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=(\pi a^5)/(30)`

Revolving about the y-axis, the volume would be


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

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


`(\pi a^5)/(30)=\frac{\pi a^4}6`

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