Calculus – Multiple integrals – Triple integrals

Let \(\rho(x,y,z)\geq0\) be a mass density function defined on a solid \(E\) in \(\mathbb{R}^3\). Our aim is to define a triple integral

\[\iiint\limits_E\rho(x,y,z)\,dV,\]

which denotes the (total) mass of the solid \(E\).

Consider the rectangular box \(B\) given by

\[B=[a,b.\times[c,d]\times[e,f]=\{(x,y,z)\,|\,a\leq x\leq b,\;c\leq y\leq d,\;e\leq z\leq f\}.\]

The first step is to divide \(B\) into sub-boxes. We do this by dividing the interval \([a,b]\) into \(\ell\) subintervals \([x_{i-1},x_i]\) of equal width \(\Delta x=(b-a)/\ell\), dividing the interval \([c,d]\) into \(m\) subintervals \([y_{j-1},y_j]\) of equal width \(\Delta y=(d-c)/m\) and dividing the interval \([e,f]\) into \(n\) subintervals \([z_{k-1},z_k]\) of equal width \(\Delta z=(f-e)/n\).

This divides the rectangular box \(B\) into sub-boxes

\[B_{ijk}=[x_{i-1},x_i]\times[y_{j-1},y_j]\times[z_{k-1},z_k].\]

Each sub-box \(B_{ijk}\) has volume \(\Delta V=\Delta x\,\Delta y\,\Delta z\).

Now we form the triple Riemann sum

\[\sum_{i=1}^{\ell}\sum_{j=1}^m\sum_{k=1}^n\rho(x_{ijk}^*,y_{ijk}^*z_{ijk}^*)\,\Delta V,\]

where the sample point \((x_{ijk}^*,y_{ijk}^*,z_{ijk}^*)\) is in \(B_{ijk}\). This leads to

Definition: The triple integral of \(\rho\) over the rectangular box \(B\) is

\[\iiint\limits_B\rho(x,y,z)\,dV=\lim\limits_{\ell\to\infty}\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} \sum_{i=1}^{\ell}\sum_{j=1}^m\sum_{k=1}^n\rho(x_{ijk}^*,y_{ijk}^*z_{ijk}^*)\,\Delta V.\]

The triple integral exists if \(\rho\) is continuous. We can choose the sample point to be any point in the sub-box, however if we choose it to be the point \((x_i,y_j,z_k)\) we get a simpler looking expression for the triple integral:

\[\iiint\limits_B\rho(x,y,z)\,dV=\lim\limits_{\ell\to\infty}\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} \sum_{i=1}^{\ell}\sum_{j=1}^m\sum_{k=1}^n\rho(x_i,y_j,z_k)\,\Delta V.\]

Fubini's theorem

Theorem: If \(\rho\) is continuous on the rectangular box \(B=[a,b]\times[c,d]\times[e,f]\), then

\[\iiint\limits_B\rho(x,y,z)\,dV=\int_e^f\int_c^d\int_a^b\rho(x,y,z)\,dx\,dy\,dz.\]

The iterated integral on the right can be written in six different orders, all with the same value.

Stewart §15.6, Example 1
Evaluate the triple integral \(\displaystyle\iiint\limits_Bxyz^2\,dV\), where \(B\) is the rectangular box given by \[B=\{(x,y,z)\,|\,0\leq x\leq1,\;-1\leq y\leq2,\;0\leq z\leq3\}.\]

Solution: Note that

\begin{align*} \iiint\limits_Bxyz^2\,dV&=\int_0^3\int_{-1}^2\int_0^1xyz^2\,dx\,dy\,dz=\left(\int_0^1x\,dx\right)\left(\int_{-1}^2y\,dy\right)\left(\int_0^3z^2\,dz\right)\\[2.5mm] &=\bigg[\frac{1}{2}x^2\bigg]_{x=0}^1\bigg[\frac{1}{2}y^2\bigg]_{y=-1}^2\bigg[\frac{1}{3}z^3\bigg]_{z=0}^3 =\frac{1}{2}\cdot\frac{3}{2}\cdot9=\frac{27}{4}. \end{align*}

A tripe integral over a general bounded region (a solid) \(E\) in \(\mathbb{R}^3\) is defined by much the same procedure that we used for double integrals. We enclose \(E\) in a rectangular box \(B\) and we define a function \(\Omega\) so that it agrees with \(\rho\) on \(E\) and is \(0\) for points in \(B\) that are outside \(E\). Then:

\[\iiint\limits_E\rho(x,y,z)\,dV=\iiint\limits_B\Omega(x,y,z)\,dV.\]

This integral exists if \(\rho\) is continuous and the boundary of \(E\) is "reasonably smooth".

Regions of type 1

A solid region \(E\) in \(\mathbb{R}^3\) is said to be of type 1 if it lies between the graphs of two continuous functions of \(x\) and \(y\):

\[E=\{(x,y,z)\,|\,(x,y)\in D,\;u_1(x,y)\leq z\leq u_2(x,y)\},\]

where \(D\) is the projection of \(E\) onto the \(xy\)-plane. Then:

\[\iiint\limits_E\rho(x,y,z)\,dV=\iint\limits_D\left(\int_{u_1(x,y)}^{u_2(x,y)}\rho(x,y,z)\,dz\right)\,dA.\]

Regions of type 2

A solid region \(E\) in \(\mathbb{R}^3\) is said to be of type 2 if it lies between the graphs of two continuous functions of \(y\) and \(z\):

\[E=\{(x,y,z)\,|\,(y,z)\in D,\;u_1(y,z)\leq x\leq u_2(y,z)\},\]

where \(D\) is the projection of \(E\) onto the \(yz\)-plane. Then:

\[\iiint\limits_E\rho(x,y,z)\,dV=\iint\limits_D\left(\int_{u_1(y,z)}^{u_2(y,z)}\rho(x,y,z)\,dx\right)\,dA.\]

Regions of type 3

A solid region \(E\) in \(\mathbb{R}^3\) is said to be of type 3 if it lies between the graphs of two continuous functions of \(x\) and \(z\):

\[E=\{(x,y,z)\,|\,(x,z)\in D,\;u_1(x,z)\leq y\leq u_2(x,z)\},\]

where \(D\) is the projection of \(E\) onto the \(xz\)-plane. Then:

\[\iiint\limits_E\rho(x,y,z)\,dV=\iint\limits_D\left(\int_{u_1(x,z)}^{u_2(x,z)}\rho(x,y,z)\,dy\right)\,dA.\]

Stewart §15.6, Example 2
Evaluate \(\displaystyle\iiint\limits_Ez\,dV\), where \(E\) is the solid tetrahedron bounded by the four planes \(x=0\), \(y=0\), \(z=0\) and \(x+y+z=1\).

Solution: Note that the projection of \(E\) on the \(xy\)-plane is

\[D=\{(x,y)\,|\,0\leq x\leq x,\;0\leq y\leq 1-x\}.\]

Using the description of \(E\) as a type 1 region, the tetrahedron is between the planes \(z=0\) and \(z=1-x-y\), we have:

\begin{align*} \iiint\limits_Ez\,dV&=\iint\limits_D\left(\int_0^{1-x-y}z\,dz\right)\,dA=\frac{1}{2}\iint\limits_D(1-x-y)^2\,dA =\int_0^1\int_0^{1-x}(1-x-y)^2\,dy\,dx\\[2.5mm] &=\int_0^1\bigg[-\frac{1}{3}(1-x-y)^3\bigg]_{y=0}^{1-x}\,dx =\frac{1}{6}\int_0^1(1-x)^3\,dx=-\frac{1}{24}(1-x)^4\bigg|_{x=0}^1=\frac{1}{24}. \end{align*}

Stewart §15.6, Example 3
Evaluate \(\displaystyle\iiint\limits_E\sqrt{x^2+z^2}\,dV\), where \(E\) is the region bounded by the paraboloid \(y=x^2+z^2\) and the plane \(y=4\).

Solution: Note that the projection of \(E\) on the \(xy\)-plane is

\[D_1=\{(x,y)\,|\,-2\leq x\leq 2,\;x^2\leq y\leq 4\}.\]

Therefore, if we consider \(E\) as a type 1 region, we obtain

\[\iiint\limits_E\sqrt{x^2+z^2}\,dV=\iint\limits_{D_1}\left(\int_{-\sqrt{y-x^2}}^{\sqrt{y-x^2}}\sqrt{x^2+z^2}\,dz\right)\,dA =\int_{-2}^2\int_{x^2}^4\int_{-\sqrt{y-x^2}}^{\sqrt{y-x^2}}\sqrt{x^2+z^2}\,dz\,dy\,dx,\]

which is extremely difficult to evaluate.

However, if we consider \(E\) as a type 3 region, the projection of \(E\) onto the \(xz\)-plane is the disk

\[D_3=\{(x,z)\,|\,x^2+z^2\leq 4\}.\]

Then we have:

\(\displaystyle\iiint\limits_E\sqrt{x^2+z^2}\,dV=\iint\limits_{D_3}\left(\int_{x^2+z^2}^4\sqrt{x^2+z^2}\,dz\right)\,dA =\iint\limits_{D_3}(4-x^2-z^2)\sqrt{x^2+z^2}\,dA\).

Using polar coordinates we now obtain

\begin{align*} \iiint\limits_E\sqrt{x^2+z^2}\,dV&=\iint\limits_{D_3}(4-x^2-z^2)\sqrt{x^2+z^2}\,dA=\int_0^{2\pi}\int_0^2(4-r^2)\,r\,r\,dr\,d\theta\\[2.5mm] &=\int_0^{2\pi}d\theta\int_0^2(4r^2-r^4)\,dr=2\pi\bigg[\frac{4}{3}r^3-\frac{1}{5}r^5\bigg]_{r=0}^2 =2\pi\left(\frac{32}{3}-\frac{32}{5}\right)=2\pi\cdot32\cdot\frac{2}{15}=\frac{128}{15}\pi. \end{align*}

Applications of triple integrals

If \(\rho(x,y,z)=1\) for all points \((x,y,z)\) in \(E\), then the (total) mass of \(E\) equals the volume of \(E\):

\[\textrm{vol}(E)=\iiint\limits_E1\,dV.\]

If \(\rho(x,y,z)\geq0\) is a mass density function defined on a solid \(E\) in \(\mathbb{R}^3\), then

\[m=\iiint\limits_E\rho(x,y,z)\,dV\]

denotes the (total) mass of \(E\). Then the integrals

\[\iiint\limits_Ex\,\rho(x,y,z)\,dV,\quad\iiint\limits_Ey\,\rho(x,y,z)\,dV\quad\textrm{and}\quad\iiint\limits_Ez\,\rho(,y,z)\,dV\]

are the moments of \(E\) about the three coordinate planes \(x=0\), \(y=0\) and \(z=0\) respectively.

The center of mass of \(E\) is located at the point \((\overline{x},\overline{y},\overline{z})\) with

\[\overline{x}=\frac{1}{m}\iiint\limits_Ex\,\rho(x,y,z)\,dV,\quad\overline{y}=\frac{1}{m}\iiint\limits_Ey\,\rho(x,y,z)\,dV \quad\textrm{and}\quad\overline{z}=\frac{1}{m}\iiint\limits_Ez\,\rho(x,y,z)\,dV,\]

where \(m=\displaystyle\iiint\limits_E\rho(x,y,z)\,dV\) denotes the (total) mass of \(E\).

If the mass density function \(\rho(x,y,z)\) is constant, then the center of mass is called the centroid of the solid \(E\).

Stewart §15.6, Example 5
Use a triple integral to find the volume of the tetrahedron \(T\) bounded by the planes \(x+2y+z=2\), \(x=2y\), \(x=0\) and \(z=0\).

Solution: The projection of the tetrahedron \(T\) on the \(xy\)-plane is

\[D=\{(x,y)\,|\,0\leq x\leq 1,\;\frac{1}{2}x\leq y\leq 1-\frac{1}{2}x\}.\]

Therefore we have

\begin{align*} \textrm{vol}(T)&=\iiint\limits_T1\,dV=\iint\limits_D\left(\int_0^{2-x-2y}dz\right)\,dA=\int_0^1\int_{\frac{1}{2}x}^{1-\frac{1}{2}x}(2-x-2y)\,dy\,dx =\int_0^1\bigg[2y-xy-y^2\bigg]_{y=\frac{1}{2}x}^{1-\frac{1}{2}x}\,dx\\[2.5mm] &=\int_0^1\left(\left(1-\frac{1}{2}x\right)^2-x+\frac{1}{2}x^2+\frac{1}{4}x^2\right)\,dx=\int_0^1(1-2x+x^2)\,dx =-\frac{1}{3}(1-x)^3\bigg|_{x=0}^1=\frac{1}{3}. \end{align*}

Stewart §15.6, Example 6
Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder \(x=y^2\) and the planes \(x=z\), \(z=0\) and \(x=1\).

Solution: The projection of the solid \(E\) on the \(xy\)-plane is

\[D=\{(x,y)\,|\,-1\leq y\leq 1,\;y^2\leq x\leq 1\}.\]

Therefore, if the mass density is \(\rho(x,y,z)=\rho\), then

\[m=\iiint\limits_E\rho\,dV=\iint\limits_D\left(\int_0^x\rho\,dz\right)\,dA=\rho\int_{-1}^1\int_{y^2}^1x\,dx\,dy =\frac{1}{2}\rho\int_{-1}^1(1-y^4)\,dy=\rho\int_0^1(1-y^4)\,dy=\rho\bigg[y-\frac{1}{5}y^5\bigg]_{y=0}^1=\frac{4}{5}\rho.\]

If the center of mass is \((\overline{x},\overline{y},\overline{z})\), then due to the symmetry of \(E\) and the constant mass density, we conclude that \(\overline{y}=0\). For the other moments we obtain

\[\iiint\limits_Ex\,\rho\,dV=\rho\iint\limits_D\left(\int_0^xx\,dz\right)\,dA=\rho\int_{-1}^1\int_{y^2}^1x^2\,dx\,dy =\frac{1}{3}\rho\int_{_1}^1(1-y^6)\,dy=\frac{2}{3}\int_0^1(1-y^6)\,dy=\frac{2}{3}\rho\bigg[y-\frac{1}{7}y^y\bigg]_{y=0}^1=\frac{4}{7}\rho\]

and

\[\iiint\limits_Ez\,\rho\,dV=\rho\iint\limits_D\left(\int_0^xz\,dz\right)\,dA=\frac{1}{2}\rho\int_{-1}^1\int_{y^2}^1x^2\,dx\,dy =\frac{1}{6}\int_{-1}^1(1-y^6)\,dy=\frac{1}{3}\int_0^1(1-y^6)\,dy=\frac{1}{3}\rho\bigg[y-\frac{1}{7}y^y\bigg]_{y=0}^1=\frac{2}{7}\rho.\]

This implies that \(\overline{x}=\displaystyle\frac{\frac{4}{7}\rho}{\frac{4}{5}\rho}=\frac{5}{7}\) and \(\overline{z}=\displaystyle\frac{\frac{2}{7}\rho}{\frac{4}{5}\rho}=\frac{5}{14}\). Hence, the center of mass is \((\frac{5}{7},0,\frac{5}{14})\).

Moments of inertia

If \(\rho(x,y,z)\geq0\) is a mass density function defined on a solid \(E\) in \(\mathbb{R}^3\), then the integrals

\[I_x=\iiint\limits_E(y^2+z^2)\rho(x,y,z)\,dV,\quad I_y=\iiint\limits_E(x^2+z^2)\rho(x,y,z)\,dV\quad\textrm{and}\quad I_z=\iiint\limits_E(x^2+y^2)\rho(x,y,z)\,dV\]

are the moments of inertia of the solid \(E\) about the \(x\)-axis, the \(y\)-axis and the \(z\)-axis respectively.

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Last modified on October 4, 2021