- #1

oteggis

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## 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

$$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta $$

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

$$\iint_{V} F.\hat n\mathrm dS = 432\pi $$

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