Find Equal Area Between y=x^2 and y=9

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Homework Statement


"Find a horizontal line y=k that divides the area between y=x^2 and y=9 into two parts"


Homework Equations





The Attempt at a Solution


Found intersection at (-3,9), (3,9)
Found total area to be 36, half(the area needed for each portion) to be 18. Don't know where to go from here.
 
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Find the two areas as a function of k.
 
christianjb said:
Find the two areas as a function of k.

Can you elaborate more on that? I'm not quite sure what you mean.
 
y=k divides the total area into two parts, A1=A1(k) and A2=A2(k). You need to find an expression for each area as a function of k and then find the value of k for which A1(k)=A2(k)
 
Don't be intimidated by the variable k -- the fact it's there changes nothing. You know how to compute areas, so compute the area of one of the portions.
 
What is the area of the region bounded by y= k, y= 9 and y= x2? For what value of k is that 18?

Even simpler: What is the area of the region bounded by y= 0, y= k and y= x2? For what value of k is that 18?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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