Question

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.

Answer 1

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 \(\left(\dfrac{a}{2},\dfrac{a^2}{4}\right)\).