How can we use a polar double integral to derive the volume of a sphere?

In summary, the conversation is about deriving the volume of a sphere using a double polar integral. The integral involves finding the volume under the positive half of the sphere and over a circle in the x-y plane, with r going from 0 to a and theta going from 0 to 2pi. The issue the speaker is having is with the conversion of variables and understanding the difference between the variable r and the radius a of the sphere.
  • #1
G01
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Hey everyone, My task this time is to derive the volume of a sphere using a polar double integral.

The sphere has radius [tex] a[/tex] we know that r goes from 0 to a in this integral.

The equation for a sphere is:

[tex] x^2 + y^2 +z^2 = r^2[/tex]
or [tex] f(x,y) = \sqrt{r^2 -x^2 -y^2}[/tex]
and it intersects the x-y plane in a circle:

[tex] x^2 + y^2 = r^2 [/tex]So if we find the volume over this circle and under the positive half of the sphere and double it we should get the volume we want.

r goes from 0 to a, and [tex]\theta[/tex] goes from 0 to[tex]2\pi[/tex].

So we get:

[tex]2 \int_0^{2\pi} \int_0^a f( r\cos\theta , r\sin\theta ) r dr d\theta [/tex]

My problem is that whenever I fill in the polar conversions for x and y my sphere equation becomes z=0. I know I'm missing something simple but I can't figure it out. Any help would be appreciated. Thank you.
 
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  • #2
Because you're confusing your r's. One of them is a variable, the other is the radius of the sphere
 
  • #3
wow! i must be tired! Thanks a lot shredder
 

1. What is a polar double integral?

A polar double integral is a type of integral used to calculate the volume between a polar curve and the x-y plane. It is represented by ∬f(r,θ)rdrdθ, where f is a function of r and θ representing the polar coordinates.

2. How is a polar double integral different from a regular double integral?

A polar double integral is different from a regular double integral in that it uses polar coordinates instead of rectangular coordinates. This means that instead of integrating over a rectangular area, it integrates over a polar area.

3. What are the steps for solving a polar double integral?

The steps for solving a polar double integral are as follows:

  1. Determine the boundaries of integration for r and θ.
  2. Convert the integrand into polar coordinates.
  3. Integrate the inner integral with respect to r, treating θ as a constant.
  4. Integrate the resulting expression from step 3 with respect to θ, using the boundaries determined in step 1.

4. How is a polar double integral used in real-world applications?

Polar double integrals are used to calculate the volume of irregularly shaped objects, such as cones, spheres, and tori. They are also used in physics to calculate the moment of inertia of objects with rotational symmetry.

5. Are there any special cases in which a polar double integral can be simplified?

Yes, there are two special cases in which a polar double integral can be simplified:

  • If the integrand is an even function, the integral can be simplified by integrating from 0 to 2π and multiplying the result by 2.
  • If the region of integration is symmetric with respect to the origin, the integral can be simplified by integrating over only one half of the region and multiplying the result by 2.

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