# Volume of Cone in R3

• May 2nd 2011, 01:15 PM
Ant
Volume of Cone in R3
I'm trying to compute the area of the Cone (Z = sqrt[ x^2 + y^2] below the plane z = 2.

Using cylindricl polars coordinates:

phi: [ -pi/2, pi/2]
r: [ 0, 2]
z: [ r, 2]

with the Jacobian being r.

I get 4/3 pi which I think is the correct answer.

However, using Spherical polar coordinates and using the following limits:

Phi: [ -pi/2, pi/2 ]
Theta: [ 0, pi/4 ]
r: [ 0, 2 ]

with the Jacobian being r^2 sin (theta) I'm getting a different answer. Could anyone tell me if my limits are correct?

Thanks!
• May 2nd 2011, 02:06 PM
TheEmptySet
Quote:

Originally Posted by Ant
I'm trying to compute the area of the Cone (Z = sqrt[ x^2 + y^2] below the plane z = 2.

Using cylindricl polars coordinates:

phi: [ -pi/2, pi/2]
r: [ 0, 2]
z: [ r, 2]

with the Jacobian being r.

I get 4/3 pi which I think is the correct answer.

However, using Spherical polar coordinates and using the following limits:

Phi: [ -pi/2, pi/2 ]
Theta: [ 0, pi/4 ]
r: [ 0, 2 ]

with the Jacobian being r^2 sin (theta) I'm getting a different answer. Could anyone tell me if my limits are correct?

Thanks!

$\int_{0}^{2\pi} \int_{0}^{2} \int _{r}^{2}rdzdrd\phi$

$\int_{0}^{2\pi} \int_{0}^{2} r(2-r)drd\phi$

$\int_{0}^{2\pi} r^2-\frac{1}{3}r^3\bigg|_{0}^{2}d\phi$

$\frac{4}{3}\int_{0}^{2\pi} d\phi=\frac{8\pi}{3}$

Remember that geometry that the volume formula is

$V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi (2)^2(2)=\frac{8\pi}{3}$

for the 2nd one r goes from

$r=0..\frac{2}{\cos(\theta)}$

why?
• May 2nd 2011, 02:11 PM
Ant
Sorry! I've missed a vital piece of information. It is given that X > 0!

That's why my limits run from -pi/2 to pi/2.
• May 2nd 2011, 02:17 PM
TheEmptySet
Quote:

Originally Posted by Ant
Sorry! I've missed a vital piece of information. It is given that X > 0!

That's why my limits run from -pi/2 to pi/2.

This will just change phi as you have them above

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} \int _{r}^{2}rdzdrd\phi$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} r(2-r)drd\phi$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} r^2-\frac{1}{3}r^3\bigg|_{0}^{2}d\phi$

$\frac{4}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\phi=\frac{4\pi}{3}$

Remember that geometry that the volume formula is

$V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi (2)^2(2)=\frac{8\pi}{3}$

for the 2nd one

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\pi}\int_ {0}^{\frac{2}{\cos(\theta)}}r^2\sin(\theta)dr d\theta d\phi$

why?
• May 2nd 2011, 02:29 PM
Ant
Was the area that I computed in the second case a portion of a sphere? So it was curved at the top were z=2 (if that makes sense) whereas I'm trying to find the area of a cone so it's a straight line.
• May 2nd 2011, 02:34 PM
TheEmptySet
Quote:

Originally Posted by Ant
Was the area that I computed in the second case a portion of a sphere? So it was curved at the top were z=2 (if that makes sense) whereas I'm trying to find the area of a cone so it's a straight line.

Yes it looks kind of like an ice cream cone.
• May 2nd 2011, 02:34 PM
Ant
I think I know where the limits comes from:

Unlike for a sphere the r is changes, as we track round with theta changing form zero to pi/4 the distance form the centre isn't constant.

Here r it is given by the hypotenuse of a triangle.

Cos (theta) = adjacent/hyp = z/r Where z =2, so we have;

r = 2/cos(theta)
• May 2nd 2011, 03:04 PM
TheEmptySet
Quote:

Originally Posted by Ant
I think I know where the limits comes from:

Unlike for a sphere the r is changes, as we track round with theta changing form zero to pi/4 the distance form the centre isn't constant.

Here r it is given by the hypotenuse of a triangle.

Cos (theta) = adjacent/hyp = z/r Where z =2, so we have;

r = 2/cos(theta)

Yes that is correct.