Wize University Calculus 3 Textbook > Multiple Integrals

Similar to double integrals, we can define triple integrals for functions of three variables. Therefore, if f(x,y,z)f(x,y,z)f(x,y,z) is defined over a rectangular box, E, where E={(x,y,z)∣ a≤x≤b, c≤y≤d, e≤z≤f}E=\{(x,y,z)|\ a\le x\le b,\ c\le y\le d,\ e\le z\le f\}E={(x,y,z)∣ a≤x≤b, c≤y≤d, e≤z≤f}, then the triple integral of f(x,y,z)f(x,y,z)f(x,y,z) over EEE is equal to:
∭Ef(x,y,z)dV=lim⁡m,n,l→∞∑i=1m∑j=1n∑k=1lf(xijk∗,yijk∗,zijk∗)ΔV\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∭E​f(x,y,z)dV=m,n,l→∞lim​i=1∑m​j=1∑n​k=1∑l​f(xijk∗​,yijk∗​,zijk∗​)ΔV
Similar to double integrals, Fubini's Theorem expresses a triple integral in terms of three single partial integrals:
∭Ef(x,y,z)dV=∫ab∫cd∫eff(x,y,z)dzdydx\color{blue}\iiint_Ef(x,y,z)dV=\int_a^b\int_c^d\int_e^ff(x,y,z)dzdydx∭E​f(x,y,z)dV=∫ab​∫cd​∫ef​f(x,y,z)dzdydx
Similar to double integrals, the integration over a variable is done by treating the rest of the variables as constants.

0:00 / 0:00

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)∈D,z1(x,y)≤z≤z2(x,y)}E = \{(x, y, z)| (x,y) \in D, z_1(x,y) \le z \le z_2(x,y) \}E={(x,y,z)∣(x,y)∈D,z1​(x,y)≤z≤z2​(x,y)}, where DDD is the projection of region EEE in the xy-plane, then the triple integral of f(x,y,z)f(x,y,z)f(x,y,z) over EEE is equal to:
∭Ef(x,y,z)dV=∬D[∫z1(z,y)z2(x,y)f(x,y,z)dz]dA\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∭E​f(x,y,z)dV=∬D​[∫z1​(z,y)z2​(x,y)​f(x,y,z)dz]dA

In general, if E={(x,y,z)∣a≤x≤b, h1(x)≤y≤h2(x), z1(x,y)≤z≤z2(x,y)}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)\}E={(x,y,z)∣a≤x≤b, h1​(x)≤y≤h2​(x), z1​(x,y)≤z≤z2​(x,y)}, then, the triple integral of f(x,y,z)f(x,y,z)f(x,y,z) over EEE is equal to:
∭Ef(x,y,z)dV=∫ab∫h1(x)h2(x)∫z1(x,y)z2(x,y)f(x,y,z)dzdydx\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∭E​f(x,y,z)dV=∫ab​∫h1​(x)h2​(x)​∫z1​(x,y)z2​(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.

0:00 / 0:00

Evaluate the triple integral ∭Exy2z2dV\iiint_Exy^2z^2dV∭E​xy2z2dV, where EEE is defined as:
E={(x,y,z)∣0≤x≤2,  −1≤y≤1,  0≤z≤2}E=\{(x,y,z)|0\leq x\leq2,\;-1\leq y\leq 1,\;0\leq z\leq2\}E={(x,y,z)∣0≤x≤2,−1≤y≤1,0≤z≤2}

I=∫02dx∫−11dy∫02xy2z2  dz=∫02dx∫−11dy  xy2z33∣02\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_0^2I=∫02​dx∫−11​dy∫02​xy2z2dz=∫02​dx∫−11​dyxy23z3​​02​

=∫02dx∫−1183xy2  dy=83∫02dx∫−11xy2  dy=83∫02dx  xy33∣−11\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=∫02​dx∫−11​38​xy2dy=38​∫02​dx∫−11​xy2dy=38​∫02​dxx3y3​​−11​

=89∫02x  dx((1)3−(−1)3)=89∫022x  dx=89x2∣02=894=329\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}=98​∫02​xdx((1)3−(−1)3)=98​∫02​2xdx=98​x2​02​=98​4=932​

0:00 / 0:00

Evaluate	∭EydV\iiint_EydV∭E​ydV where EEE is bounded by planes x=0x=0x=0, y=0y=0y=0, z=0z=0z=0, x+y+z=2x+y+z=2x+y+z=2

z=0⇒x+y=2, x≥0, y≥0z=0\Rightarrow x+y=2,\ x\geq0,\ y\geq0z=0⇒x+y=2, x≥0, y≥0
0≤x≤20\leq x\leq20≤x≤2
0≤y≤2−x0\leq y\leq2-x0≤y≤2−x
0≤z≤2−x−y0\leq z\leq2-x-y0≤z≤2−x−y



I=∭Ey  dV=∫02dx∫02−xdy∫02−x−yy  dz\displaystyle I=\iiint_{E}y\;dV=    \int_0^2dx\int_0^{2-x}dy\int_0^{2-x-y}y\;dzI=∭E​ydV=∫02​dx∫02−x​dy∫02−x−y​ydz
=∫02dx∫02−xdy  yz∣02−x−y\displaystyle=    \int_0^2dx\int_0^{2-x}dy\;yz\bigg\vert_0^{2-x-y}=∫02​dx∫02−x​dyyz​02−x−y​
=∫02dx∫02−xy(2−x−y)  dy=∫02dx∫02−x(2y−xy−y2)  dy\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=∫02​dx∫02−x​y(2−x−y)dy=∫02​dx∫02−x​(2y−xy−y2)dy
=∫02dx(y2−xy22−y33)∣02−x\displaystyle=\int_0^2dx(y^2-x\frac{y^2}{2}-\frac{y^3}{3})\bigg\vert_0^{2-x}=∫02​dx(y2−x2y2​−3y3​)​02−x​
=∫02((2−x)2−x(2−x)22−(2−x)33)dx\displaystyle=\int_0^2\left((2-x)^2-x\frac{(2-x)^2}{2}-\frac{(2-x)^3}{3}\right)dx=∫02​((2−x)2−x2(2−x)2​−3(2−x)3​)dx
=∫02(4/3+x2−2x−1/6x3)  dx=4/6=23\displaystyle=\int_0^2(4/3+x^2-2x-1/6x^3)\;dx=4/6=\frac{2}{3}=∫02​(4/3+x2−2x−1/6x3)dx=4/6=32​

Determine the volume of the region between x+y+z=8 x+y+z=8~x+y+z=8 and the yz-plane that is bounded by z=32y z=\frac{3}{2}\sqrt{y}~z=23​y​ and z=34yz=\frac{3}{4}yz=43​y.

Answer

I don't know

Evaluate ∫∫∫3−4xdVE                            \displaystyle\int\int\int3-4xdV\newline{}
\scriptsize{E}~~~~~~~~~~~~~~~~~~~~~~~~~~~~∫∫∫3−4xdVE                            , where E is between z = 4 - xy and they xy-plane such that 0≤x≤2, 0≤y≤10\leq{x}\leq{2},~0\leq{y}\leq{1}0≤x≤2, 0≤y≤1.

Answer

I don't know

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Express ∫∫∫f(x, y, z)dVE                                    \displaystyle\int\int\int f(x,~y,~z)dV\newline{}
\scriptsize{E}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~∫∫∫f(x, y, z)dVE                                      6 ways if y=4−x2−4z2 & y=0y=4-x^2-4z^2~\&~y=0y=4−x2−4z2 & y=0.

I have answered this question

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Let E be the region bounded by the planes y = 0, y = 2, y + z = 3, and the surface z=x2z=x^2z=x2. Consider 
I=∫∫∫f(x, y, z) dVI = \int\int\int f(x,~y,~z)~dVI=∫∫∫f(x, y, z) dV
Express the integral as dzdxdy, dxdydz, dydxdz.dzdxdy,~dxdydz,~dydxdz.dzdxdy, dxdydz, dydxdz.

I have answered this question