Volume in a cone, using a double integral.

[Jerbearrrrrr]

Homework Statement

Evaluate the volume under z^2 = x^2 + y^2
and the disc x^2 + y^2 < 4.

Just wondering what I should write to constitute a proper solution. Would this do?:

V=(int)(int) z dA
R is {x²+y² < 4} [context: R in other problems was the region over which integrals were performed]

(int)(int) z dA
=
(int)(int) sqrt( r² ) r drdT [T for theta]
(using x²+y²=r² and dA->r dr dT)

The integral to actually be computed is:
(int)(int) r² dr dT
with r in [0,2]
T in [0, 2pi]
= 2pi/3 whatever the hell it is.

For a 1st year student in not-mathematics (which isn't me), is that too concise?

 

[Dick]

Looks to me like it's got all the essential steps, given the abbreviations you are using like (int). If you aren't the 1st year student in not-mathematics, why are you asking?

 

[Jerbearrrrrr]

I need to know how to explain it to 1st year not-mathematics students.