fun calculus problem...need to check my answer?

Question

the two functions y=x^2 and y=9 create an area...find horizontal line h=k such that it divides the area into two equal halves

Answer 1

I'd start by finding a general equation for the area bounded by y=x² and y=k, for any given k.

Area in the rectangle bounded by y=0, y=k, x=±sqrt(k) is: k*2*sqrt(k); or k^(3/2)

Area under the parabola (down to the x-axis) is just the integral of x²dx from -sqrt(k) to +sqrt(k). This is: (2/3)k^(3/2)

So the area inside the "cup" of the parabola, is the first area minus the second area. This is:
k^(3/2) – (2/3)k^(3/2) = (1/3)k^(3/2)

When k=9, that's (1/3)9^(3/2) = 9.

To reduce that by half, you'd have to reduce k by a factor of 2^(2/3), or 4^(1/3). That is:
k′ = 9 / (4^(1/3)), or about 5.67.

That's the height of the line you're looking for.

Answer 2

The area between y = x^2 and y = k is:

integral (k - x^2) dx from x = -sqrt(k) to x = sqrt(k)

(since y = k, the intersection of the two curves happen at k = x^2, or when x = +/- sqrt(k))

That area is said to be half of the area between y = x^2 and y = 9, which is:

integral (9 - x^2) dx from x = -3 to x = 3.

Thus, we have:

integral (k - x^2) dx from x = -sqrt(k) to x = sqrt(k)
= 1/2 * integral (9 - x^2) dx from x = -3 to x = 3

=> by symmetry,

2 * integral (k - x^2) dx from x = 0 to x = sqrt(k)
= 1/2 * 2 * integral (9 - x^2) dx from x = 0 to x = 3

=>

2 * (kx - 1/3 * x^3) evaulated from x = 0 to x = sqrt(k)
= 9x - 1/3 * x^3 from x = 0 to x = 3

=>

2 * (k*sqrt(k) - 1/3 * k*sqrt(k)) = 27 - 1/3 * 27

=>

4/3 * k*sqrt(k) = 18

=>

k*sqrt(k) = 27/2

=> square both sides

k^3 = 729/4

=>

k = 8/(cuberoot 4)

=> rationalize the denominator

k = 2 * cuberoot(16)

which is approximately 5.04, which makes intuitive sense, since k should be greater than half of 9, or 4.5.

I hope it is right! A ver nice calculus problem indeed.