If the circumference of a circle is 2*pi*r, and the integral of that is its area being pi*r^2, using calculus techniques, show each stage/ step to get from the circumference equation to the volume of a sphere

To derive the volume of a sphere from the circumference of a circle using calculus techniques, we'll proceed through the following stages:

1. **Understand the Relationship Between the Circle and the Sphere:**
   - The circumference \( C \) of a circle is given by \( C = 2\pi r \).
   - The area \( A \) of a circle is given by \( A = \pi r^2 \).
   - The volume \( V \) of a sphere is given by \( V = \frac{4}{3}\pi r^3 \).

2. **Using the Circumference to Find the Area of the Circle:**
   - We start with the circumference of the circle: \( C = 2\pi r \).
   - To find the area \( A \), we recognize that the radius \( r \) is the variable. The relationship can be seen as integrating the circumference with respect to the radius.

3. **Integral of the Circumference to Find the Area:**
   - The circumference can be thought of as a derivative of the area with respect to the radius: \( \frac{dA}{dr} = C = 2\pi r \).
   - To find the area, we integrate the circumference: 
     \[
     A = \int C \, dr = \int 2\pi r \, dr = 2\pi \int r \, dr = 2\pi \left( \frac{r^2}{2} \right) = \pi r^2.
     \]

4. **Using the Area to Find the Volume of the Sphere:**
   - Consider the sphere as a series of infinitesimally thin circular disks stacked along the \( r \)-axis from \(-r\) to \(r\).
   - The volume \( V \) can be found by integrating the area of these disks across the radius.

5. **Integral of the Area to Find the Volume:**
   - The volume of a thin disk of thickness \( dr \) and radius \( r \) is \( dV = A \, dr = \pi r^2 \, dr \).
   - To find the total volume, integrate this expression over the range of \( r \):
     \[
     V = \int_{-r}^{r} \pi r^2 \, dr.
     \]
   - However, because of symmetry, we can integrate from 0 to \( r \) and double the result:
     \[
     V = 2 \int_{0}^{r} \pi r^2 \, dr.
     \]
   - We need to express the radius of the disks as a function of their position along the \( z \)-axis. For a sphere, \( r \) at any point \( z \) is given by \( r = \sqrt{R^2 - z^2} \), where \( R \) is the radius of the sphere and \( z \) varies from \(-R\) to \( R \).

6. **Integrate the Expression for Volume:**
   - Replace \( r \) with \( \sqrt{R^2 - z^2} \):
     \[
     V = 2 \int_{0}^{R} \pi (R^2 - z^2) \, dz.
     \]
   - Simplify and integrate:
     \[
     V = 2\pi \left[ \int_{0}^{R} R^2 \, dz - \int_{0}^{R} z^2 \, dz \right].
     \]
     - The first integral is straightforward:
       \[
       \int_{0}^{R} R^2 \, dz = R^2 z \Big|_{0}^{R} = R^3.
       \]
     - The second integral involves a standard formula:
       \[
       \int_{0}^{R} z^2 \, dz = \frac{z^3}{3} \Big|_{0}^{R} = \frac{R^3}{3}.
       \]
     - Combining these results:
       \[
       V = 2\pi \left( R^3 - \frac{R^3}{3} \right) = 2\pi \left( \frac{3R^3}{3} - \frac{R^3}{3} \right) = 2\pi \left( \frac{2R^3}{3} \right) = \frac{4\pi R^3}{3}.
       \]

Therefore, the volume of a sphere is \( V = \frac{4}{3} \pi R^3 \).

Would you like more details on any of these steps or have any questions?

Here are 5 questions you might want to explore:
1. How does the method of slicing relate to the calculation of volume?
2. Can you explain the concept of an infinitesimally thin disk in more detail?
3. What are the applications of integral calculus in geometry?
4. How would the process differ for finding the volume of a cylinder?
5. What are some other geometric shapes where similar calculus techniques are used?

**Tip:** When working with integrals, always ensure you understand the geometric interpretation of what you're integrating—this helps in setting up the correct bounds and integrands.

Discover how to derive the volume of a sphere from the circumference of a circle using calculus techniques. Learn step-by-step with detailed explanations and formulas, including the application of the Fundamental Theorem of Calculus. Suitable for Grades 11-12 students and calculus enthusiasts.

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