REDUCTION TO REPEATED INTEGRALS

Let be a solid in The projection of onto the -plane is denoted by The solid is then the set of all with

in and

The triple integral over can be evaluated by setting

Suppose that takes the form

itself is then the set of all with

The triple integral over can then be expressed by three ordinary integrals:

It is customary to omit the brackets and parentheses and write

Here we first integrate with respect to [from to ], then with respect to [from to ], and finally with respect to [from to ].

There is nothing sacred about this order of integration. Other orders of integration are possible and in some cases more convenient. Suppose, for example, that the projection of onto the -plane is a region of the form

If is the set of all with

then

In this case we integrate first with respect to , then with respect to , and finally with respect to . Still four other orders of integration are possible.

Problem 1. Use triple integration to find the volume of the tetrahedron bounded by the planes

Where is the centroid?

Solution. The volume of is given by the triple integral

To evaluate this triple integral we can project onto any one of the three coordinate planes. We will project onto the -plane. The base region is then the triangle

Since the inclined face is part of the plane , we have as the set of all with

It follows that

By symmetry . We can calculate as follows:

Since we have The centroid is the point

Problem 2. Find the mass of a solid right circular cylinder of radius and height given that the mass density varies directly with the distance from one of the bases.

Solution. Call the solid . We can characterize by the following inequalities:

The first two inequalities define the base region . Supposing that the density varies directly with the distance from the lower base, we have Then

Remark. In Problem 2 we would have profited by not skipping the double integral stage; namely, we could have written

Problem 3. Integrate over that part of the first octant that is cut off by the ellipsoid

Solution. Call the solid . The upper boundary of has equation

This surface intersects the -plane in the curve

We can take

as the base region and characterize as the set of all with

We can therefore calculate the triple integral by evaluating

A straightforward computation gives an answer of .

Another Solution. We return to Problem 3 but this time carry out the integration in a different order. The same solid is now projected onto the -plane. In terms of and the curved surface has equation

This surface intersects the -plane in the curve

We can take

as the base region and characterize as the set of all with

This leads to the repeated integral

which also gives

Problem 4. Use triple integration to find the volume of the solid bounded above by the parabolic cylinder and bounded below by the elliptic paraboloid .

Solution. Solving the two equations simultaneously, we have

and thus

This tells us that the two surfaces intersect in a space curve that lies along the elliptic cylinder The projection of this intersection onto the -plane is the ellipse .

The projection of onto the -plane is the region

The solid is then the set of all with

Its volume is given by

CYLINDRICAL COORDINATES

Introduction to Cylindrical Coordinates

The first two coordinates, and , of the cylindrical coordinates of a point in -space are the usual plane polar coordinates, except that is taken as nonnegative and is restricted to the interval . The third coordinate is the third rectangular coordinate . In rectangular coordinates the coordinate surfaces

are three mutually perpendicular planes. In cylindrical coordinates the coordinate surfaces take the form

The first surface is a circular cylinder of radius . The central axis of the cylinder is the -axis. The surface is a vertical half-plane hinged at the -axis. The plane stands at an angle of radians from the positive -axis. The last coordinate surface is the plane of rectangular coordinates.

The -solids easiest to describe in cylindrical coordinates are the cylindrical wedges. A wedge consists of all points that have cylindrical coordinates in the box

Evaluating Triple Integrals Using Cylindrical Coordinates

Suppose that is some basic solid in -space, not necessarily a wedge. If is the set of all with cylindrical coordinates in some basic solid in -space, then

Derivation. We will carry out the argument on the assumption that is projectable onto some basic region of the -plane. (It is for such solids that the formula is most useful.) has some lower boundary and some upper boundary is then the set of all with

The region has polar coordinates in some set (which we assume is a basic region). Then is the set of all with

Therefore

Volume Formula

If for all in , then the formula reduces to

The triple integral on the left is the volume of . In summary, if is a basic solid in -space and the cylindrical coordinates of constitute a basic solid in -space, then the volume of is given by the formula

Calculations

Cylindrical coordinates are particular]y useful when an axis of symmetry is present. The axis of symmetry is then taken as the -axis.

Problem 1. Find the mass of a solid cylinder of radius and height given that the density varies directly with the distance from the axis of the cylinder.

Solution. Place the cylinder on the -plane so that the axis of coincides with the -axis. The density function then takes the form and consists of all points with cylindrical coordinates in the set

Therefore

Problem 2. Use cylindrical coordinates to find the volume of the solid bounded above by the plane and below by the paraboloid

Solution. In cylindrical coordinates the plane has equation and the paraboloid has equation . Solving these two equations simultaneously, we have . This tells us that the two surfaces intersect in a space curve that lies along the circular cylinder . The projection of this intersection onto the -plane is the circle with polar equation . The base region is thus the set of all with polar coordinates in the set

itself is the set of all with cylindrical coordinates in the set

Therefore

Problem 3. Locate the centroid of the solid in Problem 2.

Solution. Since is symmetric about the -plane, we see that . To get we begin as usual:

Since , we have Now for :

Division by gives The centroid is thus the point

THE TRIPLE INTEGRAL AS THE LIMIT OF RIEMANN SUMS; SPHERICAL COORDINATES

The Triple Integral as the Limit of Riemann Sums

We have seen how single integrals and double integrals can be obtained as limits of Riemann sums. The same holds true for triple integrals.

Start with a basic solid in -space and decompose it into a finite number of basic solids . If is continuous on , then is continuous on each . From each pick an arbitrary point and form the Riemann sum

The triple integral over is the limit of such sums; namely, given any there exists such that, if the diameters of the are all less than , then

no matter how the are chosen within the . We express this by writing

Introduction to Spherical Coordinates

The spherical coordinates of a point in -space have the following meaning. The first coordinate, , is the distance from the origin; thus . The second coordinate, the angle , is the longitude; ranges from to . The third coordinate, the angle marked , ranges only from to is the colatitude, or more simply the polar angle. (The complement of would be the latitude on a globe.)

The coordinate surfaces

have the following meaning. The surface is a sphere; the radius is and the center is the origin. The second surface, , is the same as in cylindrical coordinates: the vertical half-plane hinged at the -axis and standing at an angle of radians from the positive -axis. The surface requires detailed explanation. If or , the surface is a nappe of a cone; it is generated by revolving about the -axis any ray that emerges from the origin at an angle of radians from the positive -axis. The surface is the -plane. (The nappe of the cone has opened up completely.) The equation gives the positive -axis, and the equation gives the negative -axis. (When or , the nappe of the cone has closed up completely.)

Rectangular coordinates are related to spherical coordinates by the following equations:

Conversely, with obvious exclusions, we have

The Volume of a Spherical Wedge

A spherical wedge consists of all points that have spherical coordinates in the box

The volume of this wedge is given by the formula

Proof. Note first that is a solid of revolution. One way to obtain is to rotate the face of , call it , about the -axis for radians. On that face and play the role of polar coordinates. The face is the set of all with polar coordinates in the set The centroid of is at a distance from the -axis where

(area of )

As the face is rotated from to the centroid travels through a circular arc of length

From Pappus formula we know that

the volume of (area of ) =

Evaluating Triple Integrals Using Spherical Coordinates

Suppose that is a basic solid in -space with spherical coordinates in some basic solid of -space. Then

Derivation. Assume first that is a spherical wedge . is then a box . Now decompose into boxes . This induces a subdivision of into spherical wedges . Writing for to save space, we have

Here

and this expression is a Riemann sum for

and, as such, will differ from that integral by less than any preassigned positive number provided only that the diameters of all the are sufficiently small. This we can guarantee by making the diameters of all the sufficiently small.

This verifies the formula for the case where is a spherical wedge.

Volume Formula

If for all in , then the change of variables formula reduces to

The integral on the left is the volume of . It follows that the volume of is given by the formula

Spherical coordinates are commonly used in applications where there is a center of symmetry. The center of symmetry is then taken as the origin.

Calculations

Problem 1. Calculate the mass of a solid ball of radius given that the density varies directly with the square of the distance from the center of the ball.

Solution. Center the ball at the origin. The ball, call it , is now the set of all with spherical coordinates in the box

Therefore

Problem 2. Find the volume of the solid enclosed by the surface

Solution. In spherical coordinates the bounding surface takes the form

This equation places no restriction on ; thus can range from to . Since remains nonnegative, can range only from to Thus the solid is the set of all with spherical coordinates in the set

The rest is straightforward:

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