What is a benchmark in mathematics
1 answer:
You might be interested in
The expression on the left side describes a parabola. Factorize it to determine where it crosses the y-axis (i.e. the line x = 0) :
-3y² + 9y - 6 = -3 (y² - 3y + 2)
… = -3 (y - 1) (y - 2) = 0
⇒ y = 1 or y = 2
Also, complete the square to determine the vertex of the parabola:
-3y² + 9y - 6 = -3 (y² - 3y) - 6
… = -3 (y² - 3y + 9/4 - 9/4) - 6
… = -3 (y² - 2•3/2 y + (3/2)²) + 27/4 - 6
… = -3 (y - 3/2)² + 3/4
⇒ vertex at (x, y) = (3/4, 3/2)
I've attached a sketch of the curve along with one of the shells that make up the solid. For some value of x in the interval 0 ≤ x ≤ 3/4, each cylindrical shell has
radius = x
height = y⁺ - y⁻
where y⁺ refers to the half of the parabola above the line y = 3/2, and y⁻ is the lower half. These halves are functions of x that we obtain from its equation by solving for y :
x = -3y² + 9y - 6
x = -3 (y - 3/2)² + 3/4
x - 3/4 = -3 (y - 3/2)²
-x/3 + 1/4 = (y - 3/2)²
± √(1/4 - x/3) = y - 3/2
y = 3/2 ± √(1/4 - x/3)
y⁺ and y⁻ are the solutions with the positive and negative square roots, respectively, so each shell has height
(3/2 + √(1/4 - x/3)) - (3/2 - √(1/4 - x/3)) = 2 √(1/4 - x/3)
Now set up the integral and compute the volume.
Substitute u = 1/4 - x/3, so x = 3/4 - 3u and dx = -3 du.
