Question 1181750
Here's the proof that the volume removed from the sphere is the difference of two spheres:

**1. Visualize the Problem:** Imagine the cylinder passing through the sphere. The removed volume is a cylindrical "tunnel" with curved ends.  We'll find this volume by integrating.

**2. Set up the Integral:**

*   Consider a thin horizontal slice of the sphere at a height *z* from the center.  This slice is a circle.
*   The radius of this circular slice (let's call it *x*) can be found using the Pythagorean theorem: x² + z² = r². So, x = √(r² - z²).
*   However, the cylinder removes a central portion of this circle. The "kept" parts of the circular slice are two segments, and the removed part is what's inside the cylinder. The width of this removed part is 2y, where y can be found using the Pythagorean theorem in the cross-section of the cylinder: y² + z² = a². So, y = √(a² - z²). Note that this equation is only valid for |z| <= a. 
*   The area of the removed part of the circular slice is 2y*√(r² - z²) = 2√(a² - z²)*√(r² - z²).
*   To find the total removed volume, we integrate this area from -a to a:

V_removed = ∫(from -a to a) 2√(a² - z²)*√(r² - z²) dz

**3. Trigonometric Substitution:**

Let z = r*sin(θ). Then dz = r*cos(θ)*dθ. When z = a, sin(θ) = a/r, so θ = arcsin(a/r). Let's call this angle θ₀. When z = -a, sin(θ) = -a/r, so θ = -arcsin(a/r) = -θ₀.

Now substitute:
V_removed = ∫(from -θ₀ to θ₀) 2√(a² - r²sin²θ)*√(r² - r²sin²θ) * r*cos(θ)*dθ
V_removed = 2r²∫(from -θ₀ to θ₀) cos²θ*√(a² - r²sin²θ) dθ

Since a = rsin(θ₀),
V_removed = 2r²∫(from -θ₀ to θ₀) cos²θ*√(r²sin²θ₀ - r²sin²θ) dθ
V_removed = 2r³∫(from -θ₀ to θ₀) cos²θ*cosθ*√(sin²θ₀ - sin²θ) dθ
V_removed = 2r³∫(from -θ₀ to θ₀) cos³θ*√(sin²θ₀ - sin²θ) dθ

This integral is quite complex. Instead, let's consider a different approach.

**4. Alternative Approach (Geometric):**

The volume removed is the volume of the cylinder plus the volume of the two spherical caps at the ends.

*   **Volume of the cylinder:** V_cyl = πa² * 2√(r² - a²) = 2πa²√(r² - a²)
*   **Volume of each spherical cap:** The volume of a spherical cap of height h and radius r is given by (πh²/3)(3r - h).
    *   The height of each cap is h = r - √(r² - a²).
    *   V_cap = (π(r - √(r² - a²))²/3)(3r - (r - √(r² - a²)))
    *   V_cap = (π(r - √(r² - a²))²/3)(2r + √(r² - a²))
    *   Total volume of the two caps = 2V_cap.

*   **Total removed volume:** V_removed = V_cyl + 2V_cap

This expression is also quite complex.

**5. The key insight:** The volume removed can also be calculated as the difference between the volume of the sphere of radius r and the volume of a sphere of radius rcosθ, where sinθ = a/r.

*   Volume of sphere of radius r: (4/3)πr³
*   Volume of sphere of radius rcosθ: (4/3)π(rcosθ)³
*   Difference: (4/3)πr³ - (4/3)πr³cos³θ = (4/3)πr³(1 - cos³θ)

This approach is significantly simpler. While the integral approach is valid, this geometric insight provides a much easier way to express the volume removed.

**Conclusion:**

The volume removed from the sphere is indeed the difference of the volumes of two spheres, one with radius *r* and the other with radius *rcosθ*, where *sinθ = a/r*. This is a result of the geometry of the intersection.