# Volume in a cone, using a double integral.

• Jerbearrrrrr
In summary, the problem is asking us to find the volume under the equation z^2 = x^2 + y^2, within the region defined by the disc x^2 + y^2 < 4. To solve this, we can use the polar coordinate system and convert the equation to z = sqrt(r^2), with r representing the radius. The integral to be computed is then (int)(int) r^2 dr dT, with r ranging from 0 to 2 and T ranging from 0 to 2pi. The final solution is 2pi/3. This explanation may be too concise for a 1st year student in a non-mathematics field.

## 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?

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?

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

## What is the formula for finding the volume of a cone using a double integral?

The formula for finding the volume of a cone using a double integral is V = ∫∫R h(x,y) dA, where R is the base of the cone and h(x,y) is the height function.

## What is the difference between using a single integral and a double integral to find the volume of a cone?

A single integral only accounts for one variable (usually the height) while a double integral takes into consideration both the height and the base of the cone, resulting in a more accurate volume calculation.

## Can a double integral be used to find the volume of any cone?

Yes, a double integral can be used to find the volume of any cone as long as the base and height are defined and the cone has a circular base.

## How can the limits of integration be determined when using a double integral to find the volume of a cone?

The limits of integration can be determined by setting up a coordinate system with the origin at the center of the base of the cone. The limits for the inner integral will be from 0 to the radius of the base, and the limits for the outer integral will be from 0 to the height of the cone.

## Are there any real-life applications of using a double integral to find the volume of a cone?

Yes, calculating the volume of a cone using a double integral can be useful in fields such as engineering, architecture, and physics, where accurate volume calculations are necessary for designing structures and analyzing physical phenomena.