## Related questions with answers

A solid is generated by revolving about the x-axis the region bounded by the graph of the positive continuous function y = f(x), the x-axis, the fixed line x = a, and the variable line x = b, b > a. Its volume, for all b, is $b^{2}-a b.$ Find f(x).

Solution

VerifiedWe have a solid generated by revolving about the $x$-axis the region bounded by the graph of the positive continuous function $y=f\left(x\right)$, the $x$-axis, the fixed line $x=a$, and the variable line $x=b,\enskip b>a$. Its volume, for all $b$, is $b^2-ab.$ We have to find $f\left(x\right)$.

To find the volume of a solid, we need only observe that the cross-sectional area $A(x)$ is the area of a disk of radius $f(x)$, the distance of the planar region's boundary from the axis of revolution. Therefore, the volume of a solid is

$\begin{align*} V=\pi\int_a^b\left[f\left(x\right)\right]^2dx. \end{align*}$

We know that volume is $b^2-ab$. So, we have

$\begin{align*} \pi\int_a^x\left[f\left(t\right)\right]^2\enskip dt=x^2-ax\quad\left(\text{for all}\enskip x>a\right). \end{align*}$

Now, we can conclude that $f\left(x\right)$ is

$\begin{align*} \pi\left[f\left(x\right)\right]^2=\dfrac{d}{dx}(x^2-ax)=2x-a\quad\Rightarrow\quad f\left(x\right)=\sqrt{\dfrac{2x-a}{\pi}}. \end{align*}$

Hence, we get

$\boxed{f(x)=\sqrt{\dfrac{2x-a}{\pi}}.}$

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