A solid shape described as the region bounded by revolved around the y-axis, needs to have 1/4 of its volume trimmed off to meet a weight restriction. This will be done by reducing the radius of the piece - in other words, the resulting shape will be described as above, but with the region also bounded by x = q. Find the resulting radius, q.

Question

A solid shape described as the region bounded by  revolved around the y-axis, needs to have 1/4 of its volume trimmed off to meet a weight restriction. This will be done by reducing the radius of the piece - in other words, the resulting shape will be described as above, but with the region also bounded by x = q. Find the resulting radius, q.

Quizplus · Accepted Answer

Let V be the volume of the original solid, and let r be its radius. Then the volume of the quarter that needs to be removed is (1/4) V. We want to find the value of r when (3/4) V is the volume of the remaining solid. 
The volume of a solid obtained by revolving a region bounded by y = f(x), x = a, x = b, and the x-axis around the y-axis is given by the integral:
V = ∫[a,b] πf(x)² dx. 
In this case, we have f(x) = sqrt(r² - x²) for -r ≤ x ≤ r, and f(x) = 0 for r ≤ x ≤ q. The interval of integration is then -r ≤ x ≤ q. Therefore, we have:
V = ∫[-r,q] πf(x)² dx 
  = π∫[-r,r] (r² - x²) dx + π∫[r,q] 0 dx  
  = 2πr³/3 + π(r² - q²)r. 
We want to solve for r such that (3/4) V = 2πr³/3 + π(r² - q²)r. We know that V = πr²(4/3)r = (4/3)πr³, so we have:
(3/4)(4/3)πr³ = 2πr³/3 + π(r² - q²)r
3πr³ = 8πr³/3 + 4π(r² - q²)r
9r³ = 8r³ + 4r²(r - q) 
r³ - 4r²q - 8r³ = 0 
r²(2r - 4q) = 0. 
Since r ≠ 0, we have:
2r - 4q = 0
r = 2q. 
Therefore, q = (1/2) r, and the correct choice is B.

A Solid Shape Described as the Region Bounded by