Find the values of c such that the area of the region bounded by the parabolas y = 4x^2 − c^2 and y = c^2 − 4x^2 is 256/3.

Question

Find the values of c such that the area of the region bounded by the parabolas y = 4x² − c² and y = c² − 4x²

 is 256/3.

Answer 1

The curves intersect when 4x² - c² = c² - 4x^2.

8x² = 2c²

x² = (1/4)c²

x = ± (1/2)c

By symmetry, the desired area = 2∫_{(from 0 to (1/2)c)} [(c² - 4x²) - (4x² - c²)]dx

= 2∫_{(from 0 to (1/2)c)} [ 2c² - 8x²]dx = 2[2c²x - (8/3)x³]_{(from 0 to (1/2)c)}

= 2c³ - (2/3)c³ = (4/3)c³ = 256/3

c³ = 64

So, c = 4.