Volume in a cone, using a double integral.

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.
  • #1
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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?
 
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  • #2
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?
 
  • #3
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.

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