Find the value of c

Question

Find the value of c for which the area enclosed by the curves y=c - x^2 and y=x^2 - c is equal to 128.

Answer 1

Find the value of c

Find the value of c for which the area enclosed by the curves y=c - x^2 and y=x^2 - c is equal to 128.

Hello Maisie,

Use the fact that the graphs are both symmetrically shaped parabolas. The only difference between the two curves is: one parabola opens downward from a point +c on the y-axis, the other parabola opens upward from a point -c on the y-axis. Set the integral equal to 128 since it is the area under the curve. I found it was easier to set up the problem using Δy on the interval from -c to +c, insteaad of using Δx on the interval from -x to x. Although, I am sure this can be solved using either Δx or Δy. Just be sure to take the upper parabola minus the lower parabola:

Using Δy in the integral, means we need to solve for x in terms of y and plug that into the area formula:

Upper parabola:

y = c - x²

x = +/-√(c - y)

Lower parabola:

y = x² - c

x = +/-√(c + y)

The following equation should be the correct expression to solve for c:

Area = ∫_-c^c [√(c - y) - √(c + y)]Δy

128 = ∫_-c^c [√(c - y) - √(c + y)]Δy

However, I am lazy and see another way to approach it.

Important Note: To make the problem simpler, I will be using only the upper parabola and integrating on the interval from y = 0, to the value +c on the y-axis. This shortcut can be done as the parabolas are equal in shape and size and meet on the x-axis. Therefore, I need only use half of the area, which is 64.

64 = ∫₀^+c [ √(c - y) ]Δy

64 = ∫₀^+c [ √(c - y) ]dy

* Using u-substitution:

Let u = c - y , then find the derivative of u or du:

d [u = c - y]

du = - dy

-du = dy .....* Now substitute into area integral:

64 = ∫₀^+c [ √(u) ] *(-du)

64 = - ∫₀^+c [ u^1/2 ]*du

64 = - 2/3 [(u)^3/2 ]₀^+c

64 = - 2/3 [(c - c)^3/2 - (c - 0)^3/2 ]

-96 = - (c)^3/2

(c)^3/2 = 96

c = 96^2/3

c = 8 * 18^1/3

The value of c = +/- 96^2/3 or 8 * 18^1/3