9.15 Triple Integrals

INTRODUCTION

The steps leading to the definition of the three-dimensional definite integral or triple integral are quite similar to the steps leading to the definition of the double integral. Obvious differences: Instead of a function of two variables we are integrating a function f of three variables, not over a region R in a coordinate plane, but over a region D of 3-space.

  1. Let w = F(x, y, z) be defined over a closed and bounded region D of space.
  2. By means of a three-dimensional grid of vertical and horizontal planes parallel to the coordinate planes, form a partition P of D into n subregions (boxes) Dk of volumes ΔVk that lie entirely in D.
  3. Let be the norm of the partition or the length of the longest diagonal of the Dk.
  4. Choose a sample point (, , ) in each subregion Dk. See FIGURE 9.15.1.
  5. Form the sum (, , ) ΔVk.
A graph. A region labeled D is graphed in a three dimensional x y z coordinate system. Several equally spaced out lines, parallel to the 3 axes, partition the coordinate plane into several identical subregions in the shape of rectangular boxes. One of the boxes inside the three dimensional region contains a sample point (x subscript k asterisk, y subscript k asterisk), and is highlighted.

FIGURE 9.15.1 Sample point in kth subregion

The type of partition used in step 2, where all the lie completely within D, is called an inner partition of D. A sum F(, , ) ΔVk, where (, , ) is an arbitrary point within each Dk and denotes the volume of each Dk, is called a Riemann sum after the German mathematician Georg Friedrich Bernhard Riemann (1826–1866). Although Riemann made significant contributions to complex analysis, he is famous mostly for the first rigorous definition of the integral of a real function on an interval, known to every student of calculus as a definite integral.

DEFINITION 9.15.1 The Triple Integral

Let F be a function of three variables defined over a closed region D of 3-space. Then the triple integral of F over D is given by

.(1)

As in our previous discussions on the integral, when F is continuous over D, the limit in (1) exists; that is, F is integrable over D.

Evaluation by Iterated Integrals

If the region D is bounded above by the graph of z = f2(x, y) and bounded below by the graph of z = f1(x, y), then it can be shown that the triple integral (1) can be expressed as a double integral of the partial integral (x, y, z) dz; that is,

where R is the orthogonal projection of D onto the xy-plane. In particular, if R is a Type I region, then, as shown in FIGURE 9.15.2, the triple integral of F over D can be written as an iterated integral:

.(2)

A graph. A three dimensional region and 2 two dimensional region is graphed on a three dimensional x y z coordinate system. The region labeled D is bound above by a surface labeled z = f subscript 2(x y), and below by a surface labeled z = f subscript 1(x y). The region D is projected vertically down on the x y plane. The projected region labeled R on the x y plane is bound by 2 curves. The first curve, labeled y = g subscript 1(x), begins at a marked point at x = a, and ends at a marked point at x = b. The second curve, labeled y = g subscript 2(x), also begins and ends at the same marked points. A right horizontal arrow inside the region R, is parallel to the y axis. The arrow is divided into several identical rectangles. A subregion in the region D, in the shape of a rectangular box, is projected vertically down on the x y plane as one of the rectangles inside the right arrow.

FIGURE 9.15.2 Geometric interpretation of (2)

To evaluate the iterated integral in (2) we begin by evaluating the partial integral

in which both x and y are held fixed.

In a double integral there are only two possible orders of integration: dy dx and dx dy. The triple integral in (2) illustrates one of six possible orders of integration:

dz dy dx, dz dx dy, dy dx dz,

dx dy dz, dx dz dy, dy dz dx.

The last two differentials tell the coordinate plane in which the region R is situated. For example, the iterated integral corresponding to the order of integration dx dz dy must have the form

.

The geometric interpretation of this integral and the region R of integration in the yz-plane are shown in FIGURE 9.15.3.

A graph. A region, resembling a cylinder with a distorted circular base and parallel to the x axis, is bound by 2 surfaces. The region labeled D is bound above by a surface labeled x = h subscript 1(y, z), and below by a surface labeled x = h subscript 2(y, z). The three dimensional region is projected on the y z plane, parallel to the x axis. The projected region labeled R on the y z plane is bound by 2 curves. The curves above is labeled z = k subscript 2(y) and the curve below is labeled z = k subscript 1(y). Both curves begin at a point on the y z plane at y = c and end at a point at y = d. A vertical up arrow inside the region R, is parallel to the z axis. The arrow is divided into several identical rectangles. A subregion in the region D, in the shape of a rectangular box, is projected horizontally on the y z plane, parallel to the x axis, as one of the rectangles inside the right arrow.

FIGURE 9.15.3 Integration: first x, then z, then y

Applications

A list of some of the standard applications of the triple integral follows:

Volume: If F(x, y, z) = 1, then the volume of the solid D is

Mass: If ρ(x, y, z) is density, then the mass of the solid D is given by

First Moments: The first moments of the solid about the coordinate planes indicated by the subscripts are given by

Center of Mass: The coordinates of the center of mass of D are given by

Centroid: If ρ(x, y, z) = a constant, the center of mass is called the centroid of the solid.

Second Moments: The second moments, or moments of inertia of D about the coordinate axes indicated by the subscripts, are given by

Radius of Gyration: As in Section 9.10, if I is a moment of inertia of the solid about a given axis, then the radius of gyration is

EXAMPLE 1 Volume of a Solid

Find the volume of the solid in the first octant bounded by the graphs of z = 1 – y2, y = 2x, and x = 3.

SOLUTION

As indicated in FIGURE 9.15.4(a), the first integration with respect to z is from 0 to 1 – y2. Furthermore, from Figure 9.15.4(b) we see that the projection of the solid D in the

2 graphs. Graph a. A quarter of a cylinder and a plane is graphed on a three dimensional x y z plane. The quarter cylinder is labeled z = 1 minus y^2. The central axis of the quarter cylinder is on the x axis. The length of the cylinder ranges from x = 0 to x = 3. A vertical plane, following the line labeled y = 2 x on the x y plane, begins on the z axis and intersected the quarter cylinder. A portion of the region of the cylinder is shaded, labeled D, and bound by a lateral surface and 4 planes. The lateral surface is bound by 2 lines and 2 curves The first line begins at the point (0, 0, 1), goes horizontally parallel to the x axis, and ends at the point (3, 0, 1). The first curve begins at the point (3, 0, 1), goes downward following a circular path, and ends at the point (3, 1, 0). The second line begins at the point (3, 1, 0), goes horizontally parallel to the x axis, and ends at the point (1 over 2, 1, 0). The second curve begins at the point (1 over 2, 1, 0), goes upward following a circular path on the graphed plane, and ends at the point (0, 0, 1). The region is also bound by the following planes: The graphed plane, the plane labeled x = 3, the plane z = 0, and the plane y = 0. The vertical projection of the region D on the x y plane is a trapezoidal region bound by vertices at the following points: (0, 0, 0), (3, 0, 0), (3, 1, 0), and (1 over 2, 1, 0). A subregion of the region D, in the form of a rectangular box, is projected vertically down on the y x plane, in the shaded trapezoidal region, as a rectangle. Graph b. A trapezoidal region is graphed on an x y coordinate plane. The trapezoid is formed by 4 lines. The first line, labeled x = 1 over 2 y, begins at the origin, goes up and to the right, and ends at the point (1 over 2, 1). The second line, labeled y = 1, begins at the point (1 over 2, 1), goes horizontally to the right, and ends at the point (3, 1). The third line, labeled x = 3, begins at the point (3, 1), goes vertically down, and ends on the x axis at the point (3, 0). The fourth line begins at the point (3, 0), goes horizontally to the left, along the x axis, and ends at the origin. The region inside the trapezoid is shaded and labeled R. A horizontal right arrow inside the shaded region R is divided into several identical rectangles. One of the rectangles is highlighted.

FIGURE 9.15.4 Solid D and region R of integration in Example 1

xy-plane is a region of Type II. Hence, we next integrate, with respect to x, from y/2 to 3. The last integration is with respect to y from 0 to 1. Thus,

≡

EXAMPLE 2 Changing the Order of Integration

Change the order of integration in

to dy dx dz.

SOLUTION

As seen in FIGURE 9.15.5(a), the region D is the solid in the first octant bounded by the three coordinate planes and the plane 2x + 3y + 4z = 12. Referring to Figure 9.15.5(b) and the table, we conclude that

2 graphs. Graph a. A shaded region is graphed on a three dimensional x y z coordinate system. The region is bound by the plane z = 0, the plane y = 0, the plane x = 0, and a graphed plane labeled z = 3 minus 1 over 2 x minus 3 over 4 y. The graphed plane is bound by 3 lines. The first line begins at the point (0, 0, 3), goes down on the x z plane, and ends at the point (6, 0, 0). The second line begins at the point (0, 0, 3), goes down on the y z plane, and ends at the point (0, 4, 0). The third line, labeled y = 4 minus 2 over 3 x, begins at the point (0, 4, 0), goes along the x y plane, and ends at the point (6, 0, 0). A right horizontal arrow inside the region on the x y plane, is parallel to the y axis. The arrow is divided into several identical rectangles. A subregion in the shaded region above the arrow, in the shape of a rectangular box, is projected vertically down on the x y plane as one of the rectangles, highlighted inside the right arrow. Graph b. A shaded region is graphed on a three dimensional x y z coordinate system. The region is bound by the plane z = 0, the plane y = 0, the plane x = 0, and a graphed plane labeled y = 4 minus 2 over 3 x minus 4 over 3 z. The graphed plane is bound by 3 lines. The first line, labeled x = 6 minus 2 z, begins at the point (0, 0, 3), goes down on the x z plane, and ends at the point (6, 0, 0). The second line begins at the point (0, 0, 3), goes down on the y z plane, and ends at the point (0, 4, 0). The third line begins at the point (0, 4, 0), goes along the x y plane, and ends at the point (6, 0, 0). A horizontal arrow inside the region on the x z plane, is parallel to the x axis. The arrow is divided into several identical rectangles. A subregion in the shaded region to the right of the arrow, in the shape of a rectangular box, is projected horizontally to the left on the x z plane as one of the rectangles, highlighted inside the arrow.

FIGURE 9.15.5 Changing order of integration in Example 2 ≡

Depending on the geometry of a region in 3-space, the evaluation of a triple integral over that region may be made easier by utilizing a new coordinate system.

Cylindrical Coordinates

The cylindrical coordinate system combines the polar description of a point in the plane with the rectangular description of the z-component of a point in space. As seen in FIGURE 9.15.6(a), the cylindrical coordinates of a point P are denoted by the ordered triple (r, θ, z). The word cylindrical arises from the fact that a point P in space is determined by the intersection of the planes z = constant and θ = constant with a cylinder r = constant. See Figure 9.15.6(b).

2 graphs. Graph a. A point is graphed on a three dimensional x y z coordinate system. The coordinates of the point labeled P are as follows: (x y z) or (r, theta, z). The point P is projected vertically down on the x y plane. The distance between the projected point and the origin O is labeled r. The distance between the projected point and the y axis is labeled x. The distance between the projected point and the x axis is labeled y. The angle formed by the x axis and the line labeled r, is labeled theta. The coordinates of the projected point is (r, theta). Graph b. A cylinder and 2 planes are graphed on a three dimensional x y z coordinate system. The central axis of the cylinder is on the vertical z axis. The first plane is horizontal, labeled z = constant, and limits the cylinder at the top. The top circular base is bound by a circle that passes through a point labeled P. The second plane, labeled theta = constant, is vertical, begins on the z axis, and ends on the lateral surface of the cylinder in the first octant at the point P.

FIGURE 9.15.6 Cylindrical coordinates

Conversion of Cylindrical Coordinates to Rectangular Coordinates

From Figure 9.15.6(a) we also see that the rectangular coordinates (x, y, z) of a point can be obtained from the cylindrical coordinates (r, θ, z) by means of

(3)

EXAMPLE 3 Cylindrical to Rectangular Coordinates

Convert (8, π/3, 7) in cylindrical coordinates to rectangular coordinates.

SOLUTION

From (3),

Thus, (8, π/3, 7) is equivalent to (4, 4, 7) in rectangular coordinates. ≡

Conversion of Rectangular Coordinates to Cylindrical Coordinates

To express rectangular coordinates (x, y, z) as cylindrical coordinates, we use

(4)

EXAMPLE 4 Rectangular to Cylindrical Coordinates

Convert () in rectangular coordinates to cylindrical coordinates.

SOLUTION

From (4) we see that

If we take r = 2, then, consistent with the fact that x < 0 and y > 0, we take θ = 3π/4.* Consequently, ( 1) is equivalent to (2, 3π/4, 1) in cylindrical coordinates. See FIGURE 9.15.7. ≡

A graph. A point is graphed on a three dimensional x y z coordinate system. The coordinates of the point are as follows: (negative sqrt(2), sqrt(2), 1) or (2, 3 pi over 4, 1). The point is projected vertically down on the x y plane. An arrow beginning at the origin O and ending at the projected point, is labeled r = 2. The angle formed by the positive x axis and the arrow labeled r = 2, is labeled theta = 3 pi over 4. The coordinates of the projected point are (negative sqrt(2), sqrt(2), 0).

FIGURE 9.15.7 Converting rectangular to cylindrical coordinates in Example 4

Triple Integrals in Cylindrical Coordinates

Recall from Section 9.11 that the area of a polar rectangle is ΔA = r* Δr Δθ, where r* is the average radius. From FIGURE 9.15.8(a) we see that the volume of a cylindrical wedge is simply ΔV = (area of base)(height) = r* Δr Δθ Δz. Thus, if F(r, θ, z) is a continuous function over the region D, as shown in Figure 9.15.8(b), then the triple integral of F over D is given by

.

2 graphs. Graph a. A cylindrical wedge is graphed on a three dimensional x y z coordinate system. The length of the cylindrical wedge is labeled delta r, and the height is labeled delta z. 2 arcs begin at the origin and extend outward on the x y plane. The length of the arcs is labeled r asterisk and the angle between them is labeled delta theta. The cylindrical wedge is projected vertically as a polar rectangle at the right end of the region between the arcs. Graph b. A region labeled D is graphed on a three dimensional x y z coordinate system. The region is a portion of a three dimensional circular sector bound by a surface above labeled z = f subscript 2(r, theta), and a surface labeled z = f subscript 1(r, theta) below. The region D is projected vertically down on the x y plane. The shaded projected surface labeled R is bound by 2 circular arcs and 2 rays. The 2 rays originate from the origin. The ray closer to the x axis is labeled theta = alpha and the ray closer to the y axis is labeled theta = beta. The circular arc closer to the origin is labeled r = g subscript 1(theta), and the second circular arc is labeled r = g subscript 2(theta). 2 more rays originate between the first 2 rays and are closer together. A subregion inside the region D in the shape of a circular wedge is projected vertically down on the x y plane as a polar rectangle, placed between the third and fourth rays.

FIGURE 9.15.8 Cylindrical wedge in (a); region D in (b)

EXAMPLE 5 Center of Mass

A solid in the first octant has the shape determined by the graph of the cone z = and the planes z = 1, x = 0, and y = 0. Find the center of mass if the density is given by ρ(r, θ, z) = r.

SOLUTION

In view of (4), the equation of the cone is z = r. Hence, we see from FIGURE 9.15.9 that

A graph. A solid is graphed in the first octant of a three dimensional x y z coordinate system. The solid is a cone bound by the planes z = 1, x = 0 and y = 0. The lateral surface of the portion of the cone is labeled z = r. The region, bound by the solid and labeled D, is projected vertically down on the x y plane, as a shaded region in the shape of quarter circle, labeled r = 1. A subregion inside the region D in the shape of a rectangular box is projected vertically down on the x y plane as a polar rectangle inside the shaded region, placed between 2 rays originating from the origin. The x axis is also labeled theta = 0 and the y axis is also labeled theta = pi over 2.

FIGURE 9.15.9 Solid in Example 5

In the integrals for Mxz and Myz we substitute y = r sin θ and x = r cos θ:

Hence,

The center of mass has the approximate coordinates (0.38, 0.38, 0.8). ≡

Spherical Coordinates

As seen in FIGURE 9.15.10(a), the spherical coordinates of a point P are given by the ordered triple (ρ, ϕ, θ), where ρ is the distance from the origin to P, ϕ is the angle between the positive z-axis and the vector , and θ is the angle measured from the positive x-axis to the vector projection of .* Figure 9.15.10(b) shows that a point P in space is determined by the intersection of a cone ϕ = constant, a plane θ = constant, and a sphere ρ = constant, whence arises the name “spherical” coordinates.

2 graphs. Graph a. A point is graphed on a three dimensional x y z coordinate system. The coordinates of the point labeled P are as follows: (x y z) or (rho, phi, theta). An arrow beginning at the origin O and ending at the point P, is labeled rho. The angle between the z axis and the arrow O P is labeled phi. The point P is projected vertically down on the x y plane. The projected point is labeled Q. The distance between the point Q and the y axis is labeled x. The distance between the point Q and the x axis is labeled y. A second arrow begins at the origin O and ends at the point Q. The angle formed by the x axis and the second arrow, is labeled theta. Graph b. A sphere and 2 planes are graphed on a three dimensional x y z coordinate system. The center of the sphere is at the origin. The first plane is horizontal, labeled phi = constant (cone), and limits the sphere at the top. The intersection of the plane and the sphere is a circle that passes through a point labeled P. The second plane, labeled theta = constant (plane), is vertical, begins on the z axis inside the sphere, and ends on the lateral surface of the sphere in the first octant, at the point P.

FIGURE 9.15.10 Spherical coordinates

Conversion of Spherical Coordinates to Rectangular and Cylindrical Coordinates

To transform from spherical coordinates (ρ, ϕ, θ) to rectangular coordinates (x, y, z), we observe from Figure 9.15.10(a) that

Since = ρ sin ϕ and = ρ, the foregoing equations become

(5)

It is customary to take ρ ≥ 0 and 0 ≤ ϕ ≤ π. Also, since = ρ sin ϕ = r, the formulas

(6)

enable us to transform from spherical coordinates (ρ, ϕ, θ) to cylindrical coordinates (r, θ, z).

EXAMPLE 6 Spherical to Rectangular and Cylindrical Coordinates

Convert (6, π/4, π/3) in spherical coordinates to rectangular coordinates and cylindrical coordinates.

SOLUTION

Identifying ρ = 6, ϕ = π/4, and θ = π/3, we find from (5) that

The rectangular coordinates of the point are

From (6) we obtain

Thus, the cylindrical coordinates of the point are ≡

Conversion of Rectangular Coordinates to Spherical Coordinates

To transform from rectangular coordinates to spherical coordinates, we use

(7)

Triple Integrals in Spherical Coordinates

As seen in FIGURE 9.15.11, the volume of a spherical wedge is given by the approximation

ΔV ≈ ρ2 sin ϕ Δρ Δϕ Δθ.

Thus, in a triple integral of a continuous spherical coordinate function F(ρ, ϕ, θ), the differential of volume dV is given by

dV = ρ2 sin ϕ dρ dϕ dθ.

A typical triple integral in spherical coordinates has the form

.

A graph. A spherical wedge is graphed on a three dimensional x y z coordinate system. 2 sets or circular arcs and 3 sets of rays and 2 sets of arcs are graphed in the first octant. Each set of circular arcs contains 2 arcs, and each set of rays contains 2 rays of length labeled rho. The 2 arcs of radius rho in the first set, begin on the z axis, extend downward and end at 2 points in the x y plane. The first set of 2 rays begins at the origin and extends out on the x y plane and end at the end points of the first 2 arcs. The angle between the closest ray to the x axis, and the x axis, is labeled theta. The angle between the rays is labeled delta theta. The 2 arcs in the second set, begin on the z axis under the first set of arcs, extend downward and end on the first 2 rays respectively. The length between the 2 sets of arcs is labeled delta rho. The second set of 2 rays begin at the origin, and extend upward passing through the second set of arcs and ending on the first set of arcs respectively. The third set of 2 rays begin at the origin, and extend upward between the second set of rays and the z axis, passing through the second set of arcs and ending on the first set of arcs respectively. The angle between the third set of rays and the z axis is labeled phi. The horizontal distance between the intersection of the third set of rays with the second set of arcs, and the z axis is labeled rho sin phi. The length between the rays in the third set on the second set of arcs, is labeled rho sin phi delta theta. The angle between the rays that are closer to the x axis in the second and third sets, is labeled delta phi. The length between the rays that are closer to the x axis in the second and third sets on the second set arcs, is labeled rho delta phi. The spherical wedge shaped region bound by the 2 sets of arcs and the second and third set of rays, is shaded.

FIGURE 9.15.11 Spherical wedge

EXAMPLE 7 Moment of Inertia

Find the moment of inertia about the z-axis of the homogeneous solid bounded between the spheres

x2 + y2 + z2 = a2 and x2 + y2 + z2 = b2, a < b.

SOLUTION

If δ(ρ, ϕ, θ) = k is the density,* then

From (5) we find x2 + y2 = ρ2 sin2 ϕ and x2 + y2 + z2 = ρ2. Thus the equations of the spheres are simply ρ = a and ρ = b. See FIGURE 9.15.12. Consequently, in spherical coordinates the foregoing integral becomes

≡

A graph. 2 concentric spheres are graphed on a three dimensional x y z coordinate system. A set of arrows go completely around the larger sphere on the x y plane, beginning on the positive y axis. The following information is given beside: theta varies from 0 to 2 pi. A second set of arrows go around the larger sphere from the top to the bottom, on the x z plane. The following information is given beside: phi varies from 0 to pi. A third set of arrows begin on the surface of the smaller sphere, go along the radius, and end on the larger sphere. The following information is given beside: rho varies from a to b.

FIGURE 9.15.12 Limits of integration in Example 7

REMARKS

Spherical coordinates are used in navigation. If we think of the Earth as a sphere of fixed radius centered at the origin, then a point P can be located by specifying two angles θ and ϕ. As shown in FIGURE 9.15.13, when ϕ is held constant, the resulting curve is called a parallel. Fixed values of θ result in curves called great circles. Half of one of these great circles joining the north and south poles is called a meridian. The intersection of a parallel and a meridian gives the position of a point P. If 0° ≤ ϕ ≤ 180° and –180° ≤ θ ≤ 180°, the angles 90° – ϕ and θ are said to be the latitude and longitude of P, respectively. The prime meridian corresponds to a longitude of 0°. The latitude of the equator is 0°; the latitudes of the north and south poles are, in turn, +90° (or 90° North) and –90° (or 90° South).

The spherical coordinate system to locate a point consists of 2 sets of intersecting circles on the surface of the earth. The first set of circles joins the north and south poles. The following information is given about one of these circles: theta = constant. The second set of circles is parallel to the largest horizontal circle at the center labeled equator, and extend from the North Pole to the South Pole. The following information is given about one of these circles: phi constant. A point labeled P is located at the intersection of 2 circles. One of the circles connecting the north and south poles on the right is labeled prime meridian. A horizontal 2 way arrow along one of the parallel circles is labeled longitude. A vertical 2 way arrow along one the circles connecting the North and South Poles, is labeled latitude.

FIGURE 9.15.13 Parallels and great circles

9.15 Exercises Answers to selected odd-numbered problems begin on page ANS-26.

In Problems 1–8, evaluate the given iterated integral.

  9. Evaluate DzdV, where D is the region in the first octant bounded by the graphs of y = x, y = x – 2, y = 1, y = 3, z = 0, and z = 5.
  10. Evaluate D(x2 + y2) dV, where D is the region bounded by the graphs of y = x2, z = 4 – y, and z = 0.

In Problems 11 and 12, change the indicated order of integration to each of the other five orders.

In Problems 13 and 14, consider the solid given in the figure. Set up, but do not evaluate, the integrals giving the volume V of the solid using the indicated orders of integration.

  1. A graph. A solid is graphed in the first octant of a three dimensional x y z coordinate system. The solid is bound by the horizontal plane z = 0 below and the plane z = 4 above, the vertical plane y = 8 on the right, and a vertical surface that follows the curve labeled y = x^3 on the x y plane.

    FIGURE 9.15.14 Solid for Problem 13

    1. dz dy dx
    2. dx dz dy
    3. dy dx dz
  2. A graph. A solid is graphed in the first octant of a three dimensional x y z coordinate system. The solid is bound by the horizontal plane z = 0, the vertical planes y = 0 on the left and y = 3 on the right, a curved surface perpendicular to the x z plane that follows the curve labeled z = sqrt(x) on the x z plane, and a plane perpendicular to the x z plane that follows the line labeled x + z = 2 on the x z plane.

    FIGURE 9.15.15 Solid for Problem 14

    1. dx dz dy
    2. dy dx dz
    3. dz dx dy

[Hint: Part (c) will require two integrals.]

In Problems 15–20, sketch the region D whose volume V is given by the iterated integral.

In Problems 21–24, find the volume of the solid bounded by the graphs of the given equations.

  1. x = y2, 4 – x = y2, z = 0, z = 3
  2. x2 + y2 = 4, z = x + y, the coordinate planes, first octant
  3. y = x2 + z2, y = 8 – x2 – z2
  4. x = 2, y = x, y = 0, z = x2 + y2, z = 0
  5. Find the center of mass of the solid given in FIGURE 9.15.14 if the density at a point P is directly proportional to the distance from the xy-plane.
  6. Find the centroid of the solid in FIGURE 9.15.15 if the density is constant.
  7. Find the center of mass of the solid bounded by the graphs of x2 + z2 = 4, y = 0, and y = 3 if the density at a point P is directly proportional to the distance from the xz-plane.
  8. Find the center of mass of the solid bounded by the graphs of y = x2, y = x, z = y + 2, and z = 0 if the density at a point P is directly proportional to the distance from the xy-plane.

In Problems 29 and 30, set up, but do not evaluate, the iterated integrals giving the mass of the solid that has the given shape and density.

  1. x2 + y2 = 1, z + y = 8, z – 2y = 2; ρ(x, y, z) = x + y + 4
  2. x2 + y2 – z2 = 1, z = –1, z = 2; ρ(x, y, z) = z2 [Hint: Do not use dz dy dx.]
  3. Find the moment of inertia of the solid in Figure 9.15.14 about the y-axis if the density is as given in Problem 25. Find the radius of gyration.
  4. Find the moment of inertia of the solid in Figure 9.15.15 about the x-axis if the density is constant. Find the radius of gyration.
  5. Find the moment of inertia about the z-axis of the solid in the first octant that is bounded by the coordinate planes and the graph of x + y + z = 1 if the density is constant.
  6. Find the moment of inertia about the y-axis of the solid bounded by the graphs of z = y, z = 4 – y, z = 1, z = 0, x = 2, and x = 0 if the density at a point P is directly proportional to the distance from the yz-plane.

In Problems 35–38, convert the point given in cylindrical coordinates to rectangular coordinates.

In Problems 39–42, convert the point given in rectangular coordinates to cylindrical coordinates.

  1. (1, –1, –9)
  2. , 2, 17)
  4. (1, 2, 7)

In Problems 43–46, convert the given equation to cylindrical coordinates.

  1. x2 + y2 + z2 = 25
  2. x + y – z = 1
  3. x2 + y2 – z2 = 1
  4. x2 + z2 = 16

In Problems 47–50, convert the given equation to rectangular coordinates.

  1. z = r2
  2. z = 2r sin θ
  3. r = 5 sec θ
  4. θ = π/6

In Problems 51–58, use triple integrals and cylindrical coordinates. In Problems 51–54, find the volume of the solid that is bounded by the graphs of the given equations.

  1. x2 + y2 = 4, x2 + y2 + z2 = 16, z = 0
  2. z = 10 – x2 – y2, z = 1
  3. z = x2 + y2, x2 + y2 = 25, z = 0
  4. y = x2 + z2, 2y = x2 + z2 + 4
  5. Find the centroid of the homogeneous solid that is bounded by the hemisphere z = and the plane z = 0.
  6. Find the center of mass of the solid that is bounded by the graphs of y2 + z2 = 16, x = 0, and x = 5 if the density at a point P is directly proportional to distance from the yz-plane.
  7. Find the moment of inertia about the z-axis of the solid that is bounded above by the hemisphere z = and below by the plane z = 2 if the density at a point P is inversely proportional to the square of the distance from the z-axis.
  8. Find the moment of inertia about the x-axis of the solid that is bounded by the cone z = and the plane z = 1 if the density at a point P is directly proportional to the distance from the z-axis.

In Problems 59–62, convert the point given in spherical coordinates to (a) rectangular coordinates and (b) cylindrical coordinates.

In Problems 63–66, convert the points given in rectangular coordinates to spherical coordinates.

In Problems 67–70, convert the given equation to spherical coordinates.

  1. x2 + y2 + z2 = 64
  2. x2 + y2 + z2 = 4z
  3. z2 = 3x2 + 3y2
  4. −x2 – y2 + z2 = 1

In Problems 71–74, convert the given equation to rectangular coordinates.

  1. ρ = 10
  2. ϕ = π/3
  3. ρ = 2 sec ϕ
  4. ρ sin2 ϕ = cos ϕ

In Problems 75–82, use triple integrals and spherical coordinates. In Problems 75–78, find the volume of the solid that is bounded by the graphs of the given equations.

  1. z = + y2 + z2 = 9
  2. x2 + y2 + z2 = 4, y = x, y = x, z = 0, first octant
  3. z2 = 3x2 + 3y2, x = 0, y = 0, z = 2, first octant
  4. Inside x2 + y2 + z2 = 1 and outside z2 = x2 + y2
  5. Find the centroid of the homogeneous solid that is bounded by the cone z = and the sphere x2 + y2 + z2 = 2z.
  6. Find the center of mass of the solid that is bounded by the hemisphere z = and the plane z = 0 if the density at a point P is directly proportional to the distance from the xy-plane.
  7. Find the mass of the solid that is bounded above by the hemisphere z = and below by the plane z = 4 if the density at a point P is inversely proportional to the distance from the origin. [Hint: Express the upper ϕ limit of integration as an inverse cosine.]
  8. Find the moment of inertia about the z-axis of the solid that is bounded by the sphere x2 + y2 + z2 = a2 if the density at a point P is directly proportional to the distance from the origin.

 

* If we use θ = tan-1(–1) = –π/4, then we can use r = –2. Notice that the combinations r = 2, θ = –π/4 and r = –2, θ = 3π/4 are inconsistent.

* θ is the same angle as in polar and cylindrical coordinates.

* We must use a different symbol to denote density to avoid confusion with the symbol ρ of spherical coordinates.