# eyeheartglitter's question at Yahoo! Answers regarding dividing a region into two equal parts

#### MarkFL

##### Administrator
Staff member
Here is the question:

Calc help! One try left!?

Find the number b such that the line y = b divides the region bounded by the curves y = 16x^2 and y = 25 into two regions with equal area.
I have posted a link there to this thread so the OP can see my work.

#### MarkFL

##### Administrator
Staff member
Hello eyeheartglitter,

The given bounded area $A$ may be expressed as:

$$\displaystyle A=2\int_0^{25} x\,dy$$

From $$\displaystyle y=16x^2$$ we find (by taking the positive root):

$$\displaystyle x=\frac{y^{\frac{1}{2}}}{4}$$

Hence:

$$\displaystyle A=\frac{1}{2}\int_0^{25} y^{\frac{1}{2}}\,dy$$

Now, we wish to find some number $b$ such that:

$$\displaystyle \frac{1}{2}\int_0^{b} y^{\frac{1}{2}}\,dy=\frac{1}{2}\int_b^{25} y^{\frac{1}{2}}\,dy$$

Multiply through by 2:

$$\displaystyle \int_0^{b} y^{\frac{1}{2}}\,dy=\int_b^{25} y^{\frac{1}{2}}\,dy$$

Apply the FTOC:

$$\displaystyle \frac{2}{3}\left[y^{\frac{3}{2}} \right]_0^b=\frac{2}{3}\left[y^{\frac{3}{2}} \right]_b^{25}$$

Multiply through by $$\displaystyle \frac{3}{2}$$:

$$\displaystyle \left[y^{\frac{3}{2}} \right]_0^b=\left[y^{\frac{3}{2}} \right]_b^{25}$$

$$\displaystyle b^{\frac{3}{2}}=5^3-b^{\frac{3}{2}}$$

$$\displaystyle 2b^{\frac{3}{2}}=5^3$$

$$\displaystyle b^{\frac{3}{2}}=\frac{5^3}{2}$$

$$\displaystyle b=\frac{25}{\sqrt{4}}$$