Volume of a Cone: Solve the Problem

[twoflower]

Hi,

I have this problem:

Compute volume of solid bounded by these planes:

<br /> z = 1<br />

<br /> z^2 = x^2 + y^2<br />

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

So after converting to cylindrical coordinates:

<br /> x = r\cos \phi<br />

<br /> y = r\sin \phi<br />

<br /> z = z<br />

<br /> |J_{f}(r,\phi,z)| = r<br />

I get

<br /> 0 \leq z \leq 1<br />

<br /> 0 \leq \phi \leq 2\pi<br />

<br /> 0 \leq r \leq 1<br />

And

<br /> V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{1} r\ dr\ dz\ d\phi<br />

But I got \pi as a result, which is obviously incorrect :(

Can you see where I am doing a mistake?

Thank you!

 

[Dale]

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]

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

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 z = r as a "right side" (which is what I got from expressing z from the original equations and using cylindrical coordinates) and z = -r[/tex] as a &quot;left side&quot;. So because 0 \leq z \leq 1 also 0 \leq r \leq 1.<br /> <br /> Why isn&#039;t it correct?<br /> <br /> EDIT: Oh, I maybe see it now..Gonna try that and possibly write again..

 

[Dale]

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.
<br /> V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{z} r\ dr\ dz\ d\phi<br />

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:

\int_0^1\pi z^2dz=\pi/3

 

[twoflower]

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.
<br /> V = \iiint_{M}\ dx\ dy\ dz\ =\ \int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{z} r\ dr\ dz\ d\phi<br />
Do you see how this is a cone and the previous integration was a cylinder?
-Dale

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

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

Thank you Krab, nice approach actually...