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 the y-axis. find the value of a

Answer 1

$ax-x^2=x(a-x)=0\implies x=0,x=a$

which means the parabola $y=ax-x^2$ intersects the x-axis at $x=0$ and $x=a$. Assume $a>0$.

Then the volume of the solid generated by revolving the region about the x-axis is

$\displaystyle\pi\int_0^a(ax-x^2)^2\,\mathrm dx=\dfrac{\pi a^5}{30}$

Revolving about the y-axis, the volume would be

$\displaystyle2\pi\int_0^ax(ax-x^2)\,\mathrm dx=\dfrac{\pi a^4}6$

The volumes are the same independent of which axis is taken as the axis of revolution, so

$\dfrac{\pi a^5}{30}=\dfrac{\pi a^4}6$

We're assuming $a>0$, so we can safely divide both sides by $\dfrac{\pi a^4}6$ to get $a=5$.