# Horizontal line that divides the area in two equal parts

• June 19th 2011, 05:44 PM
Zellator
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 $y=x^2$ and $y=9$ in two equal parts.

What I thought:
Get the bounded area.
$\int_{-3}^{3} 9-x^2$

... $=36$

Divide it by two $=18$
and calculate that with line k equaling 18
$\int_{-3}^{3} 9-x^2-k=18$
$36-3k-3k=18$
$k=3$
But, if we calculate
$\int_{-3}^{3} 3-x^2=0$
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! :)
• June 19th 2011, 05:49 PM
pickslides
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 $y=x^2$ and $x=9$ in two equal parts.

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

$\displaystyle \int_{0}^{9}x^2~dx = 243$
• June 19th 2011, 06:59 PM
Zellator
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

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

Whaat, yes you are right. But I got the x for the y there!
I'm really sorry. hahahah :D
I really didn't think there would be anyone online now.
• June 19th 2011, 07:05 PM
TKHunny
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 $\int_{-\sqrt{k}}^{\sqrt{k}}k - x^{2}\;dx = 18$?
• June 20th 2011, 04:05 AM
Zellator
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 $\int_{-\sqrt{k}}^{\sqrt{k}}k - x^{2}\;dx = 18$?

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

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

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