A particular bowler wants a "balance hole" drilled into her ball. A drill bit with a diameter of 1¼ inches is used to drill a single hole into the bowling ball, which happens to have a circumference of 27 inches.
How much of the ball's surface area is removed after drilling?
How much of the ball's surface area is removed after drilling?
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It's going to remove approximately π r² = π (5/8)² = 1.23 in²
The precise answer can be obtained by integration in spherical coordinates:
S = ∫ ∫ R² sin(ϕ) dϕ dθ
{(ϕ,θ) | 0 ≤ ϕ ≤ arcsin(5π/108) ; 0 ≤ θ ≤ 2π}
... note ϕ = arcsin(d/D) ; d = 5/4 ; D = 27/π
= 729/(2π) ∫ sin(ϕ) dϕ
{(ϕ) | 0 ≤ ϕ ≤ arcsin(5π/108)}
= 729/(2π) (1 - √(1 - (5π/108)²))
Answer: 729/(2π) (1 - √(1 - (5π/108)²)) .... ≈ 1.233744158 in²
The precise answer can be obtained by integration in spherical coordinates:
S = ∫ ∫ R² sin(ϕ) dϕ dθ
{(ϕ,θ) | 0 ≤ ϕ ≤ arcsin(5π/108) ; 0 ≤ θ ≤ 2π}
... note ϕ = arcsin(d/D) ; d = 5/4 ; D = 27/π
= 729/(2π) ∫ sin(ϕ) dϕ
{(ϕ) | 0 ≤ ϕ ≤ arcsin(5π/108)}
= 729/(2π) (1 - √(1 - (5π/108)²))
Answer: 729/(2π) (1 - √(1 - (5π/108)²)) .... ≈ 1.233744158 in²