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

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MarkFL

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Feb 24, 2012
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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.
 
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MarkFL

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Feb 24, 2012
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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 \(\displaystyle \frac{3}{2}\):

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

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

\(\displaystyle 2b^{\frac{3}{2}}=5^3\)

\(\displaystyle b^{\frac{3}{2}}=\frac{5^3}{2}\)

\(\displaystyle b=\frac{25}{\sqrt[3]{4}}\)