# Volume of a solid of revolution - Getting two different results.

The region between $$y=x^2+1$$ and $$y=-x+3$$ is rotated about the $$x$$-axis.

I have to compute the volume. The intersection between these two curves are at $$x=-2$$ and $$x=1$$.

At first, I thought about $$V(x)=\pi\displaystyle\int_{-2}^1(3-x-(x^2+1))^2dx=\frac{81\pi}{10}$$, where the radius is the difference between the two curves.

Another approach I made was $$V(x)=\pi\displaystyle\int_{-2}^1[(3-x)^2-(x^2+1)^2]dx=\frac{117\pi}{5}$$. I'm not sure why both results are different, and which one is correct.

• What's your reasoning for setting up the first integral? You have used the formula $V = \pi \int_a^b (R^2-r^2) \ dx$ incorrectly. – Toby Mak Jan 16 at 3:11
• They disagree because $a^2-b^2\ne(a-b)^2$. – Brian M. Scott Jan 16 at 3:17
• Yeah I just got it, thank you. – Fabrizio Gambelín Jan 16 at 3:19

It might be easier to think about it in steps. So, we know the formula for the volume of a solid of revolution about the $$x$$ axis from $$a$$ to $$b$$, right?

$$V = \pi \int_a^b (f(x))^2 dx$$

Think of the $$f^2$$ as serving the role of a radius, and the integral as contributing the length by summing up over all of the super-small bits of length $$dx$$ on the interval.

If you have the area between two curves $$f$$ and $$g$$, though - let $$f \ge g$$ here - you can think of it not as revolving the area between $$f$$ and $$g$$, but rather as just revolving the outer area $$f$$ and then cutting a hole from it based on $$g$$, and you want the volume left over. That is, for instance,

$$V = \underbrace{\pi \int_a^b (f(x))^2 dx}_{\text{original volume}} - \underbrace{\pi \int_a^b (g(x))^2 dx}_{\text{hole you cut out}}$$

Simplifying, we get

$$V = \pi \int_a^b \Big( (f(x))^2 - (g(x))^2 \Big) dx$$

or, if you understand that $$f^2(x)$$ means $$(f(x))^2$$, you can make this a little neater to write:

$$V = \pi \int_a^b \Big( f^2(x) - g^2(x) \Big) dx$$

Thus your second approach would be the correct one.

• I can clearly understand it now, thank you a lot. – Fabrizio Gambelín Jan 16 at 3:19