the region in the first quadrant enclosed between the graph of y=ax-x^2 and the x-axis generates the same volume whether it is revolved about the x-axis or y-axis. find the value of a.

Question

anonymous · Answer

Completing the square, you have
$$\begin{align*}ax-x^2&=-\left(x^2-ax\right)\[1ex]&=-\left(x^2-ax+\frac{a^2}{4}-\frac{a^2}{4}\right)\[1ex]&=-\left(\left(x-\frac{a}{2}\right)^2-\frac{a^2}{4}\right)\[1ex]&=\frac{a^2}{4}-\left(x-\frac{a}{2}\right)^2\end{align*}$$which is a concave parabola wth vertex $\left(\dfrac{a}{2},\dfrac{a^2}{4}\right)$.
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