Marlene

Mathematics

18 July, 12:32

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)