

\(\iint_F f dF = \int_0^{2\pi} ( \int_0^{3} r (r \cos(\varphi) + \vartheta) \ \mathrm{d} \vartheta)\mathrm{d}\varphi = 9 \pi r\)

\(\iint_F f dF = \int_0^{3} \int_0^{4} f(x(\varphi, \vartheta))|| \frac{\part x}{\part \vartheta} \times \frac{\part x}{\part \varphi} || \ \mathrm{d} \vartheta\mathrm{d}\varphi \)

\(\frac{\part x}{\part \varphi} \times \frac{\part x}{\part \vartheta} = \)\( \left( \begin{matrix} - 2 \vartheta \sin(\varphi) \\ 2 \vartheta \cos(\varphi) \\ 0 \end{matrix} \right) \times \)\( \left( \begin{matrix} 2 \cos(\varphi) \\ 2 \sin(\varphi) \\ 2 \end{matrix} \right)=\)\( \left( \begin{matrix} 4 \vartheta \cos(\varphi) \\ 4 \vartheta \sin(\varphi) \\ -4 \vartheta \sin^2(\varphi)-4\vartheta \cos^2(\varphi) \end{matrix} \right) \)

\(|| \frac{\part x}{\part \vartheta} \times \frac{\part x}{\part \varphi} || =\)\( \sqrt{16 \vartheta^2 \sin ^2(\varphi)+16 \vartheta^2 \cos ^2(\varphi)+(-4 \vartheta)^2} = \)\(\sqrt{16 \vartheta^2 (\sin ^2(\varphi)+\cos ^2(\varphi)) + 16\vartheta^2}=4 \sqrt{2} \vartheta\)

\(f(x_1,x_2,x_3)=x_1x_2-x_3= 2\vartheta(\cos(\varphi)\sin(\varphi)-1)\)

\(\iint_F f dF = \int_0^{3} ( \int_0^{4} 2\vartheta(\cos(\varphi)\sin(\varphi)-1) 4 \sqrt{2} \vartheta \ \mathrm{d} \vartheta)\mathrm{d}\varphi = \frac{1}{3} (-128) \sqrt{2} (11+\cos (6))\)

\(\iint_{A} \sqrt{x+y} dxdy = \int_0^5 \int_1^2 \sqrt{x+y}dxdy = \int_0^5 \Big( \frac{2}{3}(x+y)^{\frac{3}{2}} \; \mid _{x=1}^{2} \Big) dy = \)

\(= \frac{2}{3} \int_0^5 \Big( (y+2)^{\frac{3}{2}} - (y+1)^{\frac{3}{2}} \Big) dy\)

\(= \frac{2}{3} \cdot \frac{2}{5} \Big( (y+2)^{\frac{5}{2}} \mid _0^5 - (y+1)^{\frac{5}{2}} \mid _0^5 \Big)\)

\(= \frac{2}{3} \cdot \frac{2}{5} \Big( 7^{\frac{5}{2}} - 2^{\frac{5}{2}} -6^{\frac{5}{2}}+1 \Big) = \frac{4}{15} \Big( 7^{\frac{5}{2}} - 2^{\frac{5}{2}} -6^{\frac{5}{2}}+1 \Big) \)

\(\Bigint_{-\infty}^{\infty} \Bigint_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dxdy = \Bigint_{-\infty}^{\infty} \Bigint_{-\infty}^{\infty} e^{-(x^2 + y^2)} dxdy\)

Nun wird auf Polarkoordinaten transformiert, dabei ist zu beachten, dass gilt: \(dxdy = rdrd\varphi\)

\(\Bigint_{-\infty}^{\infty} \Bigint_{-\infty}^{\infty} e^{-(x^2 + y^2)} dxdy = \Bigint_{0}^{2\pi} \Bigint_{0}^{\infty} e^{(-r^2)} rdrd\varphi = 2\pi \Bigint_{0}^{\infty} e^{(-r^2)} rdr\)

Das verbleibende Integral kann nun durch Substitution gelöst werden:

\(r^2 = u \\ \rightarrow \frac{du}{dr}=2r \\ \rightarrow dr=\frac{du}{2r}\)

\( 2\pi \Bigint_{0}^{\infty} e^{(-r^2)} rdr = 2\pi \Bigint_{0}^{\infty} e^{-u}r \frac{du}{2r} = 2\pi \Bigint_{0}^{\infty} e^{-u}\frac{du}{2} = 2\pi \big( -e^{-u}\frac{1}{2} \big) \mid_0^{\infty} \)

\(= \pi \big( -e^{-\infty} + e^0 \big) = \pi\)

\( \iint_{A} (2x+y) d x d y = \bigint_{0}^{4} \bigint_{1}^{3} (2x+y) d x d y = \bigint_{0}^{4}\left(\left.(2 \frac{x^2}{2}+xy)\right|_{x=1} ^{ 3 }\right) d y = \\ \)

\(= \bigint_{0}^{4} (3^2+3y - 1^2 -y) d y = \bigint_{0}^{4} (8+2y) d y =(8y+\frac{2y^2}{2}) \mid_{0}^{4} = 8 \cdot 4 + 4^2 = 48 \)   \(-\)

\(\iint_{A} y \cdot \ln (x) d x d y\)

\( \bigint_{0}^{3} \bigint_{1}^{e} y \cdot \ln (x) d x d y = \bigint_{0}^{3} yx(\ln (x)-1) \mid_{x=1} ^{e}d y = \\ \)\( \bigint_{0}^{3} y \Big[ e(\ln (e)-1) - 1\cdot(\ln (1) -1) \Big] d y = \bigint_{0}^{3} y \Big[ e(1-1) - (0 -1) \Big] d y = \bigint_{0}^{3} y d y =\)

\(= \frac{y^2}{2} \; \mid_0^3 = \frac{9}{2}\)

\(\int_{0}^{\pi} \int_{-1}^{2} x^{2} \cos (y) d x d y = \bigint_{0}^{\pi} \frac{x^{3}}{3} \; \mid_{-1}^{2} \cos (y) d y =\)\(\bigint_{0}^{\pi} \Big( \frac{(-1)^{3}}{3} - \frac{2^3}{3} \Big) \cos (y) d y = -3 \cdot \bigint_{0}^{\pi}\cos (y) d y = -3 \sin (x) \; \mid_0^{\pi} = \)

\(= -3 ( \sin (\pi) - \sin (0) ) = -3 ( 0- 0 ) = 0\)

Die Integration über x kann entweder direkt durch geführt werden, oder falls Sie sich unsicher sind, kann auch x+y=u substituiert werden:

\(\bigint_{0}^{1} \bigint_{0}^{e}\frac{1}{x+y} d x d y = \bigint_{0}^{1} \bigint_{u_1}^{u_2}\frac{1}{u} d u d y = \bigint_{0}^{1} \ln (u) \; \mid_{u_1}^{u_2} d y = \)\(\bigint_{0}^{1} \ln (x+y) \; \mid_{0}^{e} d y = \bigint_{0}^{1} \big( \ln (e+y) - \ln(y) \big) d y =\)

\(= (y+e) \ln (y+e) - y\ln (y) \; \mid_0^1 = \\ = \Big( (1+e) \ln(1+e) - 1\cdot\ln(1) \Big) - \Big(e \ln (e) -0 \cdot\ln(0) \Big) = \\ \)

\(= (1+e) \ln(1+e) - e \)

\(\bigint_{-1}^{1} \bigint_{0}^{\pi / 4} x \cos (2 y) d y d x = \bigint_{-1}^{1} x \frac{\sin (2 y)}{2} \; \mid_0^{\frac{\pi}{4}} d x =\)

\(= \bigint_{-1}^{1} x \frac{1}{2} \Big( \sin (2 \frac{\pi}{4}) - \sin(0) \Big)d x = \bigint_{-1}^{1} x \frac{1}{2} \Big( 1 - 0 \Big)d x = \bigint_{-1}^{1} \frac{x}{2} dx =\)

\(= \frac{1}{2} \cdot \frac{x^2}{2} \; \mid_{-1}^1 = \frac{1}{4} \cdot (1-1) = 0\)

\(\int_{0}^{\pi} \int_{0}^{2\pi} \sin (x+y) d x d y = \int_{0}^{\pi} -\cos (x+y) \; \mid_0^{2\pi} d y = \int_{0}^{\pi} - (\cos (2\pi+y) - \cos (y) ) d y = \)

\(=\int_{0}^{\pi} (-\cos (2\pi+y) + \cos (y) ) d y \)

Da der Kosinus eine Periode von \(2\pi\) aufweist, gilt: \(\cos (2\pi+y) = \cos(y)\) und somit folgt

\(\int_{0}^{\pi} (-\cos (2\pi+y) + \cos (y) ) d y = \int_{0}^{\pi} (-\cos (y) + \cos (y) ) d y = \int_{0}^{\pi} 0 d y = 0\)

\(\bigint_A \bigint \sqrt{x \cdot y} d y d x = \bigint_0^1 \bigint_0^1 x^{\frac{1}{2}} y^{\frac{1}{2}}d y d x = \)\(\bigint_0^1 x^{\frac{1}{2}} \; \frac{2}{3} \; \big( y^{\frac{3}{2}}\big) \mid_0^1 d x = \)\(\bigint_0^1 x^{\frac{1}{2}} \; \frac{2}{3} \; \big( 1^{\frac{3}{2}} - 0\big) d x = \bigint_0^1 x^{\frac{1}{2}} \; \frac{2}{3} d x =\)

\(= \frac{2}{3} \cdot \frac{2}{3} x^{\frac{3}{2}} \; \mid_0^1 = \frac{4}{9} ( 1^{\frac{3}{2}} - 0 ) = \frac{4}{9}\)

\(\bigint_{x=0}^{1} \bigint_{y=0}^{1}\left(x^{2}+y\right) d y d x = \bigint_{x=0}^{1} \big( x^2y + \frac{y^2}{2} \big) \; \mid_0^1 d x = \bigint_{x=0}^{1} \big( x^2 + \frac{1}{2} \big) d x =\)

\(= \big( \frac{x^3}{3} + \frac{1}{2}x \big) \; \mid_0^1 = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}\)

\(\bigint_{0}^{\pi } \bigint_{0}^{\pi } \cos (x+y) d x d y = \bigint_{0}^{\pi } \sin (x+y) \; \mid_0^{\pi} d y = \bigint_{0}^{\pi } \big( \sin (\pi+y) - \sin (y) \big) d y \)

\(= \big( -\cos (\pi+y) + \cos (y) \big) \; \mid_0^{\pi} = \big( -\cos (\pi+\pi) + \cos (\pi) \big) - \big( -\cos (\pi) + \cos (0) \big)=\)

\(= -\cos (2\pi) + \cos (\pi) + \cos (\pi) - \cos (0) = -1 -1 -1 -1 = -4\)

\(\bigint_{x=0}^{1 } \bigint_{y=0}^{2 } \ln(xy) d y d x = \bigint_{x=0}^{1 } y (\ln(xy)-1) \; \mid_0^2 d x = \bigint_{0}^{1 } 2 (\ln(2x)-1) d x = \bigint_{0}^{1 } (2 \cdot \ln(2x)- 2 ) d x = \)

\(= \bigint_{0}^{1 } 2 \cdot \ln(2x) dx - \bigint_{0}^{1 }2 d x \)

Das erste Integral kann z.B. über die Substitution u=2x gelöst werden:

\(\bigint_{0}^{1 } 2 \cdot \ln(2x) dx = \bigint_{u_1}^{u_2 } 2 \cdot \ln(u) \frac{du}{2} = \bigint_{u_1}^{u_2 } \ln(u) du= u \cdot (\ln(u)-1) \; \mid_{u1}^{u_2} = \)\(=2x \cdot (\ln(2x)-1) \; \mid_{0}^{1} = 2\cdot \ln(2) -2 = \ln(2^2) -2 = \ln(4) -2\)

Das zweite Integral kann direkt gelöst werden:

\(\bigint_{0}^{1 }2 d x = 2x \; \mid_0^1 = 2\)

Insgesamt erhält man:

\(\bigint_{0}^{1 } 2 \cdot \ln(2x) dx - \bigint_{0}^{1 }2 d x = \ln(4) -2 -2 = \ln(4)-4\)

\(\bigint_{x=0}^{1 } \bigint_{y=-1}^{1 } y e^x d y d x = \bigint_{y=0}^{2 } \bigint_{x=0}^{1 } y e^x d x d y = \bigint_{y=0}^{2 } y e^x \;\mid_0^1 d y = \bigint_{y=0}^{2 } y (e^1 - e^0) d y =\)

\(= \bigint_{y=0}^{2 } y (e - 1) d y = \frac{y^2}{2} (e - 1) \; \mid_{0}^2 = \frac{2^2}{2} (e - 1) -0 = 2(e - 1)\)

\(\bigint_{y=0}^{1 } \bigint_{x=0}^{y } xy^2 d xd y = \bigint_{y=0}^{1 } y^2 \cdot \frac{x^2}{2} \; \mid_0^y d y = \bigint_{y=0}^{1 } y^2 \cdot \frac{y^2}{2}d y = \)\(\bigint_{y=0}^{1 } \frac{y^4}{2}d y = \frac{y^5}{2\cdot5} \; \mid_0^1 = \frac{1}{10}\)

\(\bigint_{y=0}^{1 } \bigint_{x=1}^{y } (x+\sqrt{y}) d x d y = \bigint_{y=0}^{1 } (\frac{x^2}{2}+x\sqrt{y}) \; \mid_1^y d y = \)\(\bigint_{y=0}^{1 } (\frac{y^2}{2}+y\sqrt{y} - \frac{1}{2}-\sqrt{y}) d y =\)

\(= \bigint_{y=0}^{1 } (\frac{y^2}{2}+y^{\frac{3}{2}} - \frac{1}{2}-y^{\frac{1}{2}}) d y =\)

\(=(\frac{y^3}{2\cdot 3}+\frac{2}{5}y^{\frac{5}{2}} - \frac{1}{2}y-\frac{2}{3}y^{\frac{3}{2}}) \; \mid_0^1 = \) \(\frac{1}{6}+\frac{2}{5}\cdot 1^{\frac{5}{2}} - \frac{1}{2} \cdot 1-\frac{2}{3} \cdot1^{\frac{3}{2}} = \frac{1}{6}+\frac{2}{5} - \frac{1}{2} -\frac{2}{3} = -\frac{3}{5}\)

\(\bigint_{x=0}^{\pi} \bigint_{y=0}^{x}(1+\sin y) d y d x = \bigint_{x=0}^{\pi} (y -\cos y) \mid_0^x d x = \bigint_{x=0}^{\pi} \big[ (x -\cos x) -(0-\cos(0)) \big] d x = \)

\(= \bigint_{x=0}^{\pi} (x -\cos x + 1) d x = \big( \frac{x^2}{2} - \sin x + x \big) \; \mid_0^{\pi} = \big( \frac{\pi^2}{2} - \sin(\pi) + \pi \big) - \big( 0 - \sin (0) + 0 \big) = \)\(\frac{\pi^2}{2} + \pi\)

Zuerst werden die kartesischen Koordinaten durch Polarkoordinaten dargestellt:

\(x = r \cdot \cos (\varphi) \\ y = r \cdot \sin(\varphi)\)

Außerdem gilt: \(dx dy = r dr d\varphi\)

Daher:

\(\iint_{A} x d x d y = \iint_{A} r \cdot \cos (\varphi) r dr d\varphi = \bigint_0^{\frac{\pi}{2}} \bigint_1^3 r \cdot \cos (\varphi) r dr d\varphi = \)

\( \bigint_0^{\frac{\pi}{2}} \bigint_1^3 r^2 \cos (\varphi) dr d\varphi = \bigint_0^{\frac{\pi}{2}} \frac{r^3}{3} \mid_1^3\cos (\varphi) d\varphi =\)\(\bigint_0^{\frac{\pi}{2}} \big(\frac{3^3}{3} -\frac{1^3}{3} \big) \cos (\varphi) d\varphi = \frac{26}{3} \bigint_0^{\frac{\pi}{2}}\cos (\varphi) d\varphi = \)

\(= \frac{26}{3} \sin(\varphi) \mid_0^{\frac{\pi}{2}} = \frac{26}{3} (1 - 0 ) = \frac{26}{3}\)

Zuerst werden die kartesischen Koordinaten durch Polarkoordinaten dargestellt:

\(x = r \cdot \cos (\varphi) \\ y = r \cdot \sin(\varphi)\)

Außerdem gilt: \(dx dy = r dr d\varphi\)

\(\iint_{A} (2x+y) d x d y = \iint_{A} (2r \cdot \cos (\varphi) + r \cdot \sin(\varphi) )r dr d\varphi = \bigint_0^{2\pi} \bigint_0^5 (2 \cdot \cos (\varphi) + \cdot \sin(\varphi) )r^2dr d\varphi = \)

\(= \Big[ \bigint_0^{2\pi} (2 \cos (\varphi) + \sin(\varphi) ) d\varphi \Big] \bigint_0^5 r^2dr =\)\(\Big[ 2 \sin (\varphi) - \cos (\varphi) \Big]_0^{2\pi} \cdot \bigint_0^5 r^2dr = \Big[ 2 \sin (2\pi) - \cos (2\pi) - \big(2 \sin (0) - \cos (0) \big) \Big]\cdot \bigint_0^5 r^2dr = \)

\(=\Big[ 0 - 1 - \big(0 - 1\big) \Big]\cdot \bigint_0^5 r^2dr =\Big[ - 1 + 1 \Big] \cdot \bigint_0^5 r^2dr = 0 \cdot \bigint_0^5 r^2dr = 0\)

Zuerst werden die kartesischen Koordinaten durch Polarkoordinaten dargestellt:

\(x = r \cdot \cos (\varphi) \\ y = r \cdot \sin(\varphi)\)

Außerdem gilt: \(dx dy = r dr d\varphi\)

\(\iint_{A} \frac{y}{r} dx dy = \iint_{A} \frac{ r \cdot \sin (\varphi) }{r} r dr d\varphi = \bigint_0^{\pi} \bigint_0^2 \sin (\varphi) r dr d\varphi = \)\(\bigint_0^{\pi} \sin (\varphi) \frac{r^2}{2} \mid_0^2 d\varphi =\)

\(= \bigint_0^{\pi} \sin (\varphi) \frac{2^2}{2} d\varphi = 2 \bigint_0^{\pi} \sin (\varphi) d\varphi = 2 \cdot ( -\cos (\varphi) ) \mid_0^{\pi} = 2 \cdot [ -\cos (\pi) - (-\cos(0)) ] = \)

\(= 2 \cdot [ -\cos (\pi) + \cos(0) ] = 2 \cdot [ -(-1) + 1 ] = 2 \cdot [ 1+ 1 ] = 4\)

Zuerst werden die kartesischen Koordinaten durch Polarkoordinaten dargestellt:

\(x = r \cdot \cos (\varphi) \\ y = r \cdot \sin(\varphi)\)

Außerdem gilt: \(dx dy = r dr d\varphi\)

\(\iint_{A} \sqrt{x^2+y^2} d x d y = \iint_{A} \sqrt{r^2 \cos^2(\varphi)+r^2 \sin^2(\varphi)} \; r dr d\varphi = \iint_{A} \sqrt{r^2 \big( \cos^2(\varphi)+\sin^2(\varphi) \big)} \; r dr d\varphi \)

\(= \iint_{A} \sqrt{r^2} \; r dr d\varphi = \bigint_0^{\frac{\pi}{4}} \bigint_2^4 r^2 dr d\varphi = \bigint_0^{\frac{\pi}{4}} d\varphi \bigint_2^4 r^2 dr = \)

\(= \varphi \mid_0^{\frac{\pi}{4}} \cdot \frac{r^3}{3} \mid_2^4 = \frac{\pi}{4} \cdot \Big( \frac{4^3}{3} -\frac{2^3}{3} \Big) = \frac{\pi}{12} \cdot 56 = \frac{56\pi}{12} = \frac{14\pi}{3} \)

Zuerst werden die kartesischen Koordinaten durch Polarkoordinaten dargestellt:

\(x = r \cdot \cos (\varphi) \\ y = r \cdot \sin(\varphi)\)

Außerdem gilt: \(dx dy = r dr d\varphi\)

\(\iint_{A} \frac{y}{x} d x d y = \iint_{A} \frac{r \cdot \sin(\varphi)}{r \cdot \cos(\varphi)} r dr d\varphi = \bigint_0^{\frac{\pi}{4}} \bigint_0^{10} \tan(\varphi)r dr d\varphi = \)\(\bigint_0^{\frac{\pi}{4}} \tan(\varphi)d\varphi \cdot \bigint_0^{10} r dr\)

Das Integral des Tangens kann z.B. in einer Integraltafel nachgeschlagen werden:

\(\bigint\tan(\varphi)d\varphi = -\ln ( \cos(\varphi)) + konst.\)

Somit erhalten Sie weiters:

\(\bigint_0^{\frac{\pi}{4}} \tan(\varphi)d\varphi \cdot \bigint_0^{10} r dr = -\ln (\cos (\varphi) ) \; \mid_0^{\frac{\pi}{4}} \cdot \frac{r^2}{2} \; \mid_0^{10} = \)\(\Big[ -\ln (\cos (\frac{\pi}{4}) - (-\ln (\cos (0) ) \big] \cdot \frac{10^2}{2} = \Big[ -\ln (\frac{\sqrt{2}}{2}) - (-\ln(1) ) \big] \cdot 50 = - 50 \cdot \ln \big(\frac{\sqrt{2}}{2} \big)\)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = 2x \; \text{ und } \; \frac{\partial f}{\partial y} = 2y\)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{1} \Bigint_{0}^{1}\left(\begin{array}{c} {x^{3}} \\ {y^{3}} \\ {\left(x^{2}+y^{2}\right)^{3}} \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-2 x} \\ {-2 y} \\ {1} \end{array}\right) d x d y\)\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \Big( -2x^4 - 2y^4 + (x^2+y^2)^3 \Big) d x d y =\)

\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \Big( -2 x^{4}-2 y^{4}+x^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}+y^{6} \Big) d x d y =\)

\(= \Bigint_{0}^{1}\ \Big( -\frac{2 x^{5}}{5}-2 y^{4} x+\frac{x^{7}}{7}+\frac{3 x^{5}}{5} y^{2}+\frac{3 x^{3}}{3} y^{4}+y^{6} x \Big) \mid_{0} ^{1} d y =\)

\(=\Bigint_{0}^{1} \left(-\frac{2}{5}-2 y^{4}+\frac{1}{7}+\frac{3}{5} y^{2}+y^{4}+y^{6}\right) d y =\)

\(=\Bigint_{0}^{1} \left(-\frac{9}{35}- y^{4}+\frac{3}{5} y^{2}+y^{6}\right) d y =\)

\(= \left( -\frac{9}{35}y- \frac{y^{5}}{5}+\frac{3}{5\cdot 3} y^{3}+\frac{y^{7}}{7} \right) \; \mid_0^1= -\frac{9}{35}- \frac{1}{5}+\frac{1}{5}+\frac{1}{7} = -\frac{4}{35} \)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = 2x \; \text{ und } \; \frac{\partial f}{\partial y} = -2y\)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{1} \Bigint_{0}^{1}\left(\begin{array}{c} {-2x^{3}} \\ {4y^{4}} \\ {\left(x^{2}-y^{2}\right)^{2}} \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-2 x} \\ {2 y} \\ {1} \end{array}\right) d x d y\)\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \Big( 4x^4 + 8y^5 + (x^2-y^2)^2 \Big) d x d y =\)

\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \Big( 4x^4 + 8y^5 + x^4 -2x^2y^2 +y^4 \Big) d x d y =\)

\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \Big( 5x^4 + 8y^5 -2x^2y^2 +y^4 \Big) d x d y =\)

\(= \Bigint_{0}^{1} \Big( \frac{5x^5}{5} + 8y^5x - \frac{2x^3}{3}y^2 +y^4x \Big) \mid_0^1d y =\)

\(= \Bigint_{0}^{1} \Big( 1 + 8y^5 - \frac{2}{3}y^2 +y^4 \Big) d y = \Big( y + \frac{8y^6}{6} - \frac{2}{3 \cdot 3}y^3 + \frac{y^5}{5} \Big) \mid_0^1 = \)

\(= 1 + \frac{8}{6} - \frac{2}{9} + \frac{1}{5} = 2 \frac{14}{45}\)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = 2 \; \text{ und } \; \frac{\partial f}{\partial y} = -8y\)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{3} \Bigint_{0}^{1}\left(\begin{array}{c} {2xy}\\ {y^{-1}} \\ {- ( 2x - 4y^2) } \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-2} \\ {8 y} \\ {1} \end{array}\right) d x d y\)\(= \Bigint_{0}^{3} \Bigint_{0}^{1} \Big( -4xy +8 - 2x + 4y^2 \Big) dx dy =\)

\(= \Bigint_{0}^{3} \Big( -\frac{4x^2}{2}y +8x - \frac{2x^2}{2} + 4y^2x \Big) \mid_0^1 dy =\)\(\Bigint_{0}^{3} \Big( -2y +8 - 1 + 4y^2 \Big) dy =\)

\(= \Bigint_{0}^{3} \Big( -2y +7 + 4y^2 \Big) dy = \Big( -\frac{2y^2}{2} + 7y + \frac{4y^3}{3} \Big) \mid_0^3 =\)

\( =-3^2 + 7\cdot3 + \frac{4\cdot 3^3}{3} = 48\)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = 2x \; \text{ und } \; \frac{\partial f}{\partial y} = 2y\)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{5} \Bigint_{0}^{2}\left(\begin{array}{c} {3x^{2}} \\ {2xy} \\ {3 (x^2 + y^2 -25)} \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-2 x} \\ {-2 y} \\ {1} \end{array}\right) d x d y =\)\(= \Bigint_{0}^{5} \Bigint_{0}^{2} \Big( -6x^3 -4xy^2+3(x^2+y^2-25) \Big) dx dy = \)

\(= \Bigint_{0}^{5} \Bigint_{0}^{2} \Big( -6x^3 -4xy^2+3x^2+3y^2-75 \Big) dx dy =\)

\(= \Bigint_{0}^{5} \Big( -\frac{6x^4}{4} -\frac{4x^2}{2}y^2+\frac{3x^3}{3}+3y^2x-75x \Big) \mid_0^2 dy =\)

\(= \Bigint_{0}^{5} \Big( -24 -8y^2+8+6y^2-150 \Big) dy = \Bigint_{0}^{5} \Big( -166 -2y^2 \Big) dy =\)

\( = \Big( -166y - \frac{2y^3}{3} \Big) \mid_0^5 = -830 - \frac{250}{3} = -913\frac{1}{3}\)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = e^x \; \text{ und } \; \frac{\partial f}{\partial y} = e^y\)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{1} \Bigint_{0}^{1}\left(\begin{array}{c} {2e^x} \\ {e^{-2y}} \\ {3(e^x+e^y)} \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-e^x} \\ {-e^y} \\ {1} \end{array}\right) d x d y\)\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \Big(-2e^{2x} - e^{-y} + 3e^x + 3e^y \Big) d x d y =\)

\(= \Bigint_{0}^{1} \Big(-e^{2x} - xe^{-y} + 3e^x + 3xe^y \Big) \mid_0^1d y =\)

\(= \Bigint_{0}^{1} \Big[ \big(-e^{2} - e^{-y} + 3e + 3e^y \big) - \big( -1 + 3 \big) \Big] d y =\)

\(= \Bigint_{0}^{1} \Big[ -e^{2} - e^{-y} + 3e + 3e^y +1 - 3 \big) \Big] d y =\)\(\Bigint_{0}^{1} \Big[ -e^{2} - e^{-y} + 3e + 3e^y -2 \big) \Big] d y =\)

\(= \Big[ -e^{2}y + e^{-y} + 3ey + 3e^y -2y \Big] \mid_{0}^{1} =\)\(\Big[ -e^{2} + e^{-1} + 3e + 3e -2 \Big] - \Big[ 0 + e^{0} + 0 + 3e^0 -0\Big] =\)

\(= \Big[ -e^{2} + e^{-1} + 3e + 3e -2 - 1- 3\Big] = -e^{2} + e^{-1} + 6e -6\)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = \frac{1}{2}x^{ -\frac{1}{2} } = \frac{1}{2 \sqrt{x}} \; \text{ und } \; \frac{\partial f}{\partial y} = \frac{1}{2}y^{ -\frac{1}{2} } = \frac{1}{2 \sqrt{y}} \)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{9} \Bigint_{0}^{1}\left(\begin{array}{c} {\sqrt{x} \; e^x} \\ {4y^2} \\ {2(\sqrt{x} + \sqrt{y}) } \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-\frac{1}{2 \sqrt{x}} } \\ {-\frac{1}{2 \sqrt{y}} } \\ {1} \end{array}\right) \)\(d x d y\)

\(=\Bigint_{0}^{9} \Bigint_{0}^{1} \big( -\frac{e^x}{2} - 2y^{ \frac{3}{2} } +2x^{ \frac{1}{2} } +2y^{ \frac{1}{2} } \big) dxdy= \)\(\Bigint_{0}^{9} \big( -\frac{e^x}{2} - 2y^{ \frac{3}{2} }x + 2 \cdot \frac{2}{3}x^{ \frac{3}{2} } +2y^{ \frac{1}{2} }x \big) \mid_0^1 dy= \)

\(=\Bigint_{0}^{9} \Big[ \big( -\frac{e}{2} - 2y^{ \frac{3}{2} } + \frac{4}{3}+2y^{ \frac{1}{2} } \big) - \big( -\frac{e^0}{2} \big) \Big] dy= \)

\(=\Bigint_{0}^{9} \Big[ -\frac{e}{2} - 2y^{ \frac{3}{2} } +2y^{ \frac{1}{2} } + \frac{11}{6} \Big] dy= \)

\(= \Big[ -\frac{e}{2}y - 2 \cdot \frac{2}{5}y^{ \frac{5}{2} } + 2 \cdot \frac{2}{3}y^{ \frac{3}{2} } + \frac{11}{6}y \Big] \mid_{0}^{9}= \)\( -\frac{9e}{2} - 2 \cdot \frac{2}{5} \cdot 9^{ \frac{5}{2} } + 2 \cdot \frac{2}{3}\cdot 9^{ \frac{3}{2} } + \frac{11}{6} \cdot 9 = \)

\(= -\frac{9e}{2} - \frac{4}{5} \cdot 243+ \frac{4}{3}\cdot 27 + \frac{99}{6} =\)\(-\frac{9e}{2} - \frac{972}{5} + 36 + \frac{33}{2} = -\frac{9e}{2} - 141.9 \)

Für das infinitesimale Flächenelement benötigen Sie die partiellen Ableitungen:

\(\frac{\partial f}{\partial x} = 1 \; \text{ und } \; \frac{\partial f}{\partial y} = 8 \frac{1}{2 \sqrt{y}} =\frac{4}{\sqrt{y}}\)

Somit erhalten Sie für das Oberflächenintegral:

\(\int_{F} \vec{A} \cdot d \vec{f}=\Bigint_{0}^{1} \Bigint_{0}^{1}\left(\begin{array}{c} {x} \\ {x \sqrt{y} } \\ {(x+ 8\sqrt{y})^2} \end{array}\right) \cdot \)\(\left(\begin{array}{c} {-1} \\ {-\frac{4}{\sqrt{y}}} \\ {1} \end{array}\right) \)\(d x d y\) \(= \Bigint_{0}^{1} \Bigint_{0}^{1} \big( -x - 4x + x^2 +16x\sqrt{y} + 64y \big) dxdy =\)

\(= \Bigint_{0}^{1} \Bigint_{0}^{1} \big(- 5x + x^2 +16x\sqrt{y} + 64y \big) dxdy = \Bigint_{0}^{1} \big( -\frac{5x^2}{2}+ \frac{x^3}{3}+ \frac{16x^2}{2}\sqrt{y} + 64yx \big) \mid_0^1 dy =\)

\(= \Bigint_{0}^{1} \big( -\frac{5}{2}+ \frac{1}{3}+ \frac{16}{2}\sqrt{y} + 64y \big) dy \)\(= \Bigint_{0}^{1} \big( -\frac{13}{6}+ 8}\sqrt{y} + 64y \big) dy =\)

\(= \big( -\frac{13}{6}y+ 8 \cdot \frac{2}{3} y^{ \frac{3}{2} } + \frac{64y^2}{2} \big) \mid_0^1 =\)\(-\frac{13}{6}+ 8 \cdot \frac{2}{3} + \frac{64}{2} = -\frac{13}{6}+\frac{16}{3} +32 = \frac{211}{6}\)

Die Oberfläche des entstehenden Drehkörpers kann mit folgendes Formel berechnet werden:

\(O = 2\pi \Bigint_{x_1}^{x_2} f(x) \sqrt{1 + (f^\prime(x))^2} dx\)

Aus \(x^2-y^2 = 9\) folgt:

\(f(x) = y = \sqrt{x^2 - 9}\) 

\(\rightarrow f^\prime(x) = \frac{1}{2} (x^2-9)^{-\frac{1}{2}} \cdot 2x = \frac{x} {\sqrt{x^2-9}}\)    \(\rightarrow (f^\prime(x))^2 = \frac{x^2} {x^2-9}\)

Nun kann alles in die Formel eingesetzt werden:

\(O = 2\pi \Bigint_{3}^{8} \sqrt{x^2 - 9}\sqrt{1 + \frac{x^2} {x^2-9}} dx =\)\(= 2\pi \Bigint_{3}^{8} \sqrt{(x^2 - 9) \cdot( 1 + \frac{x^2} {x^2-9})} dx = 2\pi \Bigint_{3}^{8} \sqrt{x^2 - 9 + \frac{x^2(x^2-9)} {x^2-9})} dx =\)

\( 2\pi \Bigint_{3}^{8} \sqrt{x^2 - 9 + x^2} dx = 2\pi \Bigint_{3}^{8} \sqrt{2x^2 - 9} dx = 2\pi \Bigint_{3}^{8} \sqrt{(\sqrt{2} \; x)^2 - 9} dx \)

Nun kann substituiert werden:

\(u = \sqrt{2} \; x \\ \rightarrow \frac{du}{dx} = \sqrt{2} \rightarrow dx = \frac{du}{\sqrt{2}}\)

Mit den Grenzen: \(u_1= 3 \cdot \sqrt{2}\)   und \(u_2= 8\cdot \sqrt{2}\)

\( 2\pi \Bigint_{3}^{8} \sqrt{(\sqrt{2} \; x)^2 - 9} dx = 2\pi \Bigint_{u_1}^{u_2} \sqrt{u^2 - 9} \; \frac{du}{\sqrt{2}} = \frac{2\pi}{\sqrt{2}}\Bigint_{u_1}^{u_2} \sqrt{u^2 - 9} \; du\)

Dieses Integral findet man üblicherweise in jeder herkömmlichen Integraltafel:

\(\Bigint \sqrt{x^2 - a^2} \; dx = \frac{1}{2} \Big[ x \sqrt{(x^2 - a^2)} \; -a^2 \cdot \ln( \mid \; x + \sqrt{x^2 - a^2} \; \mid ) \Big]\)    wobei \(a > 0\)

Somit folgt:

\(\frac{2\pi}{\sqrt{2}}\Bigint_{u_1}^{u_2} \sqrt{u^2 - 9} \; du = \)\( \frac{2\pi}{\sqrt{2}} \cdot \frac{1}{2} \Big[ u \sqrt{(u^2 - 9)} \; -9 \cdot \ln( \mid \; u + \sqrt{u^2 - 9} \; \mid ) \Big]_{u_1}^{u_2} =\)

\(= \frac{\pi}{\sqrt{2}} \Big[ u \sqrt{(u^2 - 9)} \; -9 \cdot \ln( \mid \; u + \sqrt{u^2 - 9} \; \mid ) \Big]_{3 \cdot \sqrt{2}}^{8\cdot \sqrt{2}} =\)

\(= \frac{\pi}{\sqrt{2}} \Big[ 8\sqrt{2} \cdot \sqrt{(128 - 9)} \; -9 \cdot \ln( \mid \; 8 \cdot \sqrt{2} + \sqrt{128 - 9} \; \mid ) \Big] - \)\(\frac{\pi}{\sqrt{2}} \Big[ 3\sqrt{2} \cdot \sqrt{(18 - 9)} \; -9 \cdot \ln( \mid \; 3 \cdot \sqrt{2} + \sqrt{18 - 9} \; \mid ) \Big]\)

\(= \frac{\pi}{\sqrt{2}} \Big[ \Big( 8\sqrt{2} \cdot \sqrt{119} \; - 9 \cdot \ln( \mid \; 8 \cdot \sqrt{2} + \sqrt{119} \; \mid ) \Big) - \)\(\Big( 3\sqrt{2} \cdot 3\; -9 \cdot \ln( \mid \; 3 \cdot \sqrt{2} + 3 \; \mid ) \Big) \Big]\)
  

\(= \frac{\pi}{\sqrt{2}} \Big[ \Big( 8\sqrt{238} \; - 9 \cdot \ln( \mid \; 8 \cdot \sqrt{2} + \sqrt{119} \; \mid ) \Big) - \)\(\Big( 9\sqrt{2} \; -9 \cdot \ln( \mid \; 3 \cdot \sqrt{2} + 3 \; \mid ) \Big) \Big]\)

\(\approx 223.48\)