]> Multidimensional transformation

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 4

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Multidimensional Transformation

Generalization from two dimensions

The multiple integrals of transformed variables obey the same rules as the double integrals. If the multiplicity is n , the absolute value of the Jacobian serves as a magnifier μ for converting the n-dimensional infinitesimal elements of integration to the integration measure in the transformed coordinates, namely for a transformation of n variables

$\begin{array}{l} (x_{1}, x_{2}, ....., x_{n}) \Leftrightarrow (u_{1}, u_{2}, ....., u_{n}) \\ defined by the relations \\ \begin{array}{l} x_{1} = x_{1} (u_{1}, u_{2}, ....., u_{n}) \\ x_{2} = x_{2} (u_{1}, u_{2}, ....., u_{n}) \\ ................................. \\ x_{n} = x_{n} (u_{1}, u_{2}, ....., u_{n}) \end{array}} \end{array}$	(3.2.4.1)

the relation among the infinitesimal elements of integration is

$Δ x_{1} Δ x_{2} ..... Δ x_{n} = \| \frac{D (x_{1}, x_{2}, ....., x_{n})}{D (u_{1}, u_{2}, ....., u_{n})} \| Δ u_{1} Δ u_{2} ..... Δ u_{n}$	(3.2.4.2)

with the same physical dimensions on the either side of the equation. The relation (3.2.4.2) can be proven for any n . We are not going to do that, but just to rely on the simple case of n=2 , presented at the previous page. In addition we are going to study that for a couple of 3-dimensional cases.

As in the case of two-dimensions (3.2.3.17), a multi-dimensional linear transformation is defined by a linear relation among the variables with constant factors, whose determinant is non-vanishing. As a simple example we'll use a linear transformation in 3-dimensions, in order to obtain the volume of an ellipsoid

$\begin{array}{l} by the transformation {\begin{cases} x = a \frac{u}{ρ} \\ y = b \frac{v}{ρ} \\ z = c \frac{w}{ρ} \end{cases}} \\ (a, b, c, ρ are constants) \\ the ellipsoid \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 \\ becomes a sphere u^{2} + v^{2} + w^{2} = ρ^{2} \end{array}}$	(3.2.4.3)

Since

$\begin{array}{l} the Jacobian is \frac{D (x, y, z)}{D (u, v, w)} = \| \begin{matrix} \frac{a}{ρ} & 0 & 0 \\ 0 & \frac{b}{ρ} & 0 \\ 0 & 0 & \frac{c}{ρ} \end{matrix} \| = \frac{a b c}{ρ^{3}} \\ and the volume of the sphere is \frac{4 π ρ^{3}}{3} \\ the volume of the ellipsoid becomes \frac{4 π a b c}{3} \end{array}}$	(3.2.4.4)

in agreement with previous calculations, e.g. (3.2.3.16).

Cylindrical coordinates

The cylindrical coordinates were defined previously as

$\begin{array}{l} x = r \cos φ \\ y = r \sin φ \\ z = z \\ restrictions: \\ r \geq 0 \\ 0 \leq φ < 2 π \end{array}}$	(3.2.4.5)

and were shown at Fig. Cylindrical coordinates. The Jacobian is

$\frac{D (x, y, z)}{D (r, φ, z)} = \| \begin{matrix} \cos φ & - r \sin φ & 0 \\ \sin φ & r \cos φ & 0 \\ 0 & 0 & 1 \end{matrix} \| = r$	(3.2.4.6)

exactly as that of the the polar coordinates. By looking at the Fig. Orthogonality of polar coordinates (studied previously), we'll notice that in order to obtain the infinitesimal volume element, we have to multiply the area of the basis rΔrΔφ by the perpendicular Cartesian height Δz , yielding in first approximation the volume of a cuboid rΔrΔφΔz , where μ = r is the magnifier in accordance to (3.2.4.6). A three dimensional view of the volume element is presented in Fig. Orthogonality of cylindrical coordinates.

The cylindrical coordinates displayed in the Cartesian system form an orthogonal system, since at any point the three lines corresponding to the variation of each one of the variables (by keeping the other two constant) are orthogonal. A right-handed orthogonal coordinate system was defined previously. Similarly to the Cartesian coordinates, one can choose the order of the variables of the cylindrical system, in order to represent it as a right handed coordinate system. By using the rule (as in the case of the Cartesian system) that the positive direction of a coordinate is defined by the increase of the corresponding variable, we can easily see from Fig. Orthogonality of cylindrical coordinates that the order

$r, φ, z$	(3.2.4.7)

forms a right-handed system.

The surfaces corresponding to a constant value of one of the variables shows what kind of problems are easy to be solved with that particular coordinate system. In the case of the cylindrical system:

φ=constant represents half a plane starting from the z axis and going to infinity.
The z=constant plane is known from the Cartesian coordinates.
r=constant represents an infinite cylinder, concentric about the z axis, and for this reason the cylindrical coordinates are in particular suitable for integrations over a three-dimensional region with cylindrical boundaries.

The integration of a function f of the variables over a region of a straight cylinder with radius a and length h is done with cylindrical coordinates by taking the z axis as the axis of the cylinder:

$\begin{array}{l} \int \int \int f (r, φ, z) r d r d φ d z = \int_{0}^{a} r d r \int_{0}^{2 π} d φ \int_{z_{\min}}^{z_{\max}} f (r, φ, z) d z \\ where z_{\max} - z_{\min} = h \end{array}}$	(3.2.4.8)

As an example we'll obtain the mass M of a uniform cylinder with density ρ . From (3.2.4.8) after substituting f with ρ we have

$M = ρ \int_{0}^{a} r d r \int_{0}^{2 π} d φ \int_{z_{\min}}^{z_{\max}} d z = ρ π a^{2} h$	(3.2.4.9)

As a continuation of this example we'll obtain the moment of inertia of the same cylinder about the axis of symmetry (z), by substituting f from (3.2.4.8) with $ρ (x^{2} + y^{2}) = ρ r^{2}$ :

$I_{z} = ρ \int_{0}^{a} r^{3} d r \int_{0}^{2 π} d φ \int_{z_{\min}}^{z_{\max}} d z = ρ \frac{π a^{4} h}{2} = \frac{M a^{2}}{2}$	(3.2.4.10)

The next example uses the same geometry, but the limits of z are refined. We want to calculate the moment of inertia about an axis perpendicular to the axis of the cylinder passing through the centre of mass. We know that the centre of mass is the centroid of the cylinder. The most suitable procedure is to move the coordinates, so that the centre of mass is at the origin of the coordinates, then calculate the moment of inertia about x or y . This simplifies the integration, and avoids the use of Steiner's theorem. By choosing the coordinates in this way, the limits of the integration over z are $z_{\min} = - \frac{h}{2}$ and $z_{\max} = \frac{h}{2}$ . The function for the integration about the y axis is $f = ρ (x^{2} + z^{2}) = ρ (r^{2} \cos^{2} φ + z^{2})$ :

${\begin{cases} I_{y} = ρ \int_{0}^{a} r d r \int_{0}^{2 π} d φ \int_{- \frac{h}{2}}^{\frac{h}{2}} d z (r^{2} \cos^{2} φ + z^{2}) = \\ = ρ [\int_{0}^{a} r^{3} d r \int_{0}^{2 π} d φ \cos^{2} φ \int_{- \frac{h}{2}}^{\frac{h}{2}} d z + \int_{0}^{a} r d r \int_{0}^{2 π} d φ \int_{- \frac{h}{2}}^{\frac{h}{2}} z^{2} d z] = \\ = ρ [\frac{a^{4} π h}{4} + \frac{a^{2} π h^{3}}{12}] = \frac{M}{12} (3 a^{2} + h^{2}) \end{cases}}$	(3.2.4.11)

{\begin{cases} I_{y} = ρ \int_{0}^{a} r d r \int_{0}^{2 π} d φ \int_{- \frac{h}{2}}^{\frac{h}{2}} d z (r^{2} \cos^{2} φ + z^{2}) = \\ = ρ [\int_{0}^{a} r^{3} d r \int_{0}^{2 π} d φ \cos^{2} φ \int_{- \frac{h}{2}}^{\frac{h}{2}} d z + \int_{0}^{a} r d r \int_{0}^{2 π} d φ \int_{- \frac{h}{2}}^{\frac{h}{2}} z^{2} d z] = \\ = ρ [\frac{a^{4} π h}{4} + \frac{a^{2} π h^{3}}{12}] = \frac{M}{12} (3 a^{2} + h^{2}) \end{cases}}

(3.2.4.11)

Spherical coordinates

The spherical coordinates were defined previously at (3.1.4.20) as

$\begin{array}{l} x = R \sin θ \cos φ \\ y = R \sin θ \sin φ \\ z = R \cos θ \\ restrictions: \\ R \geq 0 \\ 0 \leq θ \leq π \\ 0 \leq φ < 2 π \end{array}}$	(3.2.4.12)

and were shown in Fig. Spherical coordinates. The coordinate $φ$ is the same, as the one used in the cylindrical system, R is the length of the line connecting a point with the origin

$\begin{array}{l} R^{2} = r^{2} + z^{2} = x^{2} + y^{2} + z^{2} \\ r = R \sin θ \end{array}}$	(3.2.4.13)

and θ is the single valued angle between R and z:

$θ = acos (\frac{z}{R})$	(3.2.4.14)

called also the polar angle, or polarity. The Jacobian is

$\frac{D (x, y, z)}{D (R, θ, φ)} = \| \begin{matrix} \sin θ \cos φ & R \cos θ \cos φ & - R \sin θ \sin φ \\ \sin θ \sin φ & R \cos θ \sin φ & R \sin θ \cos φ \\ \cos θ & - R \sin θ & 0 \end{matrix} \| = R^{2} \sin θ$	(3.2.4.15)

as can be easily verified by the user. With the aid of (3.2.4.15), the volume of the infinitesimal element is

$Δ x Δ y Δ z = R^{2} Δ R \sin θ Δ θ Δ φ$	(3.2.4.16)

As in the case of the cylindrical coordinates, the spherical coordinates are also orthogonal when displayed in the Cartesian system, as shown at Fig. Orthogonality of spherical coordinates. This figure also shows the surfaces corresponding to constant value of the variables:

R=constant represents a spherical surface with its centre at the origin. This makes the spherical coordinates suitable for integration over a three-dimensional region with spherical boundaries.
θ=constant represents a conical surface of revolution about the z axis. This makes the spherical coordinates also suitable for integration over a three-dimensional region with conical boundaries.
$φ$ =constant represents the same half a plane as in the case of cylindrical coordinates.

The orthogonality of the spherical coordinates allows us to obtain (3.2.4.16) directly as a volume of a cuboid, without the aid of the Jacobian, as shown at Fig. Volume element of spherical coordinates. This figure shows also that the spherical coordinates form a right-handed system by ordering the variables in the following way:

$R, θ, φ$	(3.2.4.17)

In addition it is shown that the area of a spherical surface element (constant R) is represented by

$R^{2} \sin θ d θ d φ$	(3.2.4.18)

and the area of a conic surface element (constant θ) is represented by

$\sin θ R d R d φ$	(3.2.4.19)

Both will be discussed later in this page, in relation to surface integrals.

A simple example of spherical coordinates is an integration of a function f over a spherical region with radius a :

$\int \int \int f (R, θ, φ) R^{2} d R \sin θ d θ d φ = \int_{0}^{a} R^{2} d R \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} f (R, θ, φ) d φ$	(3.2.4.20)

For a uniform sphere with density ρ, (3.2.4.20) yields the well known mass

$M = \int_{0}^{a} R^{2} d R \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} ρ d φ = ρ \frac{a^{3} 2}{3} 2 π = ρ \frac{4 π a^{3}}{3}$	(3.2.4.21)

As a continuation of this example, for the moment of inertia about the z axis the function becomes

$f (R, θ, φ) = ρ (x^{2} + y^{2}) = ρ R^{2} \sin^{2} θ$	(3.2.4.22)

yielding

$I_{z} = ρ \int_{0}^{a} R^{4} d R \int_{0}^{π} \sin^{3} θ d θ \int_{0}^{2 π} d φ = ρ \frac{a^{5}}{5} \frac{4}{3} 2 π = \frac{2}{5} M a^{2}$	(3.2.4.23)

From the symmetry it follows that the moments of inertia about the x and the y axes should be equal to (3.2.4.23), although the function for integration is different. This is easily verifiable, e.g. for the moment of inertia about the y axis we obtain

$\begin{array}{l} f (R, θ, φ) = ρ (x^{2} + z^{2}) = ρ R^{2} (\sin^{2} θ \cos^{2} φ + \cos^{2} θ) \\ {\begin{cases} I_{y} = ρ \int_{0}^{a} R^{4} d R \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} d φ (\sin^{2} θ \cos^{2} φ + \cos^{2} θ) = \\ = ρ π \frac{a^{5}}{5} \int_{0}^{π} \sin θ (\sin^{2} θ + 2 \cos^{2} θ) d θ = \\ = ρ π \frac{a^{5}}{5} (\frac{4}{3} + 2 \times \frac{2}{3}) = ρ \frac{8 π}{5 \times 3} = \frac{2}{5} M a^{2} \end{cases} \end{array}}$	(3.2.4.24)

\begin{array}{l} f (R, θ, φ) = ρ (x^{2} + z^{2}) = ρ R^{2} (\sin^{2} θ \cos^{2} φ + \cos^{2} θ) \\ {\begin{cases} I_{y} = ρ \int_{0}^{a} R^{4} d R \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} d φ (\sin^{2} θ \cos^{2} φ + \cos^{2} θ) = \\ = ρ π \frac{a^{5}}{5} \int_{0}^{π} \sin θ (\sin^{2} θ + 2 \cos^{2} θ) d θ = \\ = ρ π \frac{a^{5}}{5} (\frac{4}{3} + 2 \times \frac{2}{3}) = ρ \frac{8 π}{5 \times 3} = \frac{2}{5} M a^{2} \end{cases} \end{array}}

(3.2.4.24)

as expected, but in a more complicated way.

As another example of spherical coordinates we'll use a cone-like body obtained by cutting it out from a solid sphere (R=a), along a conic surface of constant θ (θ=β). See Fig. Cone-like solid body for clarification. The integration of a function over such a body should be of the form of

$\int_{0}^{a} R^{2} d R \int_{0}^{β} \sin θ d θ \int_{0}^{2 π} f (R, θ, φ) d φ$	(3.2.4.25)

The volume of the body is

$V = \int_{0}^{a} R^{2} d R \int_{0}^{β} \sin θ d θ \int_{0}^{2 π} d φ = \frac{2 π}{3} a^{3} (1 - \cos β)$	(3.2.4.26)

and the z coordinate of the centroid is

$\begin{array}{l} z_{c} = \frac{1}{V} \int_{0}^{a} R^{3} d R \int_{0}^{β} \sin θ \cos θ d θ \int_{0}^{2 π} d φ = \\ = \frac{1}{V} \frac{a^{4}}{4} \frac{\sin^{2} β}{2} 2 π = \frac{3 a}{8} \frac{\sin^{2} β}{1 - \cos β} = \frac{3 a}{8} (1 + \cos β) \end{array}}$	(3.2.4.27)

The examples so far shown on this page, are about regions bounded by the natural surfaces of the coordinate systems. We already know, from integration in the Cartesian coordinate system, how to code the limits of integration of any given boundary surface. The same procedure can be applied for the spherical and the cylindrical coordinate systems. As an example we'll take a cut of a sphere obtained by the intersection of a plane. Let the radius of sphere be R=a cut by the plane z=b with 0<b<a, as illustrated at Fig. Section of sphere.

The integration over θ is between the limits 0 and θ_max , where the latter is obtained from the the relation of z with the spherical coordinates (3.2.4.12):

$z = R \cos θ \Rightarrow b = a \cos θ_{\max} \Rightarrow θ_{\max} = acos (\frac{b}{a})$	(3.2.4.28)

For the integration over R , $R_{\max} = a$ is independent of θ , but $R_{min}$ is. Its dependence is obtainable also from the relation of z with the spherical coordinates (3.2.4.12):

$z = R \cos θ \Rightarrow b = R_{\min} \cos θ \Rightarrow R_{\min} = \frac{b}{\cos θ}$	(3.2.4.29)

Finally the integral of the function f over the region of the section becomes:

$\int_{0}^{acos (\frac{b}{a})} \sin θ d θ \int_{\frac{b}{\cos θ}}^{a} R^{2} dR \int_{0}^{2 π} f (R, θ, φ) d φ$	(3.2.4.30)

The integration over the region of this example could be done also with cylindrical coordinates, although the bounding spherical surface is not natural for the system. From (3.2.4.5) the spherical surface becomes

$x^{2} + y^{2} + z^{2} = a^{2} \Rightarrow r^{2} + z^{2} = a^{2}$	(3.2.4.31)

The integration over z is from a constant $z_{\min} = b$ up to an r dependent value defined by (3.2.4.31):

$r^{2} + z_{\max}^{2} = a^{2} \Rightarrow z_{\max} = \sqrt{a^{2} - r^{2}}$	(3.2.4.32)

The integration over r is from $r_{\min} = 0$ up to the maximal value obtained from (3.2.4.31):

$r_{\max}^{2} + b^{2} = a^{2} \Rightarrow r_{\max} = \sqrt{a^{2} - b^{2}}$	(3.2.4.33)

yielding finally the integral over a function f :

$\int_{0}^{\sqrt{a^{2} - b^{2}}} r d r \int_{b}^{\sqrt{a^{2} - r^{2}}} d z \int_{0}^{2 π} f (r, φ, z) d φ$	(3.2.4.34)

The volume of this spherical section is

$V = \frac{π}{3} (2 a^{3} - 3 a^{2} b + b^{3})$	(3.2.4.35)

and the inquisitive user could compare (3.2.4.35) with his calculations, by using (3.2.4.30) and (3.2.4.34).

Simple surface integrals

We know already how to integrate over a plane surface using the Cartesian two dimensional coordinates. Very often we need to compute integrals over curved surfaces, called by the general name of surface integrals, which is subject to a special branch of mathematical analysis. However at this stage we are able to apply such integration for the limited cases of the natural curved surfaces, obtained by the cylindrical and the spherical coordinates.

The Fig. Orthogonality of cylindrical coordinates shows that the differential area element of a cylinder, symmetric about the z axis is expressed by

$d A = r d φ d z$	(3.2.4.36)

where r represents the constant value of the radius. The integration of a function f over such a cylindrical surface is of the form of

$r \int_{φ_{\min}}^{φ_{\max}} d φ \int_{z_{\min} (φ)}^{z_{\max} (φ)} f (φ, z) d z$	(3.2.4.37)

The function f can be of any origin, e.g. density of a cylindrical tube made of thin material, yielding its mass. The simplest example is the area of such a tube of radius r and hight h , yielding

$A = r \int_{0}^{2 π} d φ \int_{0}^{h} d z = 2 π r h$	(3.2.4.38)

If the density per unit area is a constant σ , then the corresponding mass is

$M = σ A = 2 π σ r h$	(3.2.4.39)

In the next example we'll calculate the moment of inertia of this tube about an axis perpendicular to z , and passing through the centre of mass. By following the steps of a the example (3.2.4.11), but applied to our case (3.2.4.38-39), we obtain

${\begin{cases} I_{y} = σ r \int_{0}^{2 π} d φ \int_{- \frac{h}{2}}^{\frac{h}{2}} d z (r^{2} \cos^{2} φ + z^{2}) = \\ = σ r h \int_{0}^{2 π} d φ (r^{2} \cos^{2} φ + \frac{2 h^{2}}{2^{3} \times 3}) = \\ = \frac{σ r h}{12} (12 π r^{2} + 2 π h^{2}) = \frac{M}{12} (6 r^{2} + h^{2}) \end{cases}}$	(3.2.4.40)

Notice the constant value of r , which corresponds to a in (3.2.4.11). For the same mass (3.2.4.40) is bigger, since the integration used only the maximal value of r , and in (3.2.4.11) r was integrated between zero and its maximal value. Physically speaking, the mass is distributed farther away from the y axis for a cylindrical tube, than that for a solid cylinder.

According to (3.2.4.19) an element of a conical surface is expressed in spherical coordinates by $\sin θ R d R d φ$ , where θ=constant . An integral of the function f over such a conical surface is expressed by

$\sin θ \int_{φ_{min}}^{φ_{max}} d φ \int_{R_{\min} (φ)}^{R_{\max} (φ)} R d R f (R, φ)$	(3.2.4.41)

As an example let's calculate the area of a conical surface symmetric about z , with its head at the origin, θ<π/2 , and R limited from above by a constant value:

$A = \sin θ \int_{0}^{2 π} d φ \int_{0}^{R_{\max}} R d R = π R_{\max}^{2} \sin θ$	(3.2.4.42)

This area is also calculable by a single integral using the method of surface of revolution. As a continuation of this example we'll obtain the z coordinate of the centre of mass, assuming uniform density per area. In such a case, by substituting z=Rcosθ we have

${\begin{cases} z_{c m} A = \sin θ \int_{0}^{2 π} d φ \int_{0}^{R_{\max}} z Rd R = \sin θ \cos θ \int_{0}^{2 π} d φ \int_{0}^{R_{\max}} R^{2} d R = \\ = 2 π \sin θ \cos θ \frac{R_{\max}^{3}}{3} = A \frac{2}{3} \cos θ R_{\max} \end{cases}}$	(3.2.4.43)

yielding finally

$\begin{array}{l} z_{c m} = \frac{2}{3} \cos θ R_{\max} = \frac{2}{3} h \\ h = height of the cone \end{array}}$	(3.2.4.44)

This result is different from that at (3.2.2.20) ( $z_{c m} = \frac{3}{4} h$ ) corresponding to a solid uniform cone, which is not surprising, since in the latter case the mass of a Δz element increases quadratically with z, while in the present case the increase is linear.

According to (3.2.4.18) the element of a spherical surface is expressed in spherical coordinates by $R^{2} \sin θ d θ d φ$ , where R=constant . An integral of the function f over such a surface is expressed by

$R^{2} \int_{φ_{\min}}^{φ_{\max}} d φ \int_{θ {}_{\min}{(φ)}}^{θ {}_{\max}{(φ)}} f (θ, φ) \sin θ d θ$	(3.2.4.45)

As an example the surface of a sphere with radius R has an area of

$A = R^{2} \int_{0}^{2 π} d φ \int_{0}^{π} \sin θ d θ = R^{2} 2 π \times 2 = 4 π R^{2}$	(3.2.4.46)

and the mass of this surface, assuming a uniform mass density of σ , becomes

$M = σ A = 4 π σ R^{2}$	(3.2.4.47)

As a continuation of this example, the moment of inertia about the z axis is

${\begin{cases} I_{z} = σ R^{2} \int_{0}^{2 π} d φ \int_{0}^{π} \sin θ (x^{2} + y^{2}) d θ = \\ = σ R^{4} \int_{0}^{2 π} d φ \int_{0}^{π} \sin^{3} θ d θ = σ R^{4} 2 π \frac{4}{3} = \frac{2}{3} M R^{2} \end{cases}}$	(3.2.4.48)

According to (3.2.4.23) a solid uniform sphere with the same mass and radius has a smaller moment of inertia about the z axis $(\frac{2}{5} M R^{2})$ , the reason being that in the case of a spherical surface the mass is distributed further away from the axis, compared to the case of a solid sphere.

Exercises

Exercise 1. A uniform solid body ( density ρ ) consists of two inseparable parts, one on top of the other according to the Fig. Sketch for exercise 1:

The bottom half is a cylinder with radius a and height h₁ ,
The top half is a cylinder with the same radius, but height - h₂ . This cylinder has a cavity of an inverted cone, with its vertex on the bottom of the cylinder, with the same radius a of the base, and height - h₂ .

The centre of mass of the whole is at the vertex of the conic cavity. Use cylindrical coordinates in order to answer numerically the following questions:

What is the ratio $\frac{h_{2}}{h_{1}}$ ?
What is the ratio $\frac{M_{2}}{M_{1}}$ , where M is the mass of the corresponding part?
What is the ratio $\frac{I_{z 2}}{I_{z 1}}$ , where $I_{z}$ is the moment of inertia about the axis of symmetry of the corresponding part?

You don't have to repeat the integrations that were done in the relevant studied examples.

Exercise 2. A spherical surface, with radius a is defined as

$x^{2} + y^{2} + {(z - a)}^{2} = a^{2}$

Draw a sketch, and check the location of the sphere's centre!
Express this surface in spherical coordinates! What is the domain of the angle θ ?
A volume is bounded by this surface and a conic surface of $θ = β$ , where β is a given constant. What is the volume?
What is the z coordinate of the volume's centroid (z_c)?
What happens with the centroid's location for β→0 ? Does it make sense, and why?

Exercise 3. Obtain the moment of inertia about the z axis (I_z) of the uniform solid ellipsoid

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$
in terms of its mass (M) and the constants a, b, c by answering the following questions. You don't have to to repeat the integrations that were done in the relevant studied examples.

Transform the Cartesian coordinates to (u, v, w) according to the relations

$x = a u y = b v z = c w$
Obtain the Jacobian, and map the ellipsoid at the new coordinate system!

Transform the (u, v, w) coordinates to spherical (s, θ, φ) according to

$\begin{array}{l} u = s \sin θ \cos φ \\ v = s \sin θ \sin φ \\ w = s \cos θ \end{array}$ and find the Jacobian $\frac{D (x, y, z)}{D (s, θ, φ)} = ?$

Express the integral for the required moment of inertia in coordinates (s, θ, φ), and make sure that the limits for the integration are correct!

Solve the integral, and make a consistency check for a=b !

Exercise 4. Answer the following related questions! You don't have to repeat the integrations that were done in the relevant studied examples.

A uniform solid body at $z \geq 0$ is bounded by the surfaces $\begin{array}{l} x^{2} + y^{2} + z^{2} = a^{2} \\ x^{2} + y^{2} + z^{2} = {(γ a)}^{2} \\ z = 0 \end{array}$ where $a > 0 and 0 \leq γ < 1$ are given constants. Find the z coordinate of the centroid (z_c) ! Check if the result for γ=0 is consistent with (3.2.4.27) for $β = \frac{π}{2}$ !

Use the surface integral to calculate the z coordinate of the centroid (z_c) of the uniform half spherical shell $x^{2} + y^{2} + z^{2} = a^{2} for z \geq 0$
Check consistency with the result of question 1 for $γ \to 1$ !

A uniform solid body at $z \geq 0$ is bounded by the surfaces $\begin{array}{l} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1 \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = γ^{2} \\ z = 0 \end{array}$ where $a > 0, b > 0, c > 0 and 0 \leq γ < 1$ are given constants. Use the transformations of exercise 3 (questions 1 and 2) for obtaining the z coordinate of the centroid (z_c) ! Check consistency with question 1, for $c = b = a$ !

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Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 4

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Multidimensional Transformation

Generalization from two dimensions

Cylindrical coordinates

Spherical coordinates

Simple surface integrals

Exercises

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Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)