# Thread: diving an area into two equal regions

1. ## diving an area into two equal regions

So, the graph is $y=4-x^{2}$ and it is to be looked at in the first quadrant. In order to find the total area, I drew a line in the graph so that it would be separated into a triangle and a semi-circle. using Area of a Circle formula (divided by 2 since its a semi-circle), I got 2pi. I added that to the Area of a triangle formula and got 10.283. Now, is there an equation I would use to find the vertical line that cuts the region into two smaller regions whose areas are equal?

2. Originally Posted by TheMathTham
So, the graph is $y=4-x^{2}$ and it is to be looked at in the first quadrant. In order to find the total area, I drew a line in the graph so that it would be separated into a triangle and a semi-circle. using Area of a Circle formula (divided by 2 since its a semi-circle), I got 2pi. I added that to the Area of a triangle formula and got 10.283. Now, is there an equation I would use to find the vertical line that cuts the region into two smaller regions whose areas are equal?
a parabola does not separate into those two geometric figures.

$\textcolor{blue}{\int_0^k 4-x^2 {dx} = \int_k^2 4-x^2 {dx}}$

3. Originally Posted by skeeter
a parabola does not separate into those two geometric figures.

$\textcolor{blue}{\int_0^k 4-x^2 {dx} = \int_k^2 4-x^2 {dx}}$
ah ok. good piece of information to know.

So I found that the $\int_{0}^{2}4-x^{2}dx$ was equal to 5.333. So by dividing by 2, i found that 2.667 was the area of each region. Then I set up $F(a_{1})-F(b_{1})=F(a_{2})-F(b_{2})$.

$(4k-\tfrac{1}{3}k^{3})-(0)=(5\tfrac{1}{3})-(4k-\tfrac{1}{3}k^{3})$. Which means that $8k-\tfrac{2}{3}k^{3}=5\tfrac{1}{3}$.

How do I solve it from there?

4. $\int_0^k 4-x^2 = \frac{8}{3}$

$\left[4x - \frac{x^3}{3}\right]_0^k = \frac{8}{3}
$

$4k - \frac{k^3}{3} = \frac{8}{3}$

$12k - k^3 = 8$

$0 = k^3 - 12k + 8$

using a calculator ... $k \approx 0.6946$

5. Originally Posted by skeeter
$\int_0^k 4-x^2 = \frac{8}{3}$

$\left[4x - \frac{x^3}{3}\right]_0^k = \frac{8}{3}
$

$4k - \frac{k^3}{3} = \frac{8}{3}$

$12k - k^3 = 8$

$0 = k^3 - 12k + 8$

using a calculator ... $k \approx 0.6946$
Is there a way to find that out non-graphically?

6. Originally Posted by TheMathTham
Is there a way to find that out non-graphically?
sure ...

The "Cubic Formula"

... if you're ready to do the extensive algebra required.

7. Originally Posted by skeeter
sure ...

The "Cubic Formula"

... if you're ready to do the extensive algebra required.
...good ol' graphs