Triple Integrals in Cylindrical Coordinates — Question 6

Question 6 – Solution

Strategy. Use $x^2+y^2=r^2$ for the squared distance to the axis and $dV=r\,dz\,dr\,d\theta$.

Step 1: Region The meridian triangle gives $0\le r\le 3$ and $0\le z\le 3-r$.

Step 2: Mass $$M=\int_0^{2\pi}\int_0^3\int_0^{3-r}2r\,dz\,dr\,d\theta =4\pi\int_0^3r(3-r)dr=\boxed{18\pi}.$$

Step 3: Moment and radius of gyration $$\begin{align*} I_z&=\iiint_E(x^2+y^2)\rho\,dV =4\pi\int_0^3r^3(3-r)dr=\boxed{\frac{243\pi}{5}},\\ k_z&=\sqrt{\frac{I_z}{M}} =\boxed{\sqrt{\frac{27}{10}}}. \end{align*}$$

Verification $k_z\approx 1.64$ lies between the minimum and maximum distances $0$ and $3$ from the axis. Also $M=2\cdot\frac 13\pi(3^2)(3)=18\pi$ agrees with the cone-volume formula.