Wize University Calculus 3 Textbook > Multiple Integrals

Triple Integrals

Similar to double integrals, we can define triple integrals for functions of three variables. Therefore, if $f(x,y,z)$ is defined over a rectangular box, E, where $E=\{(x,y,z)|\ a\le x\le b,\ c\le y\le d,\ e\le z\le f\}$ , then the triple integral of $f(x,y,z)$ over $E$ is equal to:

\color{blue}\iiint_Ef(x,y,z)dV=\lim_{m,n,l\to\infty}\sum_{i=1}^m\sum_{j=1}^n\sum_{k=1}^lf(x_{ijk}^*,y_{ijk}^*,z_{ijk}^*)\Delta V

Similar to double integrals, Fubini's Theorem expresses a triple integral in terms of three single partial integrals:

\color{blue}\iiint_Ef(x,y,z)dV=\int_a^b\int_c^d\int_e^ff(x,y,z)dzdydx

Similar to double integrals, the integration over a variable is done by treating the rest of the variables as constants.

The Fubini's theorem for regions of general shapes is similar to the one for the region with a rectangular shape. If $E = \{(x, y, z)| (x,y) \in D, z_1(x,y) \le z \le z_2(x,y) \}$ , where $D$ is the projection of region $E$ in the xy-plane, then the triple integral of $f(x,y,z)$ over $E$ is equal to:

\color{blue}\iiint_Ef(x,y,z)dV=\iint_D\bigg[\int_{z_1(z,y)}^{z_2(x,y)}f(x,y,z)dz\bigg]dA

In general, if $E=\{(x,y,z)|a\le x\le b,\ h_1(x)\le y\le h_2(x),\ z_1(x,y)\le z\le z_2(x,y)\}$ , then, the triple integral of $f(x,y,z)$ over $E$ is equal to:

\color{blue}\iiint_Ef(x,y,z)dV=\int_a^b\int_{h_1(x)}^{h_2(x)}\int_{z_1(x,y)}^{z_2(x,y)}f(x,y,z)dzdydx

Effectively, after taking integration over one of the variables, the problem is converted to a double integral and we can use the methods we learned for double integrals to do the rest of calculations.

Evaluate the triple integral $\iiint_Exy^2z^2dV$ , where $E$ is defined as:

E=\{(x,y,z)|0\leq x\leq2,\;-1\leq y\leq 1,\;0\leq z\leq2\}

$\displaystyle I=\int_0^2dx\int_{-1}^1dy\int_0^2 xy^2z^2\;dz= \int_0^2dx\int_{-1}^1dy\; xy^2\frac{z^3}{3}\bigg\vert_0^2$

$\displaystyle=\int_0^2dx\int_{-1}^1\frac{8}{3}xy^2\;dy=\frac{8}{3}\int_0^2dx\int_{-1}^1xy^2\;dy= \frac{8}{3}\int_0^2dx\;x\frac{y^3}{3}\bigg\vert_{-1}^1$

$\displaystyle=\frac{8}{9}\int_0^2x\;dx((1)^3-(-1)^3)=\frac{8}{9}\int_0^22x\;dx= \frac{8}{9}x^2\bigg\vert_0^2=\frac{8}{9}4=\frac{32}{9}$

Evaluate $\iiint_EydV$ where $E$ is bounded by planes $x=0$ , $y=0$ , $z=0$ , $x+y+z=2$

$z=0\Rightarrow x+y=2,\ x\geq0,\ y\geq0$

$0\leq x\leq2$

$0\leq y\leq2-x$

$0\leq z\leq2-x-y$

Image illustrating Example: Triple Integrals over Regions of General Shape (2)

$\displaystyle I=\iiint_{E}y\;dV= \int_0^2dx\int_0^{2-x}dy\int_0^{2-x-y}y\;dz$

$\displaystyle= \int_0^2dx\int_0^{2-x}dy\;yz\bigg\vert_0^{2-x-y}$

$\displaystyle=\int_0^2dx\int_0^{2-x}y(2-x-y)\;dy= \int_0^2dx\int_0^{2-x}(2y-xy-y^2)\;dy$

$\displaystyle=\int_0^2dx(y^2-x\frac{y^2}{2}-\frac{y^3}{3})\bigg\vert_0^{2-x}$

$\displaystyle=\int_0^2\left((2-x)^2-x\frac{(2-x)^2}{2}-\frac{(2-x)^3}{3}\right)dx$

$\displaystyle=\int_0^2(4/3+x^2-2x-1/6x^3)\;dx=4/6=\frac{2}{3}$

Determine the volume of the region between $x+y+z=8~$ and the yz-plane that is bounded by $z=\frac{3}{2}\sqrt{y}~$ and $z=\frac{3}{4}y$ .

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Evaluate $\displaystyle\int\int\int3-4xdV\newline{} \scriptsize{E}~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ , where E is between z = 4 - xy and they xy-plane such that $0\leq{x}\leq{2},~0\leq{y}\leq{1}$ .

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Express $\displaystyle\int\int\int f(x,~y,~z)dV\newline{} \scriptsize{E}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ 6 ways if $y=4-x^2-4z^2~\&~y=0$ .

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Let E be the region bounded by the planes y = 0, y = 2, y + z = 3, and the surface $z=x^2$ . Consider

I = \int\int\int f(x,~y,~z)~dV

Express the integral as $dzdxdy,~dxdydz,~dydxdz.$

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Triple Integrals

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