Find a horizontal line y=k that divides the area...?

Question

Find a horizontal line y=k that divides the area between y=x^2 and y=49 into two equal parts. Carry out all calculations exactly and round the final answer to two decimals places.
What does k equal?
Thanks so much for the help everybody, I just can't wrap my mind around this one :)

Answer 1

First, you have to find the area between the first two curves, then half it. Then set up an integral between the y=k curve and y=x^2 and set it equal to the half-area.

Integral(-7 to 7) of 49-x^2 * dx OR 2*Integral(0 to 7) of 49-x^2 * dx because these are symmetric functions (and evaluation is a breeze.

2*(49x-x^3/3) evaluated at 7 = 2*(49*4 - 7^3/3) = 2*(196-343/3) = 2*(588-343)/3 = 2*(254/3)

But we want half, so drop the two to get 254/3. Now set up a new integral and set it equal to that value.

2*Integral(0 to sqrt(k)) of k-x^2 * dx = 254/3

2*(k*x - x^3/3) evaluated at sqrt(k) = 254/3

2*(k^(3/2) - k^(3/2)/3) = 254/3
(2/3)*k^(3/2) = 254/6
k^(3/2) = (254/6)*(3/2) = 127/2
k = (127/2)^(2/3) =15.915......

So the line y= 15.915.... cuts the area between the curves y=49 and y=x^2 in half.

Viola!

Answer 2

The curves intersect at:
x^2=49
=>x=(+-)7
Therefore total area is:

A=|2∫(49-x^2)dx| ......with limits from x=-7 to x=0
=>A=|2(49x - x^3/3)|.....with limits from x=-7 to x=0
=>A=2[0-(-343 + 343/3)]=2[2*343]/3=1372/3 unit^2=457.33 unit^2


Now, y=k and the parabola will intersect at x=(+-)sqrt(k)

If this is dividing the area in half, then upper half should be equal to lower half.

Lower area=2∫(k-x^2)dx............with limits x=-7 to x=0
=2(kx- x^3/3)=-2(-7k + 343/3)

Total Area - Area of Lower half=Area of Upper Half

Condition given, 2(Area of Lower half)=Total Area

=>2(14k - 686/3) =1372/3

=>42k - 686 = 686
=>k=1372/42=98/3=32.67 units

Answer 3

in case you sketch the parabola you'll see it truly is chop up into to factors by y = 0, its line of symmetry. So ok = 0 The y = 9 area is irrelevant to the exercising. Now it that line replaced into y = 2x + 3, then it truly is going to change right into a harder exercising. fortunately for you it truly is not. ProfRay

Answer 4

i think you need to find the integral of both equations and then set them equal to each other. Not sure since it's been a couple of years since I've done calculus!