Triple Integral with Cylindrical and Spherical Coordinates

Cylindrical Volume §

$${\int }_{{\theta }_1}^{{\theta }_2}{\int }_{r_1}^{r_2}{\int }_{z_1}^{z_2}f(r\cos \theta ,r\sin \theta ,z)dzdrd\theta$$

Example 1 §

Find $\int \int \int ydV$ where Q bounded by $z=4-x^2-y^2$ in the first octant $x=0,y=0,z=0$

$$\begin{array}{rl}0\le z\le 4-r^2& \\ 0\le r\le 2& \\ 0\le \theta \le \frac{\pi }{2}& \\ {\int }_0^{\pi }{\int }_0^2{\int }_0^{4-r^2}r\sin (\theta )rdzdrd\theta & ={\int }_0^{\pi }{\int }_0^2{\int }_0^{4-r^2}{r}^{2}sin(\theta )dzdrd\theta \\ & =4.27\end{array}$$

Example 2 §

Set up $\int \int xydV$ using cylindrical cordinates where $Q:x,y,z|x^2+y^2\le 1,x\ge 1,y\le x,-1\le z\le 1$

$$\begin{array}{r}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{4}}{\int }_0^1{\int }_{-1}^1{r}^3\sin \theta \cos \theta dzdrd\theta \end{array}$$

Triple Integral Using Spherical Coordinates §

Coordinates: $(\rho ,\theta ,\varphi )$

$$\begin{array}{c}r=\rho \sin \varphi \\ x=r\cos \theta =\rho \sin \varphi \cos \theta \\ y=r\sin \theta =\rho \sin \varphi \sin \theta \\ z=\rho \cos \varphi \end{array}$$

Volume:

$$\begin{array}{c}\iiint f(\rho ,\theta ,\varphi ){\rho }^2\sin \varphi d\rho d\varphi d\theta \end{array}$$

Example 1 §

Find $\iiint \sqrt{x^2+y^2+z^2}dV$, bounded by $x^2+y^2+z^2\le 1$

$$\begin{array}{c}\rho =x^2+y^2+z^2⟹\rho \le 1\\ 0\le \varphi \le \pi \\ 0\le \theta \le 2\pi \end{array}$$ $$\begin{array}{rl}V& ={\int }_0^{2\pi }{\int }_{\varphi =0}^{\varphi =\pi }{\int }_{\rho =0}^{\rho =1}\rho \cdot {\rho }^2\sin \varphi d\varphi d\varphi d\theta \\ & ={\int }_0^{2\pi }{\int }_{\varphi =0}^{\varphi =\pi }{\int }_{\rho =0}^{\rho =1}{\rho }^3\sin \varphi d\rho d\varphi d\theta \\ & ={\int }_0^{2\pi }{\int }_{\varphi =0}^{\varphi =\pi }\frac{1}{4}\sin \varphi d\varphi d\theta \\ & ={\int }_0^{2\pi }\frac{1}{2}d\theta \\ & =\pi \end{array}$$

Example 2 §

Use spherical and cylindrical to find the volume of the solid region Q bounded by upper napped of the cone $z^2=x^2+y^2$ above by the sphere $x^2+y^2+z^2=9$

Spherical §

$$\begin{array}{c}{\rho }^2=9⟹\rho =3\\ x^2=z^2-y^2,x^2=9-(y^2+z^2)⟹2z^2=9⟹z=\frac{3}{\sqrt{2}}\\ \rho \cos \varphi =\frac{3}{\sqrt{2}}\to \varphi =\frac{\pi }{4}\end{array}$$

Bounds:

$$\begin{array}{c}0\le \rho \le 3\\ 0\le \varphi \le \frac{\pi }{4}\\ 0\le \theta \le 2\pi \end{array}$$ $$\begin{array}{rl}V& ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =0}^{\varphi =\pi /4}{\int }_{\rho =0}^{\rho =3}1\cdot {\rho }^2\sin \varphi d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =0}^{\varphi =\pi /4}9\sin \varphi d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }\left(-\frac{9\sqrt{2}}{2}+9\right)d\theta \\ & =18\pi -9\sqrt{2\pi }\end{array}$$

Cylindrical §

Bounds §

$$\begin{array}{c}z=\sqrt{x^2+y^2}⟹r\le z\\ z=\sqrt{9-x^2-y^2}⟹z\le \sqrt{0-r^2}\\ 2r^2=9⟹r=\frac{3}{\sqrt{2}}\end{array}$$

Volume §

$$\begin{array}{rl}V& ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{r=0}^{r=3/\sqrt{2}}{\int }_r^{\sqrt{9-r^2}}1\cdot rdzdrd\theta \\ & =18\pi -9\sqrt{2\pi }\end{array}$$

Example 3 §

Find the volume of the region outside the cylinder $x^2+y^2=1$ and inside the sphere $x^2+y^2+z^2=4$.

Bounds §

$$\begin{array}{c}{r}^2=1,{\rho }^2=4⟹\frac{1}{\sin \varphi }\le \rho \le 2\\ r^2=1,r^2+(\rho \cos \varphi {)}^2=4⟹\varphi =\frac{\pi }{6},\frac{5\pi }{6}\\ 0\le \theta \le \frac{5\pi }{6}\end{array}$$

Volume §

$$\begin{array}{rl}V& ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =\pi /6}^{\varphi =5\pi /6}{\int }_{\rho =\csc \theta }^{\rho =2}1\cdot {\rho }^2\sin \varphi d\rho d\varphi d\theta \\ & ={\int }_{\theta =0}^{\theta =2\pi }{\int }_{\varphi =\pi /6}^{\varphi =5\pi /6}\frac{8}{3}-{\csc }^3\theta d\varphi d\theta \end{array}$$

Bounds are Bad? §

Then change them! Changing Bounds for Integrals with the Jacobian