Question

Use cylindrical coordinates to calculate ∫∫∫W f(x,y,z) dV for the given function and region:

f(x,y,z)=z, x^2+y^2≤z≤16

Answer 1

projection of the region onto the oxy plane:

let's move on to cylindrical coordinates:

$\begin{array}{l} x = r\cos \varphi ;\,\,y = r\sin \varphi ; z=z\\ 0 \le \varphi \le 2\pi ,\,\,0 \le r \le 4,\,\,{x^2} + {y^2} \le z \le 16 \Rightarrow {r^2} \le z \le 16 \end{array}$

Then

$\begin{array}{l} \mathop {\int {\int {\int {zdV} } } }\limits_V = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^4 {rdr} \int\limits_{{r^2}}^{16} {zdz} = \frac{1}{2}\int\limits_0^{2\pi } {d\varphi } \int\limits_0^4 {r\left. {{z^2}} \right|_{{r^2}}^{16}dr} = \frac{1}{2}\left. \varphi \right|_0^{2\pi } \cdot \int\limits_0^4 {r({{16}^2} - {r^4})dr} = \\ = \frac{{2\pi }}{2}\int\limits_0^4 {\left( {256r - {r^5}} \right)} dr = \pi \left( {128\left. {{r^2}} \right|_0^4 - \left. {\frac{{{r^6}}}{6}} \right|_0^4} \right) = \pi \left( {128 \cdot 16 - \frac{{4096}}{6}} \right) = \frac{{4096\pi }}{3} \end{array}$ Answer: $\frac{{4096\pi }}{3}$