# diving an area into two equal regions

• Feb 11th 2010, 06:40 PM
TheMathTham
diving an area into two equal regions
So, the graph is $\displaystyle y=4-x^{2}$ 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?
• Feb 11th 2010, 07:29 PM
skeeter
Quote:

Originally Posted by TheMathTham
So, the graph is $\displaystyle y=4-x^{2}$ 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.

$\displaystyle \textcolor{blue}{\int_0^k 4-x^2 {dx} = \int_k^2 4-x^2 {dx}}$
• Feb 12th 2010, 08:01 AM
TheMathTham
Quote:

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

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

ah ok. good piece of information to know.

So I found that the $\displaystyle \int_{0}^{2}4-x^{2}dx$ 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 $\displaystyle F(a_{1})-F(b_{1})=F(a_{2})-F(b_{2})$.

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

How do I solve it from there?
• Feb 12th 2010, 09:12 AM
skeeter
$\displaystyle \int_0^k 4-x^2 = \frac{8}{3}$

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

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

$\displaystyle 12k - k^3 = 8$

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

using a calculator ... $\displaystyle k \approx 0.6946$
• Feb 12th 2010, 11:09 AM
TheMathTham
Quote:

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

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

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

$\displaystyle 12k - k^3 = 8$

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

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

Is there a way to find that out non-graphically?
• Feb 12th 2010, 01:48 PM
skeeter
Quote:

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.
• Feb 12th 2010, 05:07 PM
TheMathTham
Quote:

Originally Posted by skeeter
sure ...

The "Cubic Formula"

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

...good ol' graphs (Tongueout)