]> Multiple integral

### Chapter 3: Many Variables; Section 2: Integration; page 2

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# Multiple Integral

## Triple integral

The definition of a triple integral (over three variables) is very similar to that of a double integral, but just a bit more complicated. For a double integral, we used a region of the (x,y) plane, divided in rectangular cells, with "area" $\Delta x\Delta y$. In the case of a triple integral, we have to use a region in the (x,y,z) three-dimensional space, divided into cells of rectangular boxes, with "volume" $\Delta x\Delta y\Delta z$.

Formally, a triple integral of a function $f\left(x,y,z\right)$ in a region is obtained by

 $\int \int \int f\left(x,y,z\right)\text{d}x\text{d}y\text{d}z=\underset{\Delta x\to 0}{\mathrm{lim}}\text{ }\underset{\Delta y\to 0}{\mathrm{lim}}\text{ }\underset{\Delta z\to 0}{\mathrm{lim}}\sum f\left(x,y,z\right)\Delta x\Delta y\Delta z$ (3.2.2.1)

with the same requirements and restrictions, used for the definition of a double integral.

As in the case of a double integral, the region can be convex or not, by applying the same criteria. The evaluation of an integral is done again by iterations. The difference is, that the inner-most integral should have the limits defined, generally speaking by a curved surface, e.g. in the case of  z,

 ${z}_{\mathrm{min}}=z{\left(x,y\right)}_{\mathrm{min}}\text{ }\text{and}\text{ }{z}_{\mathrm{max}}=z{\left(x,y\right)}_{\mathrm{max}}$ (3.2.2.2)

As a result, one is left with a double integral in the maximal possible two dimensional region, defined by the original three dimensional one. If the first iteration was done over  z, the two dimensional region will be over  x  and  y.

As an example, for the region enclosed by the ellipsoid

 $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}+\frac{{z}^{2}}{{c}^{2}}=1$ (3.2.2.3)

we'll denote the limits of the triple integral of the function  $f\left(x,y,z\right)$

 $\begin{array}{l}\int \int \int f\left(x,y,z\right)\text{d}x\text{d}y\text{d}z=\underset{{x}_{\mathrm{min}}}{\overset{{x}_{\mathrm{max}}}{\int }}\text{d}x\underset{{y}_{\mathrm{min}}}{\overset{{y}_{\mathrm{max}}}{\int }}\text{d}y\underset{{z}_{\mathrm{min}}}{\overset{{z}_{\mathrm{max}}}{\int }}f\left(x,y,z\right)\text{d}z\\ \begin{array}{cc}{z}_{\mathrm{min}}=-c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}& \text{ }{z}_{\mathrm{max}}=c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}\\ {y}_{\mathrm{min}}=-b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}& {y}_{\mathrm{max}}=b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}\\ {x}_{\mathrm{min}}=-a& {x}_{\mathrm{max}}=a\end{array}\end{array}\right\}$ (3.2.2.4)

Different ordering of the variables for the iterations, could result in simplification or in complication of an integral's solution. It is advisable to apply some common sense, before solving a multiple (double, triple or higher) integral.

Here is a simple example for clarification of this matter. A function $f\left(x,y,z\right)$ should be integrated over a region consisting of an upside-down pyramid, with its head at the origin, and its base at  z =c>0. The base consists of a rectangle, with its corners at  $x=±a\text{ }\text{and}\text{ }y=±b$ . The pyramid is displayed at Fig. rectangular pyramid, and it is advisable to keep this figure-window open, simultaneously to the study of the example.

The first iteration over  z  gives  ${z}_{\mathrm{max}}=c$ , but  ${z}_{\mathrm{min}}$  depends on the triangular side. For instance if we take the side with the corners at

 $\left(x,y,z\right)=\left(a,b,c\right),\text{\hspace{0.28em}}\left(-a,b,c\right),\text{\hspace{0.28em}}\left(0,0,0\right)$ (3.2.2.5)

One can find the plane, where this side lays, by solving the general equation of a plane (3.1.2.36):

 $Ax+By+Cz=D$ (3.2.2.6)

for the points (3.2.2.5). The result is

 $z=\frac{c}{b}y$ (3.2.2.7)

which corresponds to the  ${z}_{\mathrm{min}}$  as function of y. If the second iteration is done over y, then

 $\left\{\begin{array}{l}{y}_{\mathrm{min}}=\frac{b}{a}|x|\\ {y}_{\mathrm{max}}=b\end{array}\right\}\text{ }\text{and}\text{ }\left\{\begin{array}{l}{x}_{\mathrm{min}}=-a\\ {x}_{\mathrm{max}}=a\end{array}\right\}$ (3.2.2.8)

Or if  x  is chosen as the second iteration, then

 $\left\{\begin{array}{l}{x}_{\mathrm{min}}=-\frac{a}{b}y\\ {x}_{\mathrm{max}}=\frac{a}{b}y\end{array}\right\}\text{ }\text{and}\text{ }\left\{\begin{array}{l}{y}_{\mathrm{min}}=0\\ {y}_{\mathrm{max}}=b\end{array}\right\}$ (3.2.2.9)

In summary, if the first iteration is done over  z, the integral over this particular part of the region, can be done in the following ways:

 $\text{or}\text{\hspace{0.28em}}\left\{\begin{array}{l}\underset{-a}{\overset{a}{\int }}\text{d}x\underset{\frac{b}{a}|x|}{\overset{b}{\int }}\text{d}y\underset{\frac{c}{b}y}{\overset{c}{\int }}f\left(x,y,z\right)\text{d}z\\ \underset{0}{\overset{b}{\int }}\text{d}y\underset{-\frac{a}{b}y}{\overset{\frac{a}{b}y}{\int }}\text{d}x\underset{\frac{c}{b}y}{\overset{c}{\int }}f\left(x,y,z\right)\text{d}z\end{array}\right\}$ (3.2.2.10)

There are three more parts of the region, each one with different limits of integration, that should be specified, in order to obtain the final expression of the integral.

On the other hand, by a simple observation of the region, one can greatly simplify the expression. The intersections with the constant  z  planes represent rectangles. For  $z={z}_{0}$ $\left(0<{z}_{0} , the corners of the rectangle are

 $x=±\frac{a}{c}{z}_{0}\text{ }\text{and}\text{ }y=±\frac{b}{c}{z}_{0}$ (3.2.2.11)

which yield finally a single expression of the integral:

 $\underset{0}{\overset{c}{\int }}\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\text{d}y\underset{-\frac{a}{c}z}{\overset{\frac{a}{c}z}{\int }}f\left(x,y,z\right)\text{d}x$ (3.2.2.12)

## Applications

In the case of  $f\left(x,y,z\right)$ = 1, the triple integral (3.2.2.1) yields the "volume" of the region. As an example let's calculate the volume of the ellipsoid (3.2.2.3), as a triple integral, by the use of the limits from (3.2.2.4):

 $V=\underset{-a}{\overset{a}{\int }}\text{d}x\underset{-b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}}{\overset{b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}}{\int }}\text{d}y\underset{-c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}}{\overset{c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}}{\int }}\text{d}z=2c\underset{-a}{\overset{a}{\int }}\text{d}x\underset{-b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}}{\overset{b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}}{\int }}\text{d}y\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}$ (3.2.2.13)

We end up with an expression, exactly the same as that obtained previously in equation (3.2.1.26), by the use of a double integral.

For another example we'll use the rectangular pyramid, defined above and shown in Fig. rectangular pyramid, for calculating its volume. According to (3.2.2.12), this volume can be written as

 $V=\underset{0}{\overset{c}{\int }}\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\text{d}y\underset{-\frac{a}{c}z}{\overset{\frac{a}{c}z}{\int }}\text{d}x=\frac{4ab}{{c}^{2}}\underset{0}{\overset{c}{\int }}{z}^{2}\text{d}z=\frac{4abc}{3}=\frac{Ac}{3}$ (3.2.2.14)

where  A  is the area of the base, and  c  is the height of the pyramid. We could actually write in this case the volume as the single integral

 $V=\underset{0}{\overset{c}{\int }}A\left(z\right)\text{d}z=\underset{0}{\overset{c}{\int }}\left[4ab{\left(\frac{z}{c}\right)}^{2}\right]\text{d}z=\underset{0}{\overset{c}{\int }}A{\left(\frac{z}{c}\right)}^{2}\text{d}z=\frac{Ac}{3}$ (3.2.2.15)

where  A(z)  is the area of the section at any z. Both the x and the y values of the section are linearly increasing with the same factor of  (z/c). The factor is squared, because the area of the section is proportional to the product of  x  and  y.

Following this study, we can state, that actually the result of (3.2.2.14-15) we obtained, is independent of the shape of the area A. The area could be bounded by any closed loop (e.g. an ellipse), and does not have to be a polygon. In other words,

 $V=\frac{Ah}{3}$ (3.2.2.16)

is applicable for the volume V of any pyramid or conical shape, where  A  is the area of the base, and  h  is the height representing the distance between the head (the vertex) and the plane of the base.

In analogy to the case of double integrals, the mean value of a function, in the case of a triple integrals is

 $〈f〉=\frac{\int \int \int f\left(x,y,z\right)\text{d}x\text{d}y\text{d}z}{\int \int \int \text{d}x\text{d}y\text{d}z}$ (3.2.2.17)

in the region of integration. The coordinates of the centroid are the mean values of the Cartesian coordinates.

The coordinates of a body's centre of mass are

 $\left\{\begin{array}{l}{x}_{cm}=\frac{\int \int \int x\rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z}{\int \int \int \rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z}\\ {y}_{cm}=\frac{\int \int \int y\rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z}{\int \int \int \rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z}\\ {z}_{cm}=\frac{\int \int \int z\rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z}{\int \int \int \rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z}\end{array}\right\}\text{\hspace{0.28em}}\text{where}\text{\hspace{0.28em}}\left\{\begin{array}{l}\rho \left(x,y,z\right)\text{\hspace{0.28em}}\text{is}\text{\hspace{0.28em}}\text{the}\text{\hspace{0.28em}}\text{density}\\ M=\left\{\begin{array}{l}\int \int \int \rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z\\ \text{mass}\text{\hspace{0.28em}}\text{of}\text{\hspace{0.28em}}\text{the}\text{\hspace{0.28em}}\text{body}\end{array}\right\}\end{array}\right\}$ (3.2.2.18)

As an example let's calculate the centre of mass of an upside down uniform rectangular pyramid, with its head at the origin, its base at  z=c>0  and with corners at  $x=±a\text{ }\text{and}\text{ }y=±b$ . Since we are familiar with such a body, we'll use the known integral limits of (3.2.2.12). For a uniform density, the centre of mass corresponds to the centroid. One obtains

 $V{x}_{cm}=\underset{0}{\overset{c}{\int }}\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\text{d}y\underset{-\frac{a}{c}z}{\overset{\frac{a}{c}z}{\int }}x\text{d}x=0\text{ }\text{\hspace{0.28em}}⇒\text{\hspace{0.28em}}\text{ }{x}_{cm}=0$ (3.2.2.19)

where V is the volume (3.2.2.14), and the integral vanishes, because of the integral of an odd function with respect to x, as known from the previous chapter at (2.2.1.19). By the same argument  ${y}_{cm}=0$

The z coordinate of the centre of mass is calculated by

 $\begin{array}{l}V{z}_{cm}=\underset{0}{\overset{c}{\int }}z\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\text{d}y\underset{-\frac{a}{c}z}{\overset{\frac{a}{c}z}{\int }}\text{d}x=\frac{4ab}{{c}^{2}}\underset{0}{\overset{c}{\int }}{z}^{3}\text{d}z=ab{c}^{2}\\ {z}_{cm}=\frac{ab{c}^{2}}{V}=\frac{ab{c}^{2}3}{4abc}=\frac{3c}{4}\end{array}\right\}$ (3.2.2.20)

where (3.2.2.14-15) were used. It is worth noticing that the result is independent of  a  and  b, and similarly to (3.2.2.15) this result is also correct for any conical shape, independently on the base's shape.

We already know, that in the case of a double integral, if a region is not convex and the function is integrable also outside of the region, one can obtain the integral as the subtraction of integrals over convex regions. The same holds also for triple (or any other multiple) integrals.

As an example, let's find the centre of mass of a uniform body, in the form of a cuboid (rectangular box), with a cuboid's cavity, laying entirely inside. Let's define first the body and the cavity as follows:

 $\begin{array}{l}\text{sides}\text{\hspace{0.28em}}\text{of}\text{\hspace{0.28em}}\text{the}\text{\hspace{0.28em}}\text{body}\left\{\begin{array}{l}0\le x\le a\\ 0\le y\le b\\ 0\le z\le c\end{array}\\ \text{sides}\text{\hspace{0.28em}}\text{of}\text{\hspace{0.28em}}\text{the}\text{\hspace{0.28em}}\text{cavity}\left\{\begin{array}{l}0\le {a}_{1}\le x\le {a}_{2}\le a\\ 0\le {b}_{1}\le y\le {b}_{2}\le b\\ 0\le {c}_{1}\le z\le {c}_{2}\le c\end{array}\end{array}\right\}$ (3.2.2.21)

From the definitions, and by simple considerations we have

 $\begin{array}{l}\text{full}\text{\hspace{0.28em}}\text{body:}\text{\hspace{0.28em}}{V}_{\text{body}}{〈x〉}_{\text{body}}={}_{\text{body}}\int \int \int x\text{d}x\text{d}y\text{d}z=abc\frac{a}{2}=\frac{{a}^{2}bc}{2}\\ \text{cavity:}\text{\hspace{0.28em}}{V}_{\text{cavity}}{〈x〉}_{\text{cavity}}={}_{\text{cavity}}\int \int \int x\text{d}x\text{d}y\text{d}z=\frac{{\left({a}_{2}-{a}_{1}\right)}^{2}\left({b}_{2}-{b}_{1}\right)\left({c}_{2}-{c}_{1}\right)}{2}\text{ }\end{array}\right\}$ (3.2.2.22)

where  V  is volume, and  <x>  - the x coordinate of the centroid. Similar relations exist for y and for z. We obtain therefore that the required centre of mass is

 $\begin{array}{l}\left\{\begin{array}{l}{x}_{cm}=〈x〉=\frac{{}_{\text{body}}\int \int \int x\text{d}x\text{d}y\text{d}z-{}_{\text{cavity}}\int \int \int x\text{d}x\text{d}y\text{d}z}{{V}_{\text{body}}-{V}_{\text{cavity}}}=\\ =\frac{{a}^{2}bc-{\left({a}_{2}-{a}_{1}\right)}^{2}\left({b}_{2}-{b}_{1}\right)\left({c}_{2}-{c}_{1}\right)}{2\left[abc-\left({a}_{2}-{a}_{1}\right)\left({b}_{2}-{b}_{1}\right)\left({c}_{2}-{c}_{1}\right)\right]}\end{array}\\ \text{similarly:}\\ {y}_{cm}=\frac{a{b}^{2}c-\left({a}_{2}-{a}_{1}\right){\left({b}_{2}-{b}_{1}\right)}^{2}\left({c}_{2}-{c}_{1}\right)}{2\left[abc-\left({a}_{2}-{a}_{1}\right)\left({b}_{2}-{b}_{1}\right)\left({c}_{2}-{c}_{1}\right)\right]}\\ {z}_{cm}=\frac{ab{c}^{2}-\left({a}_{2}-{a}_{1}\right)\left({b}_{2}-{b}_{1}\right){\left({c}_{2}-{c}_{1}\right)}^{2}}{2\left[abc-\left({a}_{2}-{a}_{1}\right)\left({b}_{2}-{b}_{1}\right)\left({c}_{2}-{c}_{1}\right)\right]}\end{array}\right\}$ (3.2.2.23)

without actually doing any integration.

## Moment of inertia

In addition to the centre of mass, the moment of inertia of a rigid body about an axis is another elementary notion of physics, that requires an integral, in order to be evaluated. A rigid body can be considered, as formed by  n  small masses  mi, each one at a distance  ri  from a given axis. Its moment of inertia, with respect to this axis, is defined by

 $I=\sum _{i=1}^{n}{m}_{i}{r}_{i}^{2}$ (3.2.2.24)

For a continuous three dimensional body, the mass of each point is expressed as a differential of the total mass

 ${m}_{i}\text{ }⇒\text{ }\text{d}m=\rho \text{d}V=\rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z$ (3.2.2.25)

where  ρ  is the density (mass/volume) and the moment of inertia becomes

 $I=\int {r}^{2}\text{d}m=\int \int \int {r}^{2}\rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z$ (3.2.2.26)

From (3.2.2.26) we see that the moments of inertia about the  x, y  and  z  axes are expressed as

 $\begin{array}{l}{I}_{x}=\int \int \int \left({y}^{2}+{z}^{2}\right)\rho \text{​}\text{d}x\text{d}y\text{d}z\\ {I}_{y}=\int \int \int \left({z}^{2}+{x}^{2}\right)\rho \text{​}\text{d}x\text{d}y\text{d}z\\ {I}_{z}=\int \int \int \left({x}^{2}+{y}^{2}\right)\rho \text{​}\text{d}x\text{d}y\text{d}z\end{array}\right\}$ (3.2.2.27)

In the case of the body's rotation about an axis by an angular velocity  ω (radians/second), the expression  r²ω²  becomes equivalent to  v², where  v  is the velocity of a point at distance  r  from the axis. Therefore the kinetic energy of rotation becomes

 $K=\frac{1}{2}\int {v}^{2}\text{d}m=\frac{{\omega }^{2}}{2}\int {r}^{2}\text{d}m=\frac{{\omega }^{2}}{2}\int \int \int {r}^{2}\rho \left(x,y,z\right)\text{d}x\text{d}y\text{d}z=\frac{I{\omega }^{2}}{2}$ (3.2.2.28)

which explains the importance of the moment of inertia. The differential of the mass  dm (3.2.2.28) is commonly used in the case of rigid bodies. It simplifies the expressions, and allows the use of the same formulae for linear, planar and three dimensional bodies:

 $\begin{array}{ccc}\underset{¯}{\text{body}\text{\hspace{0.28em}}\text{type}}& \underset{¯}{\text{density}}& \underset{¯}{\text{d}m=}\\ \text{linear,}\text{\hspace{0.28em}}\text{along}\text{\hspace{0.28em}}x& \delta \left(x\right)& \delta \text{d}x\\ \text{planar,}\text{\hspace{0.28em}}\perp \text{\hspace{0.28em}}\text{to}\text{\hspace{0.28em}}z& \sigma \left(x,y\right)& \sigma \text{d}x\text{d}y\\ 3-\text{dimensional}& \rho \left(x,y,z\right)& \rho \text{d}x\text{d}y\text{d}z\end{array}\right\}$ (3.2.2.29)

Notice that the physical dimensions of the densities are different for each type of body. By the use of equation (3.2.2.29), one could write for example the moment of inertia about the  z  axis in this form

 ${I}_{z}=\int \left({x}^{2}+{y}^{2}\right)\text{d}m$ (3.2.2.30)

For example let's calculate the moment of inertia about the  z  axis, of the uniform (constant ρ) rectangular pyramid, used above, and displayed at Fig. rectangular pyramid. We'll use the limits for integration of (3.2.2.12). From (3.2.2.27) we have

 $\left\{\begin{array}{l}{I}_{z}=\rho \underset{0}{\overset{c}{\int }}\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\text{d}y\underset{-\frac{a}{c}z}{\overset{\frac{a}{c}z}{\int }}\left({x}^{2}+{y}^{2}\right)\text{d}x=2\frac{a}{c}\rho \underset{0}{\overset{c}{\int }}\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}z\left(\frac{{a}^{2}{z}^{2}}{3{c}^{2}}+{y}^{2}\right)\text{d}y=\\ =4\frac{ab}{{c}^{2}}\rho \underset{0}{\overset{c}{\int }}{z}^{2}\left(\frac{{b}^{2}{z}^{2}}{3{c}^{2}}+\frac{{a}^{2}{z}^{2}}{3{c}^{2}}\right)\text{d}z=\frac{4ab\left({a}^{2}+{b}^{2}\right)}{3{c}^{4}}\rho \underset{0}{\overset{c}{\int }}{z}^{4}\text{d}z=\\ =\frac{4ab\left({a}^{2}+{b}^{2}\right)c\rho }{15}=\frac{V\rho \left({a}^{2}+{b}^{2}\right)}{5}=\frac{M\left({a}^{2}+{b}^{2}\right)}{5}\end{array}\right\}$ (3.2.2.31)

where the volume  V  is from (3.2.2.14), and  M  is the mass.

From the definition, one can obtain the moment of inertia for any parallel axis to the coordinate's axes. For instance the moment of inertia for an axis parallel to  z, but intersecting the (x,y) plane at the point $\left({x}_{0},{y}_{0}\right)$, from the definition of equation (3.2.2.26) should be:

 $\left\{\begin{array}{l}I=\int \left[{\left(x-{x}_{0}\right)}^{2}+{\left(y-{y}_{0}\right)}^{2}\right]\text{d}m=\\ =\int \left({x}^{2}+{y}^{2}\right)\text{d}m+\int \left({x}_{0}^{2}+{y}_{0}^{2}\right)\text{d}m-2\int \left({x}_{0}x+{y}_{0}y\right)\text{d}m=\\ ={I}_{z}+\left[{x}_{0}^{2}+{y}_{0}^{2}-2\left({x}_{0}{x}_{cm}+{y}_{0}{y}_{cm}\right)\right]M\end{array}\right\}$ (3.2.2.32)

where  Iz  from (3.2.2.30) is used, and  M  is the mass of the body. If the centre of mass (x, y) coordinates are used instead of $\left({x}_{0},{y}_{0}\right)$ , (3.2.2.32) becomes

 ${I}_{cm}={I}_{z}-\left({x}_{cm}^{2}+{y}_{cm}^{2}\right)M$ (3.2.2.33)

where Icm takes the place of  I. The subtraction of (3.2.2.33) from (3.2.2.32) yields

 $\Delta I=I-{I}_{cm}=\left[{\left({x}_{0}-{x}_{cm}\right)}^{2}+{\left({y}_{0}-{y}_{cm}\right)}^{2}\right]M={\left(\Delta {r}_{cm}\right)}^{2}M$ (3.2.2.34)

where  Δrcm  is the distance between the parallel axes of any moment of inertia, with that passing through the centre of mass. The relation (3.2.2.34) is called the theorem of Steiner, and it relates the moments of inertia with parallel axes, by means of the centre of mass. It also shows that among all the moments of inertia with parallel axes, the one with axis passing through the centre of mass is the minimal.

For example, we'll use the previous upside-down pyramid, with its centre of mass given by (3.2.2.19-20), at the  z  axis. It follows that the moment of inertia obtained at (3.2.2.31) is  Icm, therefore any moment of inertia about an axis parallel to  z, with coordinates $\left({x}_{0},{y}_{0}\right)$, is:

 $I=M\left(\frac{{a}^{2}+{b}^{2}}{5}+{x}_{0}^{2}+{y}_{0}^{2}\right)$ (3.2.2.35)

As a continuation of the same example we'll calculate the moment of inertia about the axis parallel to  x, and passing through the centre of mass. This will be done, by first calculating the moment of inertia about the  x  axis, using the integral limits (3.2.2.12):

 $\begin{array}{l}{I}_{x}=\rho \int \left({y}^{2}+{z}^{2}\right)\text{d}V=\rho \underset{0}{\overset{c}{\int }}\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\left({y}^{2}+{z}^{2}\right)\text{d}y\underset{-\frac{a}{c}z}{\overset{\frac{a}{c}z}{\int }}\text{d}x=\\ =2\frac{a}{c}\rho \underset{0}{\overset{c}{\int }}z\text{d}z\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{\int }}\left({y}^{2}+{z}^{2}\right)\text{d}y=2\frac{a}{c}\rho \underset{0}{\overset{c}{\int }}z\text{d}z\left(\frac{{y}^{3}}{3}+{z}^{2}y\right)\underset{-\frac{b}{c}z}{\overset{\frac{b}{c}z}{|}}=\\ =4\frac{ab}{{c}^{2}}\rho \underset{0}{\overset{c}{\int }}{z}^{4}\text{d}z\left(\frac{{b}^{2}}{3{c}^{2}}+1\right)=\frac{4ab{c}^{5}}{15{c}^{4}}\rho \left({b}^{2}+3{c}^{2}\right)=M\frac{\left({b}^{2}+3{c}^{2}\right)}{5}\end{array}\right\}$ (3.2.2.36)

where the mass is  $M=\rho V$  and  V  is given in equation (3.2.2.14).

From Steiner's theorem, it follows that the moment of inertia about an axis parallel to  x, and passing through the centre of mass (3.2.2.19-20) is

 ${I}_{cm}=M\left(\frac{{b}^{2}+3{c}^{2}}{5}-\frac{9{c}^{2}}{16}\right)=\frac{M}{5}\left({b}^{2}+\frac{3{c}^{2}}{16}\right)$ (3.2.2.37)

This example is symmetric under the exchange of  x  with  y, and simultaneously - of  a  with  b. In such a case of symmetry one can apply it, avoiding that way unnecessary calculations. For example, from the moment of inertia about the  x  axis, calculated at (3.2.2.36), one can obtain the moment of inertia about the  y  axis by applying the appropriate symmetry, without doing additional calculations:

 ${I}_{x}=M\frac{\left({b}^{2}+3{c}^{2}\right)}{5}\text{ }⇔\text{ }{I}_{y}=M\frac{\left({a}^{2}+3{c}^{2}\right)}{5}$ (3.2.2.38)

## Higher multiplicity

The procedure used for adding a third variable, can be applied for adding more and more variables, expanding that way the multiplicity of integrals. There is no theoretical obstacle of doing so, although we cannot visualize more than three dimensions. As far as the multiplicity of an integral is increased, generally speaking, the difficulty of its solution also increases. However such multiple integrals are a necessity in many scientific applications.

## Exercises

Exercise 1. Continue the example of the rectangular pyramid, and obtain the appropriate expressions (3.2.2.7-9), for the three sides additional to (3.2.2.5):

1.   $\left(x,y,z\right)=\left(-a,b,c\right),\text{\hspace{0.28em}}\left(-a,-b,c\right),\text{\hspace{0.28em}}\left(0,0,0\right)$
2.   $\left(x,y,z\right)=\left(-a,-b,c\right),\text{\hspace{0.28em}}\left(a,-b,c\right),\text{\hspace{0.28em}}\left(0,0,0\right)$
3.   $\left(x,y,z\right)=\left(a,-b,c\right),\text{ }\left(a,b,c\right),\text{ }\left(0,0,0\right)$

The Fig. rectangular pyramid can be very useful.

Exercise 2. A three dimensional body is enclosed between the planes

$x=0\text{\hspace{0.28em}},\text{ }y=0\text{\hspace{0.28em}},\text{ }z=0\text{\hspace{0.28em}},\text{ }\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

where a, b and c are positive constants. The density of the body is

$\rho ={\rho }_{0}\frac{z}{c}\text{ }\text{with}\text{ }{\rho }_{0}>0$.
1. What is the mass M of the body?
2. What are the coordinates of the centre of mass?
3. What is the moment of inertia about the z axis (Iz)?
4. What is the moment of inertia, about the axis parallel to z, and passing through the centre of mass?

Exercise 3. The base of a uniform pyramid with mass M is a rectangle defined by

$0\le x\le a\text{\hspace{0.28em}},\text{ }0\le y\le b\text{\hspace{0.28em}},\text{ }z=0$

$\left(x,y,z\right)=\left(0,0,c\right)\text{\hspace{0.28em}},\text{ }c>0$
1. Write the limits for integration over x, y and z, in that order of iterations, for integrating in this region!
2. What are the coordinates of the centre of mass?
3. What is the moment of inertia about the z axis (Iz)?
4. What is the moment of inertia about the axis parallel to z, and passing through the centre of the base?

Exercise 4.

1. For a planar body laying on the (x,y) plane, there is a linear expression relating Ix, Iy, and Iz. Find this expression!
2. A rectangular uniform and planar body with mass M lays at the (x,y) plane. Its corners are at xa and yb. What are Ix, Iy, and Iz (in terms of M, a and b) ?
3. The values of Ix, Iy, and Iz, of a straight rectangular and uniform pyramid are known. See (3.2.2.31), (3.2.2.36) and (3.2.2.38). If the pyramid is squashed, so that c→0, the pyramid becomes a planar body. Does the linear relation of the moments of inertia for a flat body hold?
4. For the same mass, are the moments of inertia from questions 2 and 3 the same? Give a qualitative explanation!

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