Calculating the Flux of F

1. May 24, 2004

vas85

Calculating the Flux of F (Vector Calculus)

Let S be the surface of the solid which is enclosed by the cone

\begin{align*} z = \sqrt{x^2+y^2} \end{align}

and the sphere $$x^2+y^2+(z-1)^2 = 1$$, and which lies above the
cone and below the sphere.

Let \begin{align*} \mathbf{F} = xz\mathbf{i}+yz\mathbf{j}-2\mathbf{k} \end{align}

Calculate the flux of \begin{align*} \mathbf{F} \end{align} , outwards through S

I am currently trying to work on the problem dont know were to approach it from exactly if any1 can help Great Appreciation :)

Last edited: May 24, 2004
2. May 24, 2004

vas85

I am thinking whether they want me to evaluate it as half an orange with a cone missing. any help is greatly appreciated

3. May 24, 2004

arildno

The region is the solid "ice-cream" formed.
(Letting x,y be fixed, the z-value will have its lowest value on the cone, and its highest value at a point on the spherical shell)

You should in all probability use the divergence theorem (Gauss' theorem) to evaluate this integral.

4. May 24, 2004

arildno

To help you on a bit, here's a derivation of the planes of intersection between the cone and the sphere:
$$z=\sqrt{1-(z-1)^{2}}\rightarrow{z}^{2}=2z-z^{2}\rightarrow{z}=0,1$$

Hence, the portion of the sphere directly above the cone is a hemisphere!

5. May 24, 2004

vas85

thanks Arildno

6. May 24, 2004

vas85

umm Arildno do i have to do it in TWO parts? like asin Triple Integral over V1 UNION Triple Integral over V2???
becuase of the fact that its a ICECREAM

7. May 24, 2004

arildno

Not at all!
Let z lie between the values:
$$\sqrt{x^{2}+y^{2}}\leq{z}\leq{1}+\sqrt{1-x^{2}-y^{2}}, (x^{2}+y^{2}\leq{1})$$

8. May 24, 2004

vas85

ummmm Arildno, dunno mayb i'm not a quick learner..

the DIV F i got to be 2z
umm you have told me the bounds of integration for Z but for X and Y? like can you draw down the triple integral i need to evaluate, Appreciate it

9. May 24, 2004

arildno

Well, I wrote the bounds down:
$$x^{2}+y^{2}\leq{1}$$
Intgrating 2z between limits is easy (z^{2} evaluated on the given limits),
while the disk in the x-y plane is most easily evaluated by polar coordinates.
Hence, you get to evaluate the double integral:
$$\int_{0}^{2\pi}\int_{0}^{1}((1+\sqrt{1-r^{2}})^{2}-r^{2})rdrd\theta$$