# Volume of a solid revolution 1.4 [closed]

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, kingW3Mar 25 '17 at 12:47

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• Thank you. This is edited so well by you. – Raygor Mar 24 '17 at 19:50
• Possible duplicate of Volume of a Solid 1.3 – user409521 Mar 24 '17 at 21:55
• 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. – user409521 Mar 24 '17 at 21:58

## 1 Answer

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

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