Volume of truncated cone with different methods gives different results.

Question

This is quite silly. I am computing the volume of the cone $x^2+z^2=y^2$, truncated at $y=1$ and $y=3$ with three different methods:

1. $V=V(\text{big cone})-V(\text{small cone})=9\pi - \pi/3 = 26\pi/3$, where I used the formula $V(\text{cone})=\pi r^2 h/3$, $r$ being the radius of the basis of the cone, and $h$ its height.
2. $V=\pi h/3(r^2+rR+R^2)=\pi 2/3(9+3+1)=26\pi/3$, where I used the formula of a truncated cone.
3. $V=\int_{\theta=0}^{2\pi}\int_{r=1}^3\int_{y=r}^3 r\;dydrd\theta=20\pi/3$.

I cannot find my error. Why is answer $3$ different? Thanks!

Answer 1

Your limits of integration in the third integral omit the solid cylinder $0 \leq r \leq 1$, $1 \leq y \leq 1$ of radius $1$ inside (and co-axial to) the cone.

Answer 2

Using cylindrical coordinates $x=r\cos\theta$, $z=r\sin\theta$, $y=y$, the equation of the cone is $r=y$. Each value of $y$ from $1$ to $3$ creates a circular cross-section of radius $r=y$. So the integral for your volume can be expressed $$ \int_{y=1}^3\int_{\theta=0}^{2\pi}\int_{r=0}^y r\,dr\,d\theta\,dy $$ which evaluates to $\frac{26}3\pi$.