Triple Integral for Volume with Cartesian Coordinates

Recap §

To find area under curve (2D):

$${\int }_a^{b}f(x)dx$$

To find mass of thin wire (1D):

$${\int }_a^b\rho (x)dx$$

To find volume under the surface $f(x,y)$ (3D):

$$\iint f(x,y)dA$$

Mass of a thin plate (2D):

$$\iint \rho (x)dA$$

Area of the bounded region (2D):

$$\iint 1dA$$

Measurement under $f(x,y,z)$ (4D):

$$\iiint f(x,y,z)dV$$

Mass of 3D region:

$$\iiint \rho (x,y,z)dV$$

Volume of bounded region (3D):

$$\iiint 1dV$$

Volume bounded in 3D with functions §

$${\int }_{x=a}^{x=b}{\int }_{y=g_1(x)}^{y=g_2(x)}{\int }_{z=f_1(x,y)}^{z=f_2(x,y)}f(x,y,z)dV$$

Example 1 §

Find $\iiint xydV$ where $Q$ is a tetrahedron in the first octant bounded by $2x+3y+z=6$.

$$\begin{array}{r}z=6-2x-3y\\ 0\le z\le 6-2x-3y\\ 0\le x\le 3\\ 0\le y\le -\frac{2}{3}x+2\\ {\int }_{x=0}^{x=3}{\int }_y^{y=-2/3x+2}{\int }_{z=0}^{z=6-2x-3y}xydzdydx\\ {\int }_{x=0}^{x=3}{\int }_y^{y=-2/3x+2}xy(6-2x-3y)dydx\end{array}$$

Example 2 §

Set up triple integral to find the mass of the $Q$ region: $y=x^3,y=x,z\le 2x,z\ge 0$ and the mass density given as $\rho (x,y,z)=2z$

$${\int }_0^1{\int }_{y=x^3}^{y=x}{\int }_{z=0}^{z=2x}2zdzdydx$$

Example 3 §

Set up triple integral to find the volume of the region $Q$ bounded by $x-4-y,x+z=4,x=0,z=0$

$$\begin{array}{r}0\le z\le -x+4\\ z=0⟹x=4,x=4-y^2,x=0\\ -2\le y\le 2\\ 0\le x\le 4-y^2\\ \\ {\int }_{y=-2}^{y=2}{\int }_{x=0}^{x=4-y^2}{\int }_{z=0}^{z=-x+4}dzdxdy\end{array}$$

Example 4 §

Find ${\int }_0^3{\int }_{x=0}^{x=\sqrt{9-y^2}}{\int }_{z=0}^{z=9-x^2-y^2}4xy{e}^{x^2}dzdxdy$

First change the order of integration

$$\begin{array}{rl}z& =0\to z=9-x^2-y^2\\ x& =0\to x=\sqrt{9-y^2-z^2}\\ 0\le x\le \sqrt{9-y^2-z}& \\ 0\le z\le \sqrt{9-y^2}& \\ 0\le y\le 3& \end{array}$$ $$\begin{array}{rl}{\int }_{y=0}^{y=3}{\int }_{z=0}^{z=9-y^2}{\int }_{x=0}^{x=\sqrt{9-y^2-z}}4xy{e}^{x^2}dxdzdy& \\ & ={\int }_{y=0}^{y=3}{\int }_{z=0}^{z=9-y^2}[2y{e}^{x^2}{]}_0^{\sqrt{9-x^2-z}}dzdy\\ & ={\int }_{y=0}^{y=3}{\int }_{z=0}^{z=9-y^2}(2y{e}^{9-x^2-z}-2y)dzdy\\ & ={\int }_0^3[-2y{e}^{9-x^2-z}-2yz{]}_0^{9-y^2}dx\end{array}$$

Cartesian coordinate system is too hard? §

Use Triple Integral with Cylindrical and Spherical Coordinates!