MHB Calc Help: Find b to Divide Region into 2 Equal Areas

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To find the value of b that divides the area between the curves y = 16x^2 and y = 25 into two equal parts, the area A can be expressed as A = 2∫₀²⁵ x dy. By solving for x from y = 16x², it is determined that A = (1/2)∫₀²⁵ y^(1/2) dy. The goal is to find b such that ∫₀ᵇ y^(1/2) dy equals ∫ᵇ²⁵ y^(1/2) dy, leading to the equation b^(3/2) = 125/2. Ultimately, the solution for b is b = 25/√[3]{4}.
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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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Hello eyeheartglitter,

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

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

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

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

Hence:

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

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

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

Multiply through by 2:

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

Apply the FTOC:

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

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

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

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

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

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

$$b=\frac{25}{\sqrt[3]{4}}$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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