Find the volume of the solid generated by revolving the region bounded by $y=\sqrt{9-x^2}$ and $y=0$ about the x-axis.

Question

The answer for this question is apparently $36\pi$, but I cannot get that answer. I keep getting $18\pi$, and I have a feeling my professor made a mistake on this. Can I get some clarification?

Answer 1

You may calculate the volume as follows:

$$\pi \int_{-3}^3 y^2 dx = \pi \int_{-3}^3 9 - x^2 \; dx = \pi \left[ 9x - \frac{x^3}{3}\right]_{-3}^3 = \pi \left[ 27 - 9 +27 - 9\right] = 36 \pi$$

Answer 2

You can also use symmetry to simplify the integral, i.e. when you have

$$\int_{-a}^a f(x) \ dx$$ you can rewrite it as $$2 \int_{0}^a f(x) \ dx.$$

Note: randomgirl is correct. I had mistakenly used the shell method (which is $\pi \int_a^{b} x f(x) \ dx$) rather than the disk method (which is $\pi \int_a^{b} f(x)^2 \ dx$). Sorry about that!