Find the values of c such that the area of the region bounded by the parabolas y = 16x^2 − c^2 and y = c^2 − 16x^2 is 16/3.?

y₁ = 16x² - c² ← this is a parabola

y₂ = c² - 16x² ← this is another parabola

You can see that y₁ = - y₂, so the area bounded by the parabola y₁ and the x-axis is similar to the area bounded by the parabola y₂ and the x-axis.

(from x₁ to x₂) ∫ y₁.dx = [(16/3)]/2

(from x₁ to x₂) ∫ y₁.dx = 8/3 → where x₁ & x₂ are the x-intercept of the parabola

y₁ = 16x² - c² →you can find the x-intercept when: y = 0

16x² - c² = 0

16x² = c²

x² = c²/16

x² = (± c/4)²

x = ± c/4 ← you can see that there are 2 values for x

x₁ = - c/4

x₂ = c/4

Restart:

(from x₁ to x₂) ∫ y₁.dx = 8/3

(from x₁ to x₂) ∫ (16x² - c²).dx = 8/3

[(16/3).x³ - x.c²] (x₁ → x₂) = 8/3

[(16/3).(- c/4) - (- c/4).c²] - [(16/3).(c/4) - (c/4).c²] = 8/3

[- (4/3).c + (1/4).c³] - [(4/3).c - (1/4).c³] = 8/3

- (4/3).c + (1/4).c³ - (4/3).c + (1/4).c³ = 8/3

- (8/3).c + (1/2).c³ = 8/3

- 8c + (3/2).c³ = 8

- 16c + 3c³ = 16

3c³ - 16c - 16 = 0 → let: c = (u + v)

3.(u + v)³ - 16.(u + v) - 16 = 0

3.[(u + v)².(u + v)] - 16.(u + v) - 16 = 0

3.[(u² + 2uv + v²).(u + v)] - 16.(u + v) - 16 = 0

3.[u³ + u²v + 2u²v + 2uv² + uv² + v³] - 16.(u + v) - 16 = 0

3.[u³ + v³ + 3u²v + 3uv²] - 16.(u + v) - 16 = 0

3.[(u³ + v³) + 3uv.(u + v)] - 16.(u + v) - 16 = 0

3.(u³ + v³) + 9uv.(u + v) - 16.(u + v) - 16 = 0 → you factorize (u + v)

3.(u³ + v³) + (u + v).(9uv - 16) - 16 = 0 → suppose that: (9uv - 16) = 0 ← equation (1)

3.(u³ + v³) + (u + v).(0) - 16 = 0

3.(u³ + v³) - 16 = 0 ← equation (2)

You can get a system of 2 equations:

(1) : (9uv - 16) = 0

(1) : 9uv = 16

(1) : uv = 16/9

(1) : u³v³ = (16/9)³

(2) : 3.(u³ + v³) - 16 = 0

(2) : 3.(u³ + v³) = 16

(2) : u³ + v³ = 16/3

Let: U = u³

Let: V = v³

You can get a new system of 2 equations:

(1) : UV = (16/9)³ ← this is the product P

(2) : U + V = 16/3 ← this is the sum S

You know that the values S & P are the solutions of the following equation:

x² - Sx + P = 0

x² - (16/3).x + (16/9)³ = 0

Δ = (16/3)² - [4 * (16/9)³]

Δ = (256/9) - (16384/729)

Δ = 4352/729

Δ = 17 * (256/729)

Δ = 17 * (16/27)²

x₁ = [(16/3) + (16/27).√17]/2 = (8/3) + (8/27).√17 ← this is U

x₂ = [(16/3) - (16/27).√17]/2 = (8/3) - (8/27).√17 ← this is V

Recall: u³ = U → u = U^(1/3)

u = [(8/3) + (8/27).√17]^(1/3)

Recall: v³ = V → v = V^(1/3)

v = [(8/3) - (8/27).√17]^(1/3)

Recall: c = (u + v)

c = [(8/3) + (8/27).√17]^(1/3) + [(8/3) - (8/27).√17]^(1/3)

c ≈ 1.57248899865682 + 1.13055021643796

c ≈ 2,70303921509478