Application of surface integral

**shoplifter** · January 7th 2010, 10:16 AM

Let $C$ be the solid cylinder in $\mathbb{R}^3$ defined by the equations

$x_1^2 + x_2^2 \leq 1, \text{ } 0 \leq x_2 \leq 1 $

and let $\partial C$ be its boundary with the usual orientation; thus $\partial C$ consists of a cylindrical surface segment $C_0$ together with the top disc $D_1$ and the bottom disc $D_2$ , each with the appropriate orientation. In addition consider the vector field

$\mathbf{f}(\mathbf{x}) = \left( \begin{array}{c} x_1 \\ x_2^2 \\ x_1^2 \end{array} \right) $

and the differential form

$\omega(\mathbf{x}) = x_3 dx_1 \wedge dx_2 + x_1 x_3 dx_2 \wedge dx_3 + x_3 dx_1 \wedge dx_3$ .

1. Calculate $\int_{\partial C} \mathbf{f} \cdot \mathbf{\nu} dS$ , where $\nu$ is the outward unit normal vector to the boundary $\partial C$ and $dS$ is the element of surface area, by using Stokes's Theorem.

2. Calculate $\int_{D_2} \mathbf{f} \cdot \mathbf{\nu} dS$ .

3. Calculate $\int_{C_0} \omega$ .

Note especially that the normal in part (2) is not $(0,0,1)$ , but $(0,0,-1)$ .

**Abu-Khalil** · January 7th 2010, 12:09 PM

Maybe $-1\leq x_2\leq 1$ .

Your normal vectors are $\hat k,-\hat k$ and $(\cos\theta,\sin\theta),\theta\in[0,2\pi).$

1) Using divergence:

$\int_{\partial C}f\cdot d\vec S=\int_C \nabla \cdot f dV, \nabla \cdot f = 1+2x_2.$

With cilindrical coordinates: $\int_0^{2\pi}\int_0^1 r+2r^2\sin \theta drd\theta=\pi.$

2)

$\int_{D_2}f\cdot d\vec S = \int_{D_2} -x_1^2 dA=\int_0^{2\pi}\int_0^1 -r^3\cos^2\theta drd\theta=-\frac{\pi}{4}.$

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