In the next step of the derivation, we passed a horizontal plane through the sphere b up from the center. We let the radius of the cross section be x and formed a right triangle with hypotenuse r, as shown below.
The cross-sectional area of the shaded circle is πx2. Using the Pythagorean theorem for the right triangle, we get x2 + b2 = r2. Now solve for x2 and substitute it into the area expression.
What is the result?
π(r2 – b2)
π(r2 + b2)
π(b2 – r2)
2π(b2 + r2)
