use disk method to find volume of solid?

Question

x^2+y^2=1 and the line y=2

Answer 1

This has to be done as two integrals, because x² + y² = 1 can't be expressed as a single function of y:

y² = 1 - x²
y = ±√(1 - x²)

The "outer radius" would be 2 + √(1 - x²), and the "inner radius" will be 2 - √(1 - x²).

1                          ...    1
∫ π(2 + √(1 - x²))² dx - ∫ π(2 - √(1 - x²))² dx =
-1                         ...  -1

1                                     .           1
∫ π(4 + 2√(1 - x²) + (1 - x²)) dx - ∫ π(4 - 2√(1 - x²) + (1 - x²)) dx
-1                                   .           -1

There's no point in evaluating all that, because the 4s and the 1 - x²s will cancel each other in the end, so...

1                           1
∫ 2π√(1 - x²) dx + ∫ 2π√(1 - x²) dx =
-1                         -1

1
∫ 4π√(1 - x²) dx
-1

According to my chart of integrals, this integral becomes:

4π[x/2√(1 - x²) + 1/2 arcsin(x)] evaluated from -1 to 1 =

4π[1/2√(1 - 1²) + 1/2 arcsin(1) - (-1/2√(1 - 1²) + 1/2 arcsin(-1))] =

4π[1/2 arcsin(1) - 1/2 arcsin(-1)] =

4π[1/2 (π/2) - 1/2 (-π/2)] =

4π[π/4 + π/4] =

4π[π/2] = 2π²

If I didn't mess up somewhere.