Find the volume of the solid obtained by rotating the region bounded by the curves $y=x^2$, $0<x<3$, $y=9$ and $x=0$ about the $y$-axis


I apologize in advance for not explaining the Method of Cylindrical Shells (Shell Method) in depth when it comes to revolutions of solids around axes. However, to those familiar with the Shell Method, we need to recall that the differential volume element of a shell is given by the relation:

$dV = 2\pi r h t$, where $dV$ is the differential volume of a given shell; $r$ is that shell's radius; $h$ is the height of the shell (usually corresponding to the function under which the region is generated; and $t$ is the thickness of said shell (always a differential length with respect to the axis of rotation).

For this problem, $r = x$ because you are revolving the region around the line $x=0$, more commonly known as the $y$-axis. $h = f(x) = x^2$ because your region is from the $x$-axis up to the function $y=x^2$; $t = dx$ because the differential thickness of a given shell is $\Delta x$.

Altogether, we have $dV = 2\pi r h t = 2 \pi (x) (x^2) (dx) = 2 \pi x^3 dx$. We may now set up the integral where $x$ varies from 0

$\int dV$ = $\int_0^3 2 \pi x^3 dx$ = $2 \pi \left[ \dfrac{x^4}{4} \right]^{x=3}_{x=0}$ = $\dfrac{\pi}{2} (3^4 - 0^4)$ = $\dfrac{81\pi}{2}$

The Shell Method is my favorite method for rotating regions around axes for volumes because it is so much more reliable than the Disk or Washer Methods, though at the cost spending more effort to set up the correct integral.

In general, if the function $h(x)$ is rotated about the $y$-axis (vertical axis) from $x=a$ to $x=b$, the volume of the region rotated will be $$V = 2 \pi \int_a^b xh(x) \, dx$$ If the function $h(y)$ is rotated about the $x$-axis (horizontal axis) from $y=c$ to $y=d$, the volume of the region rotated will be $$V = 2 \pi \int_c^d yh(y) \, dy$$.

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Undergraduate student majoring in Biology. My goal is to continue into medical school.

Updated on October 19, 2020


    JXXII about 3 years

    I got an answer of $\displaystyle\frac{9\pi}{2}$. Is this correct?

    • Prahlad Vaidyanathan
      Prahlad Vaidyanathan about 10 years
      These forums are not meant to check your HW answers - they are meant to help you when you get stuck (conceptually) when trying to solve said homework.
    JXXII about 10 years
    Great! This helps a whole lot. No need to apologize for that explanation, instead I thank you for that, I needed it.