# Find the number b such that the line y=b divides the region...?

## Find the number $b$ such that the line y=b divides the region bounded by the curves $y = {x}^{2}$ and $y = 4$ into two regions with equal area.

Sep 19, 2017

Find the area of the region first.

#### Explanation:

The area of the region is given by
${\int}_{-} {2}^{2} \left(4 - {x}^{2}\right) \mathrm{dx} = 2 {\int}_{0}^{2} \left(4 - {x}^{2}\right) \mathrm{dx}$
$= 2 {\left(4 x - {x}^{3} / 3\right)}_{0}^{2}$
$= \frac{32}{3}$

Second, $y = b$ intersects the curve $y = {x}^{2}$ when
$x = \pm \sqrt{b}$.
Third, we want to find b such that
${\int}_{-} {\sqrt{b}}^{\sqrt{b}} \left(b - {x}^{2}\right) \mathrm{dx} = \frac{16}{3}$
This will occur if and only if
${\int}_{0}^{\sqrt{b}} \left(b - {x}^{2}\right) \mathrm{dx} = \frac{8}{3}$
Integrate:
${\int}_{0}^{\sqrt{b}} \left(b - {x}^{2}\right) \mathrm{dx} = {\left(b x - {x}^{3} / 3\right)}_{0}^{\sqrt{b}}$
$= b \sqrt{b} - \frac{b \sqrt{b}}{3}$
$= \frac{2 b \sqrt{b}}{3}$

Now set this equal to $\frac{8}{3}$.
$\frac{2 b \sqrt{b}}{3} = \frac{8}{3}$
$b \sqrt{b} = 4$
${b}^{\frac{3}{2}} = 4$
$b = \sqrt[3]{16}$
or $b = 2 \sqrt[3]{2}$