Calculating Flux: Homework Statement

[oteggis]

Homework Statement

Let S be the surface of a solid R , which lies inside the cylinder:
$x^2+y^2=16$
and between the plane

where x=0 and z=5

There is also defined a vector field F by:
$\begin{align}F(x,y)=(-x^3i-y^3j+3z^2k)\end{align}$

(a) Calculate : $$\iint_{T} F.\hat n\mathrm dS$$with T = {$(x,y,5)∈ℝ | x^2+y^2≤16$}(b) Calculate DivF and $$\iint_{S} F.\hat n\mathrm dS$$

with n the outward pointing unit normal.

(c) Calculate: $$\iint_{V} F.\hat n\mathrm dS$$

with V = {$(x,y,z)∈ℝ |x^2+y^2≤16 and 0≤z≤5$} and the unit normal $\hat n\$points out of the solid R

Homework Equations

$$\iint_{V} F.\hat n\mathrm dS$$

The Attempt at a Solution

(a) On the top surface of the Cylinder z = 5, $\hat n$= $\hat k$
$F.\hat n = [-x^3i +-y^3j +3z^2k].[k]=3z^2$
$\iint_{S} F.\hat n \mathrm dS = \iint_{S} 3z^2\mathrm dS$
$\iint_{S} F.\hat n \mathrm dS = \iint_{S} 3(5^2)\mathrm dS$
$\iint_{S} F.\hat n \mathrm dS = 75\iint_{S}\mathrm dS$
The area enclosed by the circle is $16\pi$
since the radius of the circle is 4.
Therefore
$\iint_{S} F.\hat n \mathrm dS = 75(16\pi) = 1200\pi$(b) $DivF = \nabla.F = [i\frac{\partial }{\partial x}+ j\frac{\partial }{\partial y} +k\frac{\partial }{\partial z}].[-x^3i +-y^3j +3z^2k]= -3x^2-3y^2 +6z$
DivF = ∇.F = -3(x^2+y^2-2z)
$\iint_{S} F.\hat n \mathrm dS =\iiint_{V} \operatorname{div} F dV -\iint_{S_1} F.\hat n \mathrm dS$
$\iint_{S_1} F.\hat n \mathrm dS = 0$ since z=0, then
$\iint_{S} F.\hat n \mathrm dS = \iiint_{V} \operatorname{div} F dV$**How do I get $\hat n$ in this case?**

(c) From my understanding, I have to use Divergence Theorem here
$$\iint_{V} F.\hat n\mathrm dS = \iiint_{V} \nabla.F \mathrm dV $$
$$\iint_{V} F.\hat n\mathrm dS = \iiint_{V} -3(x^2+y^2-2z) \mathrm dV $$
$$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} (x^2+y^2-2z) \mathrm dV $$

Using Cylindrical coordinates
$$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta $$
$$\iint_{V} F.\hat n\mathrm dS = -3\int_0^{2\pi} \int_0^4 \int_0^5 [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta $$
$$\iint_{V} F.\hat n\mathrm dS = 432\pi $$

 

[BvU]

Hello oteggis,

Could you read what you posted and fix the typesetting ? for in-line math you want to use ## ... ## delimiters and for displayed math $$ ... $$

I take it the solid $R$ completely fills the cylinder ? ('lies inside' can apply to a very tiny fraction of the space). And not $x=0$ but $z=0$ ?

 

[Chestermiller]

On the curved cylinder surface, $$\hat{n}=\hat{r}=\frac{x\hat{i}+y\hat{j}}{4}$$

 

[oteggis]

On the curved cylinder surface, $$\hat{n}=\hat{r}=\frac{x\hat{i}+y\hat{j}}{4}$$

so in Polar coordinates: $\hat n$ = $cos\theta i + sin\theta j$
But I got $\hat n$ = $-cos\theta i-sin\theta j$ instead

 

[Chestermiller]

so in Polar coordinates: $\hat n$ = $cos\theta i + sin\theta j$
But I got $\hat n$ = $-cos\theta i-sin\theta j$ instead

That's the inward-directed normal, not the outward normal required by the divergence theorem.