x^2+y^2=r^2, dV=rdzdrd\theta, 영역: 0\le r\le 3, 0\le\theta\le 2\pi, 0\le z\le 5

\displaystyle\iiint_Er^2rdzdrd\theta=\displaystyle\int_0^{2\pi}\int_0^{3}\int_0^{5}r^3dzdrd\theta=\displaystyle\int_0^{2\pi}d\theta\cdot\int_0^{3}r^3dr\cdot\int_0^{5}dz

=(2\pi)\cdot\bigg[\dfrac{1}{4}r^4\bigg]_0^3\cdot[z]_0^5=2\pi\cdot\dfrac{81}{4}\cdot5=\dfrac{405\pi}{2}이다.