Triple Integrals in Cylindrical Coordinates — Question 9

Question 9 – Solution

Strategy. Approximate a small cylindrical cell by a rectangular box whose angular side is an arc of length $r\,d\theta$.

Step 1: Jacobian factor A small cell has radial length $dr$, vertical length $dz$, and circumferential length approximately $r\,d\theta$.

Consequently, $$dV\approx(dr)(r\,d\theta)(dz)=r\,dz\,dr\,d\theta,$$ and the approximation becomes exact in the integral limit.

Step 2: Apply Since $x^2+y^2=r^2$, $$\begin{align*} I&=\int_0^{\pi/3}\int_1^2\int_0^4r^2(r\,dz\,dr\,d\theta)\\ &=\left(\frac\pi 3\right)(4)\left[\frac{r^4}{4}\right]_1^2 =\boxed{5\pi}. \end{align*}$$

Verification The factors scale as angular width times height times radius to the fourth power. That produces length$^5$, the correct units for integrating a squared distance over volume.