examples of surface integral

Surface Integral: Exercises and Complete Solutions

This companion develops computational and physical skill with scalar surface integrals and flux integrals.

The exercises are ordered approximately from introductory to more advanced.

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

Part I: Exercises

Exercise 1: surface area of a plane patch

The surface is the graph

(1)

above the rectangle

(2)

Find the area of the surface patch.

Exercise 2: mass of a curved thin shell

A thin shell occupies the portion of

(3)

above the disk

(4)

Its surface mass density is

(5)

with constant σ₀.

Find the shell mass in terms of σ₀.

Exercise 3: flux through a horizontal rectangle

Let

(6)

Find the flux through

(7)

with upward orientation.

Then repeat for downward orientation.

Exercise 4: flux through a tilted plane

Let

(8)

The surface is

(9)

above the triangle

(10)

Find the upward flux.

Exercise 5: surface area of a sphere

Use a spherical parameterization to derive the surface area of a sphere of radius R.

Do not quote the result directly.

Exercise 6: flux of a radial field through a sphere

Let

(11)

Find the outward flux through a sphere of radius R centered at the origin.

Then verify the result using the divergence theorem.

Exercise 7: inverse-square law and luminosity

An isotropic star has luminosity

(12)

Find the radiative flux at

(13)

using

(14)

Then show directly from the surface integral that doubling the distance reduces the flux by a factor of four while leaving the total luminosity unchanged.

Exercise 8: electric flux of a point charge

A point charge

(15)

is at the center of a sphere.

Use the point-charge field and a surface integral to calculate the electric flux through the sphere.

Use

(16)

Explain why the answer does not depend on the sphere radius.

Exercise 9: closed flux through a cube

Let

(17)

Find the total outward flux through the cube

(18)

in two ways:

1. by summing the flux through all six faces;
2. by using the divergence theorem.

Exercise 10: cylindrical side-surface flux

Let

(19)

Find the outward flux through only the curved side of the cylinder

(20)

Then use the divergence theorem on the complete closed cylinder to explain why the top and bottom surfaces make no contribution.

Exercise 11: orientation and sign

Let

(21)

Consider the disk

(22)

Find the flux for:

1. upward orientation;
2. downward orientation.

Explain physically why the answers differ only by sign.

Exercise 12: Stokes theorem

Let

(23)

Let S be the unit disk in the plane z = 0, oriented upward.

Evaluate

(24)

and independently evaluate

(25)

Verify Stokes’ theorem.

Exercise 13: Poynting flux and radiated power

Far from a source, suppose the time-averaged Poynting vector is radial and has magnitude

(26)

where r is measured in meters.

Find the total radiated Power through any centered sphere.

Exercise 14: inverse-square singularity

Consider

(27)

Show that:

1. ∇⋅ F = 0 for r≠0;
2. the outward flux through any sphere enclosing the origin is 4πC;
3. the two facts do not contradict the divergence theorem.

State the distributional identity that represents the point source.

Part II: Complete Solutions

Solution 1

For

(28)

the graph-surface area element is

(29)

Here

(30)

Therefore

(31)

The projected rectangle has area

(32)

Hence

A	= ∫ 02 ∫ 03 dy dx	(33)
	= 6.	(34)

Thus

(35)

The result is larger than the projected area 6 because the plane is tilted.

Solution 2

The mass is

(36)

For

(37)

(38)

Use polar coordinates:

(39)

Then

M	= σ0 ∫ 02π ∫ 01r dr dϕ	(40)
	= 2πσ0 ∫ 01r dr.	(41)

Let

(42)

Therefore

M	= ∫ 15w1∕2 dw	(43)
	= .	(44)

Hence

(45)

Solution 3

The upward unit Normal is

(46)

Thus

(47)

The area is

(48)

Therefore

(49)

For downward orientation,

(50)

so

(51)

Changing orientation reverses the sign but not the magnitude.

Solution 4

The surface is

(52)

For upward orientation, the vector area element is

(53)

Since

(54)

we have

(55)

Then

F ⋅ dA	= (2, 3, 4) ⋅ (1, 1, 1)dxdy	(56)
	= 9 dxdy.	(57)

The projected triangle has area

(58)

Therefore

(59)

Solution 5

Parameterize the sphere by

(60)

Its area element is

(61)

Thus

A	= ∫ 02π ∫ 0πR2 sin 𝜃 d𝜃 dϕ	(62)
	= R2	(63)
	= R2(2π)(2).	(64)

Therefore

(65)

Solution 6

On the sphere r = R,

(66)

The outward normal is

(67)

Hence

Φ	= ∮ SF ⋅ ndS	(68)
	= kR∮ SdS	(69)
	= kR(4πR2).	(70)

Thus

(71)

Now use the divergence theorem.

Since

(72)

(73)

Therefore

Φ	= V 3k dV	(74)
	= 3k	(75)
	= 4πkR3.	(76)

The two methods agree.

Solution 7

For isotropic radiation,

(77)

Substitute

L	= 3.828 × 1026 W,	(78)
r	= 1.495978707 × 1011 m.	(79)

Then

F	=	(80)
	≈ 1.361 × 103 W m−2.	(81)

The surface-integral statement is

(82)

By spherical symmetry,

(83)

At radius 2r,

(84)

Set this equal to

(85)

(86)

Therefore

(87)

The flux density falls by four, while the integrated luminosity remains L.

Solution 8

The electric field of a point charge is

(88)

On a centered sphere,

(89)

Thus

ΦE	= ∮ SE ⋅ dA	(90)
	= (4πr2)	(91)
	= .	(92)

With

(93)

ΦE	=	(94)
	≈ 2.259 × 102 N m2 C−1.	(95)

The radius cancels because the field decreases as 1∕r² while the sphere area grows as r².

Solution 9

The field is

(96)

Consider the six cube faces.

On x = a,

(97)

The face area is a², so

(98)

On x = 0,

(99)

Thus

(100)

On y = a,

(101)

so

(102)

The y = 0 face contributes zero.

On z = a,

(103)

so

(104)

The z = 0 face contributes zero.

Adding,

(105)

Now use the divergence theorem:

∇⋅ F	= 1 + 2 + 3	(106)
	= 6.	(107)

The cube volume is a³.

Therefore

(108)

Solution 10

On the curved side of the cylinder,

(109)

Also,

(110)

At ρ = R,

(111)

The side area element is

(112)

Thus

Φside	= ∫ 0H ∫ 02πR	(113)
	= R2(2π)H.	(114)

Therefore

(115)

Now

(116)

The closed-cylinder volume is

(117)

So the divergence theorem gives

(118)

Because F has no z component, the top and bottom normals ±e_z satisfy

(119)

Therefore all of the closed flux passes through the curved side.

Solution 11

On the disk,

(120)

so

(121)

For upward orientation,

(122)

and

(123)

The disk area is

(124)

Therefore

(125)

For downward orientation,

(126)

so

(127)

The same physical field crosses the same geometric disk.

Only the bookkeeping convention for positive crossing direction has changed.

Solution 12

The field is

(128)

Its curl is

∇× F	=	(129)
	= (0, 0, 2).	(130)

For the upward disk,

(131)

Thus

S(∇× F) ⋅ ndS	= S2 dS	(132)
	= 2π.	(133)

Now parameterize the boundary circle counterclockwise:

(134)

Then

(135)

On the circle,

(136)

Therefore

F ⋅ dr	= dϕ	(137)
	= dϕ.	(138)

Hence

∮ ∂SF ⋅ dr	= ∫ 02πdϕ	(139)
	= 2π.	(140)

Thus

(141)

Stokes’s theorem is verified.

Solution 13

The time-averaged Poynting vector is radial:

(142)

The total radiated power through a sphere is

P	= ∮ S⟨S⟩⋅ dA	(143)
	= (4πr2)	(144)
	= 4π(2.50 × 104).	(145)

Therefore

(146)

The radius cancels.

That cancellation is exactly what one expects for conserved outward power in an inverse-square field.

Solution 14

Write

(147)

For r≠0, a direct calculation gives

(148)

Now calculate the flux through a sphere of radius R.

On the sphere,

(149)

Therefore

Φ	= ∮ SF ⋅ dA	(150)
	= (4πR2)	(151)
	= 4πC.	(152)

The apparent puzzle is that the divergence is zero everywhere away from the origin, yet the closed flux is nonzero.

The resolution is that the field is singular at

(153)

The ordinary divergence theorem requires the field to be sufficiently smooth throughout the enclosed volume.

A sphere containing the origin violates that hypothesis.

In distribution notation,

(154)

Therefore

(155)

The delta function represents the point source responsible for the nonzero flux.

Part III: Compact Formula Sheet

For a parameterized surface,

(156)

For a graph

(157)

(158)

For an oriented surface,

(159)

For flux,

(160)

For a sphere,

(161)

For the divergence theorem,

(162)

For Stokes’s theorem,

(163)

For isotropic luminosity,

(164)

References

References