Surface integral of circle
i want to calculate the surface of this function: $x^2+y^2=9$ ; $z=0$ for $x\ge0, y\ge0, z\ge0$ and this vektor field: $F=(-y, x, z)$.
I have chosen this parameterization: $\phi(r,\theta)=(r*cos\theta, r*sin\theta, 0)$.
Now i can calculate my $d\overrightarrow{o}=\frac{\delta\phi}{\delta r} x \frac{\delta\phi}{\delta\theta}=(0,0,r*cos(\theta)^2+r*sin(\theta)^2$
Definition of the surface integral: $\int_B\! V* \, \mathrm{d}\overrightarrow{o}=\int_0^{2\pi} \! \int_0^3 \! (-r*sin\theta, r*cos\theta,0)*(0,0,r*cos^2\theta + r*sin^2\theta) \, \mathrm{d}r \, \mathrm{d}\theta.=0$
Is this correct?
Best wishes
$\endgroup$1 Answer
$\begingroup$Write your vectorfield this way: $$\begin{gathered} F = - ydx + xdy + zdz \hfill \\ dF = - dy \wedge dx + dx \wedge dy = 2dx \wedge dy \hfill \\ \end{gathered}$$
This is given region: $$\begin{gathered} {x^2} + {y^2} = 9 \hfill \\ z = 0 \hfill \\ \end{gathered} $$
Use polar coordinates: $$\left. \begin{gathered} x = r\cos (\varphi ) \hfill \\ y = r\sin (\varphi ) \hfill \\ \end{gathered} \right\}r = 3$$
Transform volume-element: $$dx \wedge dy = rdr \wedge d\varphi$$
and integrate:
$$\int\limits_A {dF} = \int\limits_A {2dx \wedge dy} = 2\int\limits_0^{2\pi } {\int\limits_0^3 {rdrd\varphi } } = 2 \cdot \frac{9}{2}\int\limits_0^{2\pi } {d\varphi = 18\pi } $$
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