Find the volume generated by rotating the region bounded by the given curves about the specified axis. The curve is given by,

$$y = x^3$$

with bounds $y=8$ and $x=0$, and about the rotating axis $ x=3$.

Here is my current attempt:

$$V= 2\pi \int_0^8 (x-3)x^3=2\pi \int_0^8 (x^4 - 3x^3) dx$$

$$= 2\pi \left( \frac{x^5}{5} -\frac{3x^4}{4} \right)_0^8$$

The answer I get is a ridiculous number. I am unsure on how to get the limits of integration from these question in general so that makes it more challenging.

Question: 1. How do I find the limits of integration in questions such as these? 2. What am I doing wrong in this question?

  • $\begingroup$ Just click on "Edit" to see the MathJax code that corresponds to the equations. For reference see MathJax quick guide. $\endgroup$
    – user21820
    Commented Sep 15, 2019 at 14:43

2 Answers 2


Let $u= x-3$ and the curve becomes


So, the volume is equivalent to that of rotating the above curve around $u=0$. Since $y(-1) = 8$, the bounds for integration is $(-3,-1)$. The corresponding integral is

$$V = \int_{-3}^{-1} 2\pi |u|(8-y)du$$ $$=-\int_{-3}^{-1} 2\pi u[8-(u+3)^3]du=\frac{264}{5}\pi$$

Of course, you could integrate without the above variable change. Then, the correct integral expression should be,

$$V = \int_{0}^{2} 2\pi (3-x)(8-x^3)dx=\frac{264}{5}\pi$$

Note the upper bound is $x=2$ since the volume is capped at $y=x^3=8$ and the volume is between $y=8$ and $y=x^3$, hence $8-x^3$ in the integrand.

  • $\begingroup$ First of, thank you for changing my numbers and letters into an equation in my question. How did you do that if you don't mind me asking? $\endgroup$
    – Aeryes
    Commented Sep 14, 2019 at 20:06
  • $\begingroup$ I added some explanation in the answer to address the problem in your approach. $\endgroup$
    – Quanto
    Commented Sep 14, 2019 at 20:08
  • $\begingroup$ Why do you have 8-x^3 when (2)^3 is 8. Do you subtract 8 from the y function in order to account for the volume cap being 8? $\endgroup$
    – Aeryes
    Commented Sep 14, 2019 at 20:08
  • $\begingroup$ that is correct $\endgroup$
    – Quanto
    Commented Sep 14, 2019 at 20:09
  • $\begingroup$ Im honestly confused in general on how to select the limits of integration. I am doing another problem similar to this one and followed the steps you did in this problem but I keep getting an incorrect answer. $\endgroup$
    – Aeryes
    Commented Sep 14, 2019 at 20:17

I'm thinking you should integrate from $x=0$ to $x=2$. Drawing a picture should help.

So $2\pi\int_0^2(3-x)(8-x^3)\operatorname dx=2\pi\int_0^2(24-8x-3x^3+x^4)\operatorname dx=2\pi[24x-4x^2-\frac{3x^4}4+\frac{x^5}5]_0^2=2\pi(48-16-12+\frac{32}5)=\frac{264\pi}5$.

  • $\begingroup$ I got that answer of 56pi/5 after trying again by using 2 before you commented here. However the book suggests that the correct answer is 264pi/5 $\endgroup$
    – Aeryes
    Commented Sep 14, 2019 at 20:05
  • 1
    $\begingroup$ You forgot to do (8-x^3) $\endgroup$
    – Aeryes
    Commented Sep 14, 2019 at 20:20
  • $\begingroup$ It should be $8-x^3$. I made a mistake. Wrong region. $\endgroup$
    – user403337
    Commented Sep 14, 2019 at 20:21

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