What is the volume of the solid generated by revolving about the $y$-axis, the area bounded by the parabola $y=(x-1)^2-1, y=2$, and the $y$-axis?

I'm confused with the outer and inner radius. I am a little bit skeptical in using the formula of the volume of solid revolution. I don't know where to substitute those values in the formula.


closed as off-topic by Namaste, Claude Leibovici, Henrik, mathreadler, kingW3 Mar 25 '17 at 12:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Claude Leibovici, Henrik, mathreadler, kingW3
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Welcome to math.SE: I have tried to improve the readability of your question by introducing Tex. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 24 '17 at 19:47
  • 1
    $\begingroup$ Thank you. This is edited so well by you. $\endgroup$ – Raygor Mar 24 '17 at 19:50
  • $\begingroup$ Possible duplicate of Volume of a Solid 1.3 $\endgroup$ – user409521 Mar 24 '17 at 21:55
  • $\begingroup$ Most likely duplicate of Volume of a solid 1.3 which received a good answer and Volume of a solid revolution 1.5 which was closed as a duplicate of the first link. $\endgroup$ – user409521 Mar 24 '17 at 21:58

In order to gain some intuition about this problem, have a look at the following figure (produced by Wolfram alpha).

solid of revolution

What is interesting about this problem is that the area enclosed between the two functions, namely $f_1(x) = (x-1)^2 - 1$ and $f_2(x) =2$, intersects with the $y$ axis which is the axis of rotation.

We therefore need to revolve the surface enclosed between $f_1$ and $f_2$ and $x\geq 0$ with area

$$A = \int_0^{1+\sqrt{3}}(f_2(x) - f_1(x))\mathrm{d}x$$

This can be easily seen to be

$$V = \int_0^{1+\sqrt{3}}2\pi (f_2(x) - f_1(x))x \mathrm{d}x $$

If, on the other hand, we assume that the revolution creates a cavity in the solid body, then we need to subtract the following volume

$$ \Delta V = \int_0^{1+\sqrt{3}}2\pi (f_2(x) - f_1(-x))x \mathrm{d}x $$


Not the answer you're looking for? Browse other questions tagged or ask your own question.