triple integeration

7/27/2019 Triple Integeration

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Definition and Properties of Triple Integrals

Definition o f Triple Integral

We can introduce the triple integral similar to double integral as a limit of a Riemann sum. We start from the simplest case wh

of integration Uis a rectangular box (Figure 1).

Fig.1

Let the set of numbers {x0,x1, ...,xm} be a partition of [a, b] into small intervals so that the following relations are valid:

Similarly, we can construct partitions of the segment [c, d] along they-axis and the segment [p, q] along thez-axis:

TheRiemann sum of the functionf(x,y,z) over the partition of is defined by

Here (ui , vj , wk) is some point in the rectangular box (xi1,xi)(yi1,yi)(zi1,zi), and the differences are

The triple integralof a functionf(x,y,z) in the parallelepiped is defined as a limit of the Riemann su

maximum values of the differences xi, yj and zkapproach zero:

To define the triple integral over a general region U, we choose a rectangular box containing the giv


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Then we introduce the functiong(x,y,z) so that

Then the triple integral of the function f(x,y,z) over a general region Uis defined as

Propert ies of Triple Integrals

Letf(x,y,z) andg(x,y,z) be functions which are integrable in the region U. Then the following properties are valid:

1.

2.

3. , where kis a constant;

4. If at any point of the region U, then

5. If the region Uis a union of two non-overlapping regions U1 and U2, then

;

6. Let m be the minimum andMbe the maximum value of a continuous functionf(x,y,z) in the region U. Thenfollowing estimate is valid for the triple integral:

where Vis the volume of the integration region U.


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Solution.

First we calculate the volume of the region of integration U:

The estimate of the integral is defined by the inequality

Here the minimum value m of the integrand is

Accordingly, the maximum valueMis

Thus, the estimate of the integral is

Triple Integrals in Cartesian Coordinates

Calculation of a triple integral in Cartesian coordinates can be reduced to the consequent calculation of three integrals of one v

Consider the case when a three dimensional region U is a type I region, i.e. any straight line parallel to the z-axis intersects the

the region U in no more than 2 points. Let the region U be bounded below by the surface z = z1(x,y), and above by the surface

z2(x,y) (Figure 1). The projection of the solid U onto the xy-plane is the region D (Figure 2). We suppose that the

functions z1(x,y) and z2(x,y) are continuous in the region D.

Fig.1 Fig.2

Then for any function f (x,y,z) continuous in the region U we can write the relation:


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Fig.5 Fig.6

Solution.

The projection of the solid region U onto the xy-plane looks as shown in Figure 6. Taking this into account, we findthecorresponding iterated integral:

Example 4

Express the triple integral through iterated integrals in six different ways. The region U lies in the first octant an

by the cylinder x2 + z2 = 4 and the plane y = 3 (Figure 7). Find the value of the integral.


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Fig.7 Fig.8

Solution.

If the order of integration is "z-y-x", then the iterated integral can be written as

Similarly, we can write the iterated integral to carry out the integration in the order "z-x-y":

Now we consider the case of "x-y-z", i.e. when the first inner integral is taken over the variable x.

Then

Since the projection of the solid onto the yz-plane is a rectangle (Figure 8), then by changing the order of integration over yan

Finally we can write the iterated integral in the order "y-x-z" (starting from the inner integral):

The last 6th way looks as follows:

We can use any of the iterated integrals to calculate the value of the initial triple integral. Taking the last one, we get

Make the substitution:


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Here means the absolute value of the Jacobian.

Triple integrals are often easier to evaluate in the cylindrical or spherical coordinates. The corresponding examples are consid

pages

Triple Integrals in Cylindrical Coordinates Triple Integrals in Spherical Coordinates

Some examples of using other transformations of coordinates are given below.

Example 1

Find the volume of the region U defined by the inequalities

Solution.

Obviously, this region is a parallelepiped. It's convinient to change variables in such a way to transform the parallelepiped into

box. In this case the triple integral becomes the product of three integrals of one variable.

Make the following replacement:

The region of integration U' in the new variables u, v, w is defined by the inequalities

The volume of the solid is

Calculate the Jacobian of this transformation. In order to not express the old variables x, y, z through the new ones u, v, w, we

Jacobian of the inverse transformation:

Then

Hence, the volume of the solid is

Example 2

Find the volume of the parallelepiped defined by the inequalities
http://www.math24.net/triple-integrals-in-cylindrical-coordinates.htmlhttp://www.math24.net/triple-integrals-in-cylindrical-coordinates.htmlhttp://www.math24.net/triple-integrals-in-spherical-coordinates.htmlhttp://www.math24.net/triple-integrals-in-spherical-coordinates.htmlhttp://www.math24.net/triple-integrals-in-spherical-coordinates.htmlhttp://www.math24.net/triple-integrals-in-cylindrical-coordinates.html


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Solution.Introduce the new variables

Calculate the Jacobian of the inverse transformation:

By expanding the determinant along the third row, we find its value to be

Then the absolute value of the Jacobian of the direct transformation is

Now it is easy to calculate the volume of the solid:

Triple Integrals in Cylindrical Coordinates

The position of a point M(x,y,z) in the xyz-space in cylindrical coordinates is defined by three numbers: , , z , where is ththe radius vector of the point M onto the xy-plane, is the angle formed by the projection of the radius vector with the x-axis the projection of the radius vector on the z-axis (its value is the same in Cartesian and cylindrical coordinates).


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Fig.1The relationship between cylindrical and Cartesian coordinates of a point is given by

We assume here that

The Jacobian of transformation from Cartesian to cylindrical coordinates is

Then the formula of change of variables for this transformation can be written in the form

Transition from clindrical coordinates makes calculation of triple integrals simpler in those cases when the region of integratio

a cylindrical surface.

Example 1

Evaluate the integral

where the region U is bounded by the surface x2 + y2 1 and the planes z = 0, z = 1 (Figure 2).


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Fig.2 Fig.3

Solution.

It is more convenient to calculate this integral in cylindrical coordinates. Projection of the region of integration ontothe xy-placircle x2 + y2 1 or0 1 (Figure 3).

Notice that the integrand can be written as

Then the integral becomes

The second integral contains the factor which is the Jacobian of transformation of the Cartesian coordinates into cylindrical All the three integrals over each of the variables do not depend on each other. As a result the triple integral is easy to calculate

Example 2

Find he integral

where the region U is bounded by the surfaces x2 + y2 = 3z, z = 3 (Figure 4).


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Fig.4 Fig.5

Solution.

The region of integrations is shown in Figure 4. To calculate the integral we convert it to cylindrical coordinates:

The differential of this transformation is

The equation of the parabolic surface becomes

The projection of the region of integration U onto the xy-plane is the circle x2 + y2 9 with radius = 3 (Figure 5). The coorfrom 0 to 3, the angle ranges from 0 to 2, and the coordinate z ranges from 2/3 to 3. As a result, the integral becomes

Example 3

Using cylindrical coordinates evaluate the integral


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Fig.6 Fig.7

Solution.

The region of integration U is shown in Figure 6. Its projection on the xy-plane is the circle x2 + y2 = 22 (Figure 7).

The new variables in the cylindrical coordinates range within the limits:

Substituting x = cos and y = sin , we find the value of the integral:

Example 4

Calculate the integral using cylindrical coordinates:

The region U is bounded by the paraboloid z = 4 x2 y2, by the cylinder x2 + y2 = 4 and by the planes y = 0, z = 0(Figure


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It is easier to calculate triple integrals in spherical coordinates when the region of integration U is a ball (or some portion of it)

the integrand is a kind of f (x2 + y2 + z2).

It is sometimes more convenient to use so-called generalized spherical coordinates, related to the Cartesian coordinates by the

In this case the Jacobian is

Example 1

Evaluate the integral , where the region of integration U is the ball given by the equationx2 + y2

Solution.

As the region U is a ball and the integrand is expressed by a function depending on f (x2 + y2 + z2), we can convert the triple

spherical coordinates. Make the substitution:

The new variables range within the limits:

Taking into account the Jacobian 2sin, we can write the integral as follows:

Example 2

Calculate the integral

where the region U is the unit ball x2 + y2 + z2 1.

Solution.

This ball is centered at the origin. Hence, the region of integration U in spherical coordinates is described by the inequalities

Writing the integral in spherical coordinates, we have


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As can be seen, the triple integral is transformed into the product of three single-valued integrals, each of which can be calcula

independently. As a result, we obtain

Example 3

Evaluate the integral xyzdxdydz, where the region U is a portion of the ball x2 + y2 + z2 R2 lying in the first octantx

Solution.

We convert the integral to spherical coordinates. Change the variables:

The new variables range within the limits:

Then the integral in spherical coordinates becomes


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Example 4

Find the triple integral

where the region U is bounded by the ellipsoid

Solution.

To calculate the integral we use generalized spherical coordinates by making the following change of variables:

The absolute value of the Jacobian of the transformation is |I| = abc2sin. Therefore, the following relation is valid forthe di

The integral in the new coordinates becomes


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The region of integration U' (being an ellipsoid) in the new coordinates is defined by the inequalities

Then the triple integral can be written as

Example 5

Evaluate the integral

using spherical coordinates.

Solution.

The region of integration is a portion of the ball lying in the first octant (Figures 2,3) and, hence, it is bounded bythe inequaliti


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Fig.2 Fig.3

Taking into account that the integrand is

and the differential is expressed by the formula

we have

Calculation of Volumes Using Triple Integrals

The volume of a solid U in Cartesian coordinates xyz is given by

In cylindrical coordinates, the volume of a solid is defined by the formula

In spherical coordinates, the volume of a solid is expressed as


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Example 1

Find the volume of the cone of height H and base radius R (Figure 2).

Solution.

Fig.1

The cone is bounded by the surface and the plane z = H (Figure 1). Its volume in Cartesian coordinates is

the formula

Calculate this integral in cylindrical coordinates that range within the limits:

As a result, we obtain (do not forget to include the Jacobian ):

Then the volume of the cone is

Example 2


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Find the volume of the ball x2 + y2 + z2 R2.

Solution.

We calculate the volume of the part of the ball lying in the first octant (x 0, y 0, z 0), and then multiply the result by 8. T

As a result, we get the well-known expression for the volume of the ball of radius R.

Example 3

Find the volume of the tetrahedron bounded by the planes passing through the points A (1;0;0), B (0;2;0), C (0;0;3) andthe

coordinate planes Oxy, Oxz, Oyz (Figure 2).

Fig.2 Fig.3

Solution.

The equation of the straight line AB in the xy-plane (Figure 3) is written as y = 2 2x. The variable x ranges here in the interand the variable y ranges in the interval 0 y 2 2x.

Write the equation of the plane ABC in segment form. Since the plane ABC cuts the line segments 1, 2, and 3, respectively, o

and z-axis, then its equation can be written as


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In general case the equation of the plane ABC is written as

Hence, the limits of integration over the variable z range in the interval from z = 0 to . Now we can calcula

of the tetrahedron:

Example 4

Find the volume of the tetrahedron bounded by the planes x + y + z = 5, x = 0, y = 0, z = 0 (Figure 4).

Solution.

The equation of the plane x + y + z = 5 can be rewritten in the form

By setting z = 0, we get


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Fig.4 Fig.5

Hence, the region of integration D in the xy-plane is bounded by the straight line y = 5 x as shown in Figure 5.

Representing the triple integral as an iterated integral, we can find the volume of the tetrahedron:

Example 5

Find the volume of the solid formed by two paraboloids:


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Fig.6 Fig.7

Solution.

Investigate intersection of the two paraboloids (Figure 6). Since 2 = x2 + y2, the equations of the paraboloids can be written

By setting z1 = z2 for the intersection curve, we obtain

For this value of (Figure 7), the coordinate z is

The volume of the solid is expressed through the triple integral as

This integral in cylindrical coordinates becomes


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Example 6

Calculate the volume of the ellipsoid

Solution.

It is easier to calculate the volume of the ellipsoid using generalized spherical coordinates. Let

Since the absolute value of the Jacobian for transformation of Cartesian coordinates into generalized spherical coordinates is

hence,

The volume of the ellipsoid is expressed through the triple integral:

By symmetry, we can find the volume of 1/8 part of the ellipsoid lying in the first octant (x 0, y 0, z 0), and then multip8. The generalized spherical coordinates will range within the limits:

Then the volume of the ellipsoid is

Example 7

Find the volume of the solid bounded by the sphere x2 + y2 + z2 = 6 and the paraboloid x2 + y2 = z.

Solution.

We first determine the curve of intersection of these surfaces. Substituting the equation of the paraboloid into the equation of t

find:

The second root z2 = 3 corresponds to intersection of the sphere with the lower shell of the paraboloid. So we do not consideThus, intersection of the solids happens at z = 2. Obviously, the projection of the region of integration on the xy-plane is the c

defined by the equation x2 + y2 = 2.


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Solution.

First we investigate intersection of the two surfaces. By equating the coordinates z, we get the following equation:

Let x2 + y2 = t2. Then

Only the root t = 1 has the sense in the context of the given problem, i.e.

Thus, both the surfaces intersect at z = 1, and the intersection is a circle (Figure 10).

Fig.10 Fig.11

The region of integration is bounded from above by the paraboloid, and from below by the cone (Figure 11).

To calculate the volume of the solid we use cylindrical coordinates:

As a result, we find:

Physical Applications of Triple Integrals

Mass and Static Moments of a Solid

Suppose we have a solid occupying a region U. Its volume density at a point M(x,y,z) is given by the function (x,y,z).Then the mass of the solid m is expressed through the triple integral as


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The static moments of the solid about the coordinate planes Oxy, Oyz, Oxz are given by the formulas

The coordinates of the center of gravity of the solid are described by the expressions:

If a solid is homogeneous with density (x,y,z) = 1 for all points M(x,y,z) in the region U, then the center of gravity ofthe solionly by the shape of the solid and is called the centroid.

Moments of Inertia of a Solid

The moments of inertia of a solid about the coordinate planes Oxy, Oxz, Oyz are given by

and the moments of inertia of a solid about the coordinate axes Ox, Oy, Oz are expressed by the formulas

As seen, the following properties are valid:

The moment of inertia about the origin is called the integral

The moment of inertia about the origin can be expressed through the moments of inertia about the corodinate planes as follow

Tensor of Inertia

Using the 6 numbers considered above: Ix, Iy, Iz, Ixy, Ixz, Iyz, we can construct the so-called matrix of inertia or the tensor of

solid:

This tensor is symmetric and, hence, it can be transformed to a diagonal view by choosing the appropriate coordinate axesOx'

values of the diagonal elements (after transforming the tensor to a diagonal form) are called the main moments of inertia, and

directions of the axes are called the eigenvalues or the principal axes of inertia of the body.

If a body rotates about an axis which does not coincide with a principal axis of inertia, it will experience vibrations at the highspeeds. Therefore, when designing such devices it is necessary the axis of rotation to be coinciding with one of the principal a

For example, when replacing car tires, it's often necessary to balance the wheels by attaching small lead weights to ensure the

the rotation axis with the principal axis of inertia and to eliminate vibration.

Gravitational Potential and Attraction Force

The Newton potential of a body at a point P(x,y,z) is called the integral


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where (,,) is the density of the body and .

The integration is performed over the whole volume of the body. Knowing the potential, one can calculate the force of attracti

material point of mass m and the distributed body with the density (,,) by the formula

where G is the gravitational constant.

Example 1

Find the centroid of a homogeneous half-ball of radius R.

Solution.

Introduce the system of coordinates in such a way that the half-ball is located at z 0 and centered at the origin (Figure 1).

Fig.1 Fig.2Using this system of coordinates, we find the centroid (the center of gravity) of the solid.

Obviously, by symmetry,

Calculate the coordinate of the centroid by the formula

Since the half-ball is homogeneous, we set (x,y,z) =0. Then

The symbol V in the denominator denotes the volume of the solid, which is equal to


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It remains to compute the triple integral . For this, we pass to spherical coordinates. In this case, the radial coor

denoted by r in order not to be confused with the density . As a result, we have

Thus, the coordinate of the center of gravity is

Example 2

Determine the mass and coordinates of the center of gravity of the unit cube with the density (x,y,z) = x + 2y + 3z(Figure 2).

Solution.

First we calculate the mass of the cube:

Now we calculate the static moments Mxy, Mxz, Myz.


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Similarly, we find the moments Mxz and Myz:

Calculate the coordinates of the center of gravity of the cube:

Example 3

Find the mass of a ball of radius R whose density is proportional to the squared distance from the center.

Solution.

According to the condition of the problem, the density is given by = ar2, where a is a constant, r is the distance fromthe ceconvenient to calculate the mass of the ball in spherical coordinates:

Example 4

Find the moment of inertia of a right circular homogeneous cone about its axis. The cone has base radius R, height H andthe


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total mass m (Figure 3).

Fig.3

Solution.

The moment of inertia of a bodt about the z-axis is expressed by the formula

Because the cone is homogeneous, then the density (x,y,z) =0 can be taken outside the integral sign:

We pass to cylindrical coordinates by replacing

The new variables range within

Then the moment of inertia is

We express the density 0 through the known mass of the cone m. Since

hence,


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Finally we obtain

It is interesting that the moment of inertia of the cone does not depend on its height.

Example 5

With what force does a homogeneous ball of mass M attract a material point of mass m, located at distance a from the center o

R)?

Solution.

Without sacrificing generality, the material point can be placed on the z-axis (Figure 4), so that its coordinate is (0, 0, a).

Fig.4 Fig.5

We solve the problem as follows. First, we calculate the potential of the ball, and then find the force of attraction of the point

find the potential of the ball, it is more convenient to first determine the potential of the sphere (through the surface integral) in

calculating the triple integral, and then get the result for the ball (by performing one more integration).

So, calculate the potential of the sphere of arbitrary radius r (r R). Consider a small area element dS on the sphere as shownThe mass of this area element is

where (r) is the sphere density and dr is its width. This sphere creates the potential at the point P, which is equal to

where the distance from the area element dS to the point P is expressed by the cosine theorem through a, r, .


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Taking into account that the area element is , we get

Calculate separately the integral over the variable . We make the following substitution: Let

Then

As a result, we find the integral

Thus, the potential of the sphere of radius r is given by

Now we can calculate the potential of the ball of radius R. We suppose for simplicity that the density of the ball is constant an

This yields

In this expression 4/3R3 = V is the volume of the ball, and 0V = M is the mass of the ball. Thus, we have proved thatthegravitational potential of the ball at the distance a from the center of the ball (a > R) is expressed by the formula

Further, it is easy to find the force of attraction between the ball and the material point. Since

the force is

The "minus" sign means that the force is directed opposite to the z-axis, i.e. it is the force of attraction.

As can be seen, the attractive force of the ball and the point has the same form as the force of attraction between two point maone of the fundamental results in astrophysics and celestial mechanics. Because of this, the planets and stars can often be cons

material points in the description of their movement. To prove this result, Isaac Newton was even forced to postpone publicati

masterpiece on astronomy. Perhaps the difficulties were related to the fact that he did not use spherical coordinates to solve th

Example 6

Suppose that a planet has a radius R and its density is expressed by the formula


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Find the mass of the planet.

Solution.

Consider in detail the law of density variation. If = R, then

where 0 is a surface density of the planet. The value of as 0 (Figure 6).

Fig.6

We calculate the mass of the planet using triple integral by the formula:

By converting to spherical coordinates, we have

Since the volume of the planet is 4/3R3, the answer can be written in the form:

As can be seen, the mass of the planet is 25% more compared with the case when the density is distributed uniformly.

triple integeration

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