Volume of a cone

twoflower

Hi,

I have this problem:

Compute volume of solid bounded by these planes:

[tex]
z = 1
[/tex]

[tex]
z^2 = x^2 + y^2
[/tex]

When I draw it, it's cone standing on its top in the origin and cut with the [itex]z = 1[/itex] plane.

So after converting to cylindrical coordinates:

[tex]
x = r\cos \phi
[/tex]

[tex]
y = r\sin \phi
[/tex]

[tex]
z = z
[/tex]

[tex]
|J_{f}(r,\phi,z)| = r
[/tex]

I get

[tex]
0 \leq z \leq 1
[/tex]

[tex]
0 \leq \phi \leq 2\pi
[/tex]

[tex]
0 \leq r \leq 1
[/tex]

And

[tex]
V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{1} r\ dr\ dz\ d\phi
[/tex]

But I got [itex]\pi[/itex] as a result, which is obviously incorrect :(

Can you see where I am doing a mistake?

Thank you!

DaleSpam

Your limits of integration on r are wrong. Note that your maximum r changes as a function of z. What you are (correctly) calculating here is the volume of a unit cylinder, not a cone.

-Dale

twoflower

I thought so...anyway, I still can't see what's wrong. When I draw it in x-z 2D plane, I see that the cone is bounded by curve [itex]z = r[/itex] as a "right side" (which is what I got from expressing z from the original equations and using cylindrical coordinates) and [itex]z = -r[/tex] as a "left side". So because [itex]0 \leq z \leq 1[/itex] also [itex]0 \leq r \leq 1[/itex].

Why isn't it correct?

EDIT: Oh, I maybe see it now..Gonna try that and possibly write again..

DaleSpam

Well, you probably got it now anyway, but just in case I will tell you before I have to go for the evening. Basically instead of integrating r over 0 to 1 you need to integrate r over 0 to z.
[tex]
V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{z} r\ dr\ dz\ d\phi
[/tex]

Do you see how this is a cone and the previous integration was a cylinder?

-Dale

krab

Divide the cone into discs of thickness dz. The radius of each is equal to z, so area is piz^2. Now just integrate:

[tex]\int_0^1\pi z^2dz=\pi/3[/tex]

twoflower

Yes, that's exactly how I finally did it. Thank you DaleSpam!

Thank you Krab, nice approach actually...