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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

User Monta
by
7.7k points

1 Answer

3 votes

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.
User Eugene Tolmachev
by
8.5k points
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