Problem of the week #37 - December 10th, 2012

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Jameson

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Thank you to MarkFL for this problem! It does require calculus so this week will be for upperclassmen in secondary school who are taking calculus.

What condition must hold on $\displaystyle a$ and $\displaystyle b$ so that the area bounded by the $\displaystyle x$-axis and the parabola $\displaystyle y=ax(b-x)$ which is revolved separately about both axes, will have equivalent resulting volumes.
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Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) soroban
2) Sudharaka

Once again, thank you to MarkFL for this interesting problem

Solution (from soroban):
The region $R$ is bounded by $y \,=\,ax(b-x)$ and the x-axis.

Assuming $a,b > 0$, the graph looks like this:
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|   .*.
|.*:::::*. b
- - - * - - - - * - -
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*|          *
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$R$ revolved about $x$-axis:

$\displaystyle V_x \;=\;\pi\int^b_0[ax(b-x)]^2\,dx \;=\;a^2\pi\int^b_0(b^2x^2 - 2bx^3 + x^4)\,dx$
$V_x \;=\;a^2\pi \left(\frac{b^2}{3}x^3 - \frac{1}{2}bx^4 + \frac{1}{5}x^5\right)\bigg]^b_0 \;=\;a^2\pi \left(\frac{b^5}{3} - \frac{b^5}{3} + \frac{b^5}{5}\right)$
Hence: .$V_x \;=\;\frac{\pi}{30}a^2b^5$

$R$ revolved about $Y$-axis:

$\displaystyle V_y \;=\;2\pi\int^b_0 x\cdot ax(b-x)\,dx \;=\;2a\pi\int^b_0(bx^2 - x^3)\,dx$

$V_y \;=\;2\pi a\left(\frac{b}{3}x^3 - \frac{1}{4}x^4\right)\bigg]^b_0 \;=\;2\pi a\left(\frac{b^4}{3}-\frac{b^4}{4}\right)$

Hence: .$V_y \;=\;\frac{\pi}{6}ab^4$

Since $$V_x = V_y$$, we have: .$\frac{\pi}{30}a^2b^5 \:=\:\frac{\pi}{6}ab^4$

Therefore: .$\boxed{ab \:=\:5}$

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