Volume integral of a cone

Question

I do have a little problem that just drives me crazy. It is about calculating the volume of a truncated cone. I have two solutions for it.

The object is limited by three areas: $A_1: z=\sqrt{x^2+y^2}-2$, $A_2: z=-1$ and $A_3: z=1$. I tried to solve it with the formula for a volume found on wikipedia ([Link])1 with $h=2$, $r_1=1$ and $r_2=3$. The result was $\frac{26}{3}\pi$.

I want to calculate this with a volume integral and it delivers in cylindrical coordinates $$V = \int_{\varphi=0}^{2\pi} \int_{r=1}^{3} \int_{z=-1}^{r-2} r \mathrm{d}z \mathrm{d}r \mathrm{d}\varphi = \frac{28}{3}\pi$$

Can anybody see my fault? I didn't find similar questions for this, other cones were always centered in the origin. Thanks a lot!

Answer 1

It should be

$$V = \int_{0}^{2\pi} {d}\varphi\int_{-1}^{1} \mathrm{d}z \int_{0}^{z+2} r \mathrm{d}r \mathrm = \frac{26}{3}\pi$$

Answer 2

Note that $$\int_{\varphi=0}^{2\pi} \int_{r=1}^{3} \int_{z=-1}^{r-2} r \mathrm{d}z \mathrm{d}r \mathrm{d}\varphi = \frac{20\pi}{3}$$ When you add the volume of the middle cylinder to this result you get the correct answer of $26\pi /3$