Horizontal line that divides the area in two equal parts

Hi Forum!

I have found this question:

*Find the horizontal line y=k that divides the area bounded by the curves *$\displaystyle y=x^2$* and *$\displaystyle y=9$* in two equal parts*.

What I thought:

Get the bounded area.

$\displaystyle \int_{-3}^{3} 9-x^2$

... $\displaystyle =36$

Divide it by two $\displaystyle =18$

and calculate that with line k equaling 18

$\displaystyle \int_{-3}^{3} 9-x^2-k=18$

$\displaystyle 36-3k-3k=18$

$\displaystyle k=3$

But, if we calculate

$\displaystyle \int_{-3}^{3} 3-x^2=0$

This is because we are going to subtract the same area more than once, right?

The answer is probably something around 3 and 6 then.

I am not sure how to continue.

Thanks! :)

Re: Horizontal line that divides the area in two equal parts

Quote:

Originally Posted by

**Zellator** Hi Forum!

I have found this question:

*Find the horizontal line y=k that divides the area bounded by the curves *$\displaystyle y=x^2$* and *$\displaystyle x=9$* in two equal parts*.

If this is the way you want to solve the problem then I think you mean

$\displaystyle \displaystyle \int_{0}^{9}x^2~dx = 243$

Re: Horizontal line that divides the area in two equal parts

Quote:

Originally Posted by

**pickslides** If this is the way you want to solve the problem then I think you mean

$\displaystyle \displaystyle \int_{0}^{9}x^2~dx = 243$

Whaat, yes you are right. But I got the x for the y there!

I'm really sorry. hahahah :D

Thanks for the reply though,

I really didn't think there would be anyone online now.

Re: Horizontal line that divides the area in two equal parts

Maybe **y** = 9? If so, why would the limits be [-3,3] when 'k' is introduced?

Maybe $\displaystyle \int_{-\sqrt{k}}^{\sqrt{k}}k - x^{2}\;dx = 18$?

Re: Horizontal line that divides the area in two equal parts

Quote:

Originally Posted by

**TKHunny** Maybe **y** = 9? If so, why would the limits be [-3,3] when 'k' is introduced?

Maybe $\displaystyle \int_{-\sqrt{k}}^{\sqrt{k}}k - x^{2}\;dx = 18$?

Hi TKHunny!

I just solved it, it was a lot of work but I arrived at $\displaystyle \frac{9}{\sqrt[3]4}$

$\displaystyle \frac{\sqrt[3]{54^2}}{\sqrt[3]{16}} * \frac{\sqrt[3]{16}}{\sqrt[3]{16}}$

...$\displaystyle \frac{9}{\sqrt[3]4}$

Thanks for your help :)

All the best