1. ## Parameterisation variables

Let S be the surface of the solid region formed from the intersection of the solid region underneath the cone z = 2 - 3((x^2 + y^2))^(0.5) and above the region above the plane z = 0.

I suppose I use cylindrical coordinates to parameterise.

What would be the value of p (i.e rho)? Is it 1/3? And what is the domain of the parameters?

And are there 3 outward normals?

Someone please tell me if I am right or wrong. Thanks.

2. Originally Posted by maibs89
Let S be the surface of the solid region formed from the intersection of the solid region underneath the cone z = 2 - 3((x^2 + y^2))^(0.5) and above the region above the plane z = 0.

I suppose I use cylindrical coordinates to parameterise.

What would be the value of p (i.e rho)? Is it 1/3? And what is the domain of the parameters?

And are there 3 outward normals?

Someone please tell me if I am right or wrong. Thanks.
Greetings maibs89

Changing to polar coordinates we get (remember: $\displaystyle x^2+y^2=r^2$)

$\displaystyle z=2-3r$

setting z=0 gives

$\displaystyle 0=2-3r \iff 3r=2 \iff r=\frac{2}{3}$

This tells us how far the radius goes out in the xy plane (z=0)
So the paramters are

$\displaystyle \theta \in [0,2\pi] \mbox{ and } r \in [0,\frac{2}{3}]$

To find the normal vector we need to take the gradient of F(x,y,z)=0

$\displaystyle z=2-3\sqrt{x^2+y^2} \iff 3\sqrt{x^2+y^2}+z-2=0=F(x,y,z)$

$\displaystyle \vec n = \nabla F = \frac{3x}{\sqrt{x^2+y^2}}\vec i +\frac{3y}{\sqrt{x^2+y^2}} \vec j +\vec k$

Note: there is not a normal vector at the tip of the cone x=y=0

I hope this helps.

Good luck.

3. Hmmm,

This is a bit puzzling.

Isn't the required region a cylinder? I want to use cylindrical coords, if that's possible.

4. Originally Posted by maibs89
Hmmm,

This is a bit puzzling.

Isn't the required region a cylinder? I want to use cylindrical coords, if that's possible.
Hello maibs89,

$\displaystyle z=2-3r$ IS in cylindrical coordinates.

r goes from zero to 2/3 theta from 0 to 2 Pi as I stated before.

The graph is a cone. Here are a few graphs to help you see it.

First in rectangular coordinates with the plane z=0 for clarity.

Now just the cone plotted in cylindrical coordinates.

As for the normal vector to the cone it is the one given in my first post.

The normal to the bottom of the cone(z=0) plane with be $\displaystyle - \vec k$

I Hope this clears it up.

5. Ok...

But isnt't the cone shifted upwards by 2?

I could be wrong on this.