Area and Volume

Let $A_{y} = \large \int_0^{y} f(y) \, dy$ and $V_{y} = \large \int_0^{y} \pi \cdot f(y)^2 \, dy$ . If $V_y = A_y + A_{y}^2$ , then let us solve for the function $f(y)$ by differentiating both sides with respect to $y:$

$\frac{\partial}{\partial y} \cdot V_y = \frac{\partial}{\partial y} \cdot (A_y + A_{y}^2) \Rightarrow \pi f(y)^2 = f(y) + 2f(y) \cdot \large \int_0^{y} f(y) \, dy$ ;

or $\pi f(y) = 1 + 2 \cdot \large \int_0^{y} f(y) \, dy$ ;

or $\frac{\pi f(y) - 1}{2} = \large \int_0^{y} f(y) \, dy$ (i); with the initial condition $f(0) = \frac{1}{\pi}$ at $y = 0.$

Now differentiating (i) with respect to $y$ yields $\frac{\pi}{2} f'(y) = f(y) \Rightarrow \frac{f'(y)}{f(y)} = \frac{2}{\pi} \Rightarrow ln|f(y)| = \frac{2y}{\pi} + C$

and the initial condition $f(0) = \frac{1}{\pi} \Rightarrow C = ln(\frac{1}{\pi})$ , which finally results in $f(y) = \frac{1}{\pi} e^{\frac{2y}{\pi}}$ .

We now solve for $\frac{V_y}{A_y}$ at $y = \frac{\pi}{2}$ , or:

$\frac{V_y}{A_y} = \frac{A_y + A_{y}^2}{A_y} = 1 + A_y = 1 + \large \int_0^{\frac{\pi}{2}} \frac{1}{\pi} e^{\frac{2y}{\pi}} \, dy$ ;

or $1 + \frac{1}{2} e^{\frac{2y}{\pi}}$ from $0 \le y \le \frac{\pi}{2}$ ;

or $\boxed{\frac{e + 1}{2}} \approx 1.859$