examples of volume integral

Volume Integral: Exercises and Complete Solutions

This companion develops practical skill with volume integrals and their physical applications.

The exercises begin with direct triple integration and coordinate changes, then progress through mass density, center of mass, moments of inertia, charge, field energy, quantum probability, conservation laws, and astrophysical luminosity.

All exercises appear first. Complete solutions follow in a separate section.

Part I: Exercises

Exercise 1: a scalar integral over a rectangular volume

Let

(1)

Evaluate

(2)

over the box

(3)

Also calculate the average value

(4)

Exercise 2: volume beneath a paraboloid

Find the volume enclosed between

(5)

and

(6)

Use cylindrical coordinates and derive the integration bounds from the geometry.

Exercise 3: ellipsoid volume from a Jacobian

Consider the ellipsoid

(7)

Use the transformation

(8)

to derive its volume.

Then evaluate the result for

(9)

Exercise 4: mass of a radially graded cylinder

A solid cylinder has radius R, height H, and mass density

(10)

where s is cylindrical distance from the axis.

Find:

1. the total mass;
2. the average mass density.

Exercise 5: mass of a radially graded sphere

A sphere of radius R has density

(11)

Find:

1. the total mass;
2. the average density;
3. the density at the center and at the surface.

Exercise 6: center of mass of a graded block

A rectangular block occupies

(12)

Its density is

(13)

Derive the center of mass.

Then evaluate

(14)

for

(15)

Exercise 7: moment of inertia of a uniform solid cylinder

A uniform solid cylinder has mass M, radius R, and height H.

Use a volume integral to derive its moment of inertia about its symmetry axis.

Do not quote the standard result.

Exercise 8: charge from a spherical volume charge density

A spherical region of radius R contains charge density

(16)

Find the total charge.

Evaluate numerically for

k	= 1.00 × 10−6 C m−4,	(17)
R	= 0.100 m.	(18)

Exercise 9: electromagnetic energy in a capacitor

An ideal parallel-plate capacitor has:

A	= 1.00 × 10−2 m2,	(19)
d	= 1.00 × 10−3 m,	(20)
E	= 2.00 × 105 V m−1.	(21)

Ignore fringing.

Using

(22)

calculate the total electromagnetic energy by integrating over the field-filled volume.

Then independently verify the answer with

(23)

Use

(24)

Exercise 10: normalization of a three-dimensional quantum state

Consider the spherically symmetric wavefunction

(25)

where a > 0.

Normalize the wavefunction over all space:

(26)

Then calculate the probability that the particle lies inside the sphere

(27)

Exercise 11: a vector-valued volume integral and electric dipole moment

A sphere of radius R contains charge density

(28)

The electric dipole moment is

(29)

Find:

1. the total charge;
2. the dipole moment vector.

Use symmetry wherever possible before integrating.

Exercise 12: divergence theorem for a nonlinear radial field

Let

(30)

For a sphere of radius R centered at the origin:

1. calculate the outward flux directly from the surface integral;
2. calculate ∇⋅ F for r > 0;
3. evaluate the corresponding volume integral;
4. verify the divergence theorem.

Exercise 13: steady source and outward flux

A sphere of radius R contains a uniform volumetric production rate

(31)

Assume steady state and spherical symmetry.

The outward flux field inside the sphere is

(32)

Use the integral conservation law on an arbitrary sphere of radius r ≤ R to find J(r).

Then show that the total outward flow through the outer surface equals the total production throughout the volume.

Exercise 14: luminosity from a stellar volume emissivity

A spherical star of radius R has a simplified bolometric volume emissivity

(33)

where 𝜖 has units of power per unit volume.

Assume all generated radiation ultimately escapes.

Find:

1. the total luminosity L;
2. the radiative flux measured at distance d > R, assuming isotropic propagation outside the star.

Explain how this problem connects a volume integral inside the source to a surface flux outside the source.

Part II: Complete Solutions

Solution 1

The volume is

(34)

Now

I

= ∫ 02 ∫ 01 ∫ 03 dz dy dx.

(35)

It is convenient to integrate each term separately.

For the x term,

V xdV	=	(36)
	= 2(1)(3)	(37)
	= 6.	(38)

For the 2y term,

V 2y dV	=	(39)
	= 2(1)(3)	(40)
	= 6.	(41)

For the 3z term,

V 3z dV	=	(42)
	= 2(1)	(43)
	= 27.	(44)

Therefore

(45)

The average value is

(46)

As a check, because f is linear, the average also equals the value at the geometric center:

(47)

Solution 2

The upper surface is

(48)

It meets z = 0 when

(49)

Thus

0	≤ r ≤ 2,	(50)
0	≤ ϕ < 2π,	(51)
0	≤ z ≤ 4 − r2.	(52)

Using

(53)

𝒱	= ∫ 02π ∫ 02 ∫ 04−r2 r dz dr dϕ	(54)
	= 2π ∫ 02r(4 − r2) dr	(55)
	= 2π02	(56)
	= 2π(4).	(57)

Therefore

(58)

Solution 3

Use

(59)

The ellipsoid becomes

(60)

The Jacobian matrix is diagonal:

(61)

Therefore

𝒱ellipsoid	= unit sphereabcdudv dw	(62)
	= abc.	(63)

Hence

(64)

For

(65)

𝒱	= π(24)	(66)
	= 32π	(67)
	≈ 100.531.	(68)

Solution 4

Use cylindrical radius s so that the density symbol ρ_m remains unambiguous.

The volume element is

(69)

The mass is

M	= ∫ 0H ∫ 02π ∫ 0Rρ 0sdsdϕdz	(70)
	= 2πHρ0 ∫ 0Rds	(71)
	= 2πHρ0	(72)
	= πρ0R2H.	(73)

The geometric volume is

(74)

Therefore

(75)

This lies between the axis density ρ₀ and the surface density 2ρ₀, as expected.

Solution 5

The mass is

M	= ∫ 02π ∫ 0π ∫ 0Rρ 0r2 sin 𝜃 dr d𝜃 dϕ	(76)
	= 4πρ0 ∫ 0Rdr	(77)
	= 4πρ0	(78)
	= ρ0R3.	(79)

The sphere volume is

(80)

Hence

⟨ρm⟩	=	(81)
	= ρ0.	(82)

At the center,

(83)

At the surface,

(84)

The average density is much less than the central density because the outer layers occupy most of the sphere’s volume.

Solution 6

By symmetry,

(85)

The total mass is

M	= abρ0 ∫ 0Hdz	(86)
	= abρ0H.	(87)

The first moment about the xy plane is

V zρm dV	= abρ0 ∫ 0Hzdz	(88)
	= abρ0H2.	(89)

Therefore

(90)

For

(91)

	=	(92)
	=	(93)
	=	(94)
	≈ 0.5833.	(95)

The center of mass lies above the geometric midpoint because density increases with height.

Solution 7

Let the uniform mass density be

(96)

About the symmetry axis,

(97)

Thus

Iz	= V s2ρ 0 dV	(98)
	= ρ0 ∫ 0H ∫ 02π ∫ 0Rs2	(99)
	= ρ0H(2π)	(100)
	= ρ0πR4H.	(101)

Since

(102)

we obtain

(103)

Solution 8

The total charge is

Q	= V ρq dV	(104)
	= ∫ 02π ∫ 0π ∫ 0Rkrdr d𝜃 dϕ	(105)
	= 4πk ∫ 0Rr3 dr	(106)
	= 4πk.	(107)

Therefore

(108)

For

k	= 1.00 × 10−6 C m−4,	(109)
R	= 0.100 m,	(110)

Q	= π(1.00 × 10−6)(0.100)4	(111)
	= 3.142 × 10−10 C.	(112)

Solution 9

The electric-energy density is

uE	= 𝜖0E2	(113)
	= (8.8541878128 × 10−12)(2.00 × 105)2	(114)
	≈ 0.177084 J m−3.	(115)

The field-filled volume is

(116)

Therefore

U	= uE dV	(117)
	= uEAd	(118)
	≈ 1.771 × 10−6 J.	(119)

Now verify using circuit quantities.

The capacitance is

C	=	(120)
	= 8.8541878128 × 10−11 F.	(121)

The potential difference is

(122)

Then

CV 2	= (8.8541878128 × 10−11)(200)2	(123)
	= 1.771 × 10−6 J.	(124)

The field-energy volume integral and circuit formula agree.

Solution 10

The normalization condition is

(125)

Because the state is spherically symmetric,

1

= 4πA2 ∫ 0∞r2e−2r∕a dr.

(126)

Use

(127)

With

(128)

(129)

Therefore

1	= 4πA2	(130)
	= πA2a3.	(131)

Hence

(132)

Now calculate the probability inside r ≤ a:

P(r ≤ a)

= 4πA2 ∫ 0ar2e−2r∕a dr.

(133)

Let

(134)

Then

(135)

so

P(r ≤ a)

= 4πA2 ∫ 02x2e−x dx.

(136)

The definite integral is

(137)

Using

(138)

we get

P(r ≤ a)	=	(139)
	= 1 − 5e−2	(140)
	≈ 0.3233.	(141)

Thus about 32.3 percent of the probability lies within one scale length a.

Solution 11

First calculate total charge:

(142)

The spherical region is symmetric under

(143)

while the integrand changes sign.

Therefore

(144)

Now

(145)

The x component contains xz, which is odd under x →−x.

The y component contains yz, which is odd under y →−y.

Thus

(146)

The z component is

(147)

In spherical coordinates,

(148)

Therefore

pz	= k ∫ 02π ∫ 0π ∫ 0Rr2 cos 2𝜃r2 sin 𝜃 dr d𝜃 dϕ	(149)
	= k .	(150)

The angular integral is

(151)

Hence

(152)

Thus

(153)

The distribution has zero net charge but a nonzero dipole moment.

Solution 12

On the sphere r = R,

(154)

Thus

Φ	= ∮ SF ⋅ dA	(155)
	= R2(4πR2)	(156)
	= 4πR4.	(157)

For a radial field

(158)

the divergence for r > 0 is

(159)

Here

(160)

so

∇⋅ F	=	(161)
	= 4r.	(162)

Now integrate through the sphere:

V ∇⋅ FdV	= ∫ 02π ∫ 0π ∫ 0R4rr2 sin 𝜃 dr d𝜃 dϕ	(163)
	= 4(2π)(2)	(164)
	= 4πR4.	(165)

Therefore

(166)

Solution 13

At steady state,

(167)

Apply the integral conservation law to a sphere of radius r:

(168)

By spherical symmetry,

(169)

and J(r) is constant over the spherical surface.

Thus

4πr2J(r)

= s 0.

(170)

Solving,

(171)

At the outer surface,

(172)

The total outward flow is

Qout	= 4πR2J(R)	(173)
	= 4πR2	(174)
	= πR3s 0.	(175)

The total production throughout the sphere is

(176)

Therefore

(177)

Solution 14

The luminosity is the volume integral of power generated per unit volume:

(178)

Using spherical coordinates,

L	= ∫ 02π ∫ 0π ∫ 0R𝜖 0r2 sin 𝜃 dr d𝜃 dϕ	(179)
	= 4π𝜖0 ∫ 0Rdr	(180)
	= 4π𝜖0	(181)
	= 𝜖0R3.	(182)

Outside the source, assume isotropic free propagation.

The same luminosity crosses every centered sphere:

(183)

Thus

F(d)	=	(184)
	=	(185)
	= .	(186)

This problem joins two kinds of integral.

Inside the star,

(187)

adds distributed power production throughout a volume.

Outside the star,

(188)

expresses conservation of that total power as a surface flux.

The volume source therefore produces the luminosity, while the surface integral describes how that luminosity crosses surrounding spheres.

Part III: Compact Formula Sheet

For a scalar field,

(189)

The volume itself is

(190)

In Cartesian coordinates,

(191)

In cylindrical coordinates,

(192)

In spherical coordinates,

(193)

For a general coordinate change,

(194)

Mass is

(195)

Center of mass is

(196)

Moment of inertia is

(197)

Charge is

(198)

Energy from energy density is

(199)

Quantum probability is

(200)

The divergence theorem is

(201)

A local conservation law gives

(202)

References

References