Example: 3.1

Find the area in the positive quadrant bounded by

y =1

4x and y = x3

First find the points of intersection of the two curves: clearly the curves intersect at (0, 0) and at

1

4x = x3 ⇒ x =

1

2, y =

1

8

Select a ∆x strip at an arbitrary location of x. The area of the strip is given as

∆A =

(1

4x− x3

)∆x (this is the upper curve minus the lower curve)

y y

x x

y=x3 x=y

1/3

y=1/4 xx=4y

(1/2, 1/8) (1/2, 1/8)

Now take the limit ∆x→ dx, and add up all the ∆A strips

A =∫ 1/2

x=0

(1

4x− x3

)dx =

(1

4·x2

2−x4

4

)∣∣∣∣∣1/2

x=0

=1

64

For some functions, it may be more convenient to sum up ∆y strips.

Rearrange the equations of the curves in terms of y

y = x3 or x = y1/3

1

and

y =1

4x or x = 4y

Select a ∆y strip at an arbitrary location of y. The area of the strip is given as

∆A =(y1/3 − 4y

)∆y

and the total area is

A =∫ 1/8

y=0

(y1/3 − 4y

)dy =

(3

4y4/3 − 4

y2

2

)∣∣∣∣∣1/8

y=0

=1

64

2

Example: 3.2

Find the volume of a cone with base radius R and height h, rotated about the x axis usingthe disk method.

y y

z

x

x

yR

h

Take a thin solid disc at arbitrary x, with volume

∆V = πy2︸ ︷︷ ︸area

∆x

The volume of the cone is

V =∫ hx=0

πy2dx

=∫ h0π

(R−

R

hx

)2

dx

=∫ h0π

[R2 −

2R2x

h+R2x2

h2

]dx

= π

[R2x−

R2x2

h+

1

3

R2x3

h2

]h0

= π

[R2h−R2h+

1

3R2h

]

=1

3πR2h

3

Example: 3.3

Find the volume of a cone with base radius R and height h, rotated about the x axis usingthe shell method.

y

z

x

y

y

ring

Build up the volume by constructing thin, circular annular shells with radius y and thickness ∆y

The ring area of the shell is given by

ring area = 2πy∆y

and the differential shell volume is

∆V = x(2πy∆y)

The total volume of the cone is given as

V =∫ Ry=0

2πxydy

but y can be expressed in terms of x as

y = R−R

hx ⇒ x = h−

h

Ry

4

Therefore

V = 2π∫ Ry=0

y

(h−

h

Ry

)dy

= 2π

[1

2hy2 −

1

3

h

Ry3

]R0

=1

3πR2h

5

Example: 3.4

Find the surface area of a cone with base radiusR and height h, rotated about the x axis.

y y

y

z

x

x

As

x

s

s

y=R-R xh

Recall that the arc length for y(x) is given as

ds =

√√√√1 +

(dy

dx

)2

dx

To find the cone surface area, at an arbitrary x, take a ∆x slice that has an arc length of ∆s.

The surface area for this slice is

∆AS = 2πy∆s

The total surface area is given by

As =∫ hx=0

2πyds =∫ hx=0

2πy

√√√√1 +

(dy

dx

)2

dx

y = R−R

hx ⇒

dy

dx= −

R

h

6

Therefore

As =∫ h0

2πy

√√√√1 +

(−R

h

)2

dx

=∫ h0

2π

(R−

R

hx

)(1 +

R2

h2

)1/2

dx

= 2π

[h2 +R2

h2

]1/2 ∫ h0

(R−

R

hx

)dx

= 2π

[h2 +R2

h2

]1/2 [Rx−

1

2

R

hx2

]h0

= 2π

[h2 +R2

h2

]1/2 [Rh−

1

2Rh

]︸ ︷︷ ︸

1

2Rh

= πR√R2 + h2

7

Example: 3.5a

Find the value for

I =∫ 2

0x4dx

using Trapezoidal rule and Simpson’s rule

We know the exact answer is

Iexact =x5

5

∣∣∣∣∣2

0

=25

5= 6.40

Choose n = 8 panels of equal width in the range 0→ 2. Therefore h = 1/4

Trapezoidal Rule

h = 1/4 b = 2

f(x) = x4 n = 8 panels

a = 0

I =h

2

[f(a) + f(b) + 2

n−1∑i=1

f(xi)

]

=1/4

2

04 + 24 + 2

(1

4

)4

+

(2

4

)4

+

(3

4

)4

+

(4

4

)4

+

(5

4

)4

+

(6

4

)4

+

(7

4

)4

= 6.566406

The error is ε = Itrap − Iexact = 6.566406− 6.40 = +0.166406

Simpson’s Rule

I =h

3

f(a) + f(b) + 4n−1∑

i=1,3,5,...

f(xi) + 2n−2∑

i=2,4,6,...

f(xi)

8

=1/4

4

04 + 24 + 4

(1

4

)4

+

(3

4

)4

+

(5

4

)4

+

(7

4

)4 + 2

(2

4

)4

+

(4

4

)4

+

(6

4

)4

= 6.401042

The error is ε = Isimp − Iexact = 6.401042− 6.40 = +0.001042

Simpson’s rule is clearly better for the same value of h. This is usually the case.

9

Example: 3.5b

Find the value for

I =∫ 2

0x4dx

using Romberg integration

From the trap rule

h n I

1/2 4 I1=7.06251/4 8 I2=6.56641/8 16 I3=6.4417

h h2 h4 h6

1/2 7.06254(6.5664)− 7.0625

3= 6.4010

1/4 6.566416(6.4001)− 6.4010

15= 6.4000

4(6.4417)− 6.5664

3= 6.4001

1/8 6.4417

where

• 2nd column is the result from the Trapezoidal rule with ε ∝ h2

• 3rd column is the first extrapolation of the elements with ε ∝ h2, this gives an error ofε ∝ h4, since we have eliminated the h2 terms

• the last column is the second extrapolation using the ε ∝ h4 terms to give a result with anderror where ε ∝ h6

This leads to a very accurate answer. Can keep going to the right to give

(h8) →64( )− ( )

63(h10)→

256( )− ( )

255

26 − 1 28 − 1

10

Example: 3.6

Find the area in the +’ve quadrant bounded by 2 circles

1

1-1 0 2

area A to be

found -

x2 + y2 = 1

(x− 1)2 + y2 = 1

Integration in an x− y coordinate plane would be difficult here due to the shape. r − θ is mucheasier.

1st transform the equations of the circles over to Polar coordinates

x2 + y2 = 1 ⇒ r2 (cos2 θ + sin2 θ)︸ ︷︷ ︸=1

= 1

where

x = r cos θ ⇒ x2 = r2 cos2 θ

y = r sin θ ⇒ y2 = r2 sin2 θ

(x− 1)2 + y2 = 1 ⇒ x2 − 2x+ 1 + y2 = 1

r2 cos2 θ︸ ︷︷ ︸x2

− 2r cos θ︸ ︷︷ ︸2x

+1 + r2 sin2 θ︸ ︷︷ ︸y2

= 1

where

x = r cos θ ⇒ x2 = r2 cos2 θ

2x = 2r cos θ

y = r sin θ ⇒ y2 = r2 sin2 θ

11

r2(cos2 θ + sin2 θ)︸ ︷︷ ︸=1

−2r cos θ = 0

r2 = 2r cos θ

r = 2 cos θ ⇐

y

x

r=1

r=1

r=2 cos

1 2

P

3

The intersection point is at r = 1. Therefore at r = 1 find θ. Set r = 2 cos θ. Therefore atpoint P , r = 1 and θ = π/3.

2 cos θ = 1

cos θ =1

2→ θ = 600 =

π

3radians

The bounds of integration are therefore 0 ≤ θ ≤ π/3 and 1 ≤ r ≤ 2 cos θ

Start with

dA = r dr dθ at arbitrary (r, θ)

Build up a “wedge” slice

(∫ 2 cos θ

r=1rdr

)θ fixed

dθ

add up wedges

A =∫ π/3θ=0

(∫ 2 cos θ

r=1rdr

)dθ =

∫ π/30

1

2r2∣∣∣∣∣2 cos θ

1

dθ =1

2

∫ π/30

(4 cos2 θ − 1)dθ

A =1

2

∫ π/30

(4 cos2 θ − 1)dθ

=4

2

[1

2θ +

1

4sin 2θ

]∣∣∣∣∣π/3

0

−1

2θ

∣∣∣∣∣π/3

0

12

=π

3+

1

2sin

2π

3−π

6

=π

6+

1

2sin

(2π

3

)= 0.957

where from the integral tables (Schaums 17.18.9)

∫cos2 θ =

θ

2+

1

4sin 2θ

As a rough check -1

4· πr2 ≈

π

4≈ 0.79 OK

13

Example: 3.7

Find the surface area in the +’ve octant for z = f(x, y) = 4− x− 2y.

x

y

4

4

2

plane

y = 2-x/2

xy

Let

F = z − f(x, y) = z − (4− x− 2y)

where the limits of integration are

x : 0 → 4

y : 0 → 2− x/2

The partial derivatives are

∂f

∂x= −1

∂f

∂y= −2

dS =√

1 + 1 + 4dxdy =√

6dxdy

The surface strip area is given by

√6

(∫ 2−x/2

y=0dy

)dx

14

The surface area is then

S =√

6∫ 4

x=0

∫ 2−x/2

y=0dydx =

√6∫ ∫Rxy

Therefore S = 4√

6 which is approximately equal to 9.8m2.

15

Example: 3.8

Given the sphere, x2 + y2 + z2 = a2, derive the formula for surface area.

Let

F = z − f(x, y) = z − (a2 − x2 − y2)1/2︸ ︷︷ ︸f(x,y)

x

y

z

region xy

x + y + z = a

x + y = a

2

2

2

2

2

2

2

dS

dA

Consider the 1/8 sphere in the +’ve octant.The partial derivatives are

∂f

∂x=

1(−2x)

2√a2 − x2 − y2

∂f

∂y=

1(−2y)

2√a2 − x2 − y2

and the bounds of integration are

y : 0 → (a2 − x2)1/2

x : 0 → a

Let the area in the x− y plane (Rxy) be dA = dxdy and the surface above be dS.

dS =

√√√√1 +x2

a2 − x2 − y2+

y2

a2 − x2 − y2dxdy

=

√√√√a2 − x2 − y2 + x2 + y2

a2 − x2 − y2dxdy

= a

√√√√ 1

a2 − x2 − y2dxdy

While we can sum in either direction, we will sum over y first

The strip surface area is

∫ √a2−x2

y=0a

√√√√ 1

a2 − x2 − y2dy

dx16

with x fixed.

Then we sum up the strips over x

S =∫ ax=0

∫ √a2−x2

y=0a

√√√√ 1

a2 − x2 − y2dy

dx

Solving the inner integral first, where we let c2 = a2 − x2. (Schaum’s 18.22)

S = a∫ cy=0

√√√√ 1

c2 − y2dy = a sin−1

(yc

)∣∣∣∣c0

= a[sin−1(1)− sin−1(0)

]= a

[π2

]

Therefore the surface area is

S =∫ ax=0

(aπ

2

)dx = a

π

2x∣∣∣∣a0

=πa2

2⇐ 1/8 of a sphere

The total sphere area is

S = 8 ·πa2

2= 4πa2

17

Example: 3.9

Find the volume formed in the +’ve octant between the coordinate planes and the surface

z = f(x, y) = 4− x− 2y

x

y

4

4

2

plane

y = 2-x/2

xy

We will determine the volume of the solid, i.e. the solid under the f(x, y) surface. We willidentify the solid region asRxy which is the projection of the surface f(x, y) downward onto thex, y plane.

Start with a column at arbitrary (x, y) inRxy, where the volume is zdA

base area ∆x∆yheight z = f(x, y) = 4− x− 2y

The volume of the column is then

V = (4− x− 2y)︸ ︷︷ ︸height

·∆x ·∆y︸ ︷︷ ︸dA

We then sum over y (with x fixed) to get a slice in y− direction

slice volume =

{∫ 2−x/2

y=0(4− x− 2y)dy

}x−fixed

dx

18

Sum up the y− slices over x

V =∫ 4

x=0

(∫ 2−x/2

y=0(4− x− 2y)dy

)dx

The volume under the surface z = f(x, y) is then

V =∫ ∫Rxy

f(x, y)dydx

The order of integration can be reversed, it depends on which ever is easier.

Evaluation of the volume integral gives

V =∫ 4

x=0

(4y − xy − y2

)∣∣∣∣2−x/20

dx

=∫ 4

0

[4(2−

x

2

)− x

(2−

x

2

)−(2−

x

2

)2]dx

=∫ 4

0

[8− 2x− 2x+

x2

2−(

4− 2x+x2

4

)]dx

=∫ 4

0

(4− 2x+

1

4x2

)dx

= 4x− x2 +x3

12

∣∣∣∣∣4

0

= 16− 16 +64

12=

16

3m3

19

Example: 3.10a

Find the mean value of y = f(x) = sinx in the domain x = 0 to x = π.

We can think of the average value as f , where the area under the curve is f · π.

y

0

f

Therefore,

fπ =∫ π0f(x)dx

which means

f =1

π

∫ π0f(x)dx

or more specifically

f =1

π

∫ π0

sinxdx =− cosπ + cos 0

π= 0.637

20

Example: 3.10b

Find the mean value of temperature for T = f(x, y) = 4− x− 2y.

x

y

4

2

T of plate at (x,y)

= f(x,y)

plate x=0, y=0, y=2-x/2

fARxy =∫ ∫Rxy

f(x, y)dydx

f

A

Therefore

f =1

ARxy

∫ ∫Rxy

f(x, y)dydx

ARxy =1

2· 2 · 4 = 4m2

∫ ∫Rxy

→16

3(from previous example)

f =1

4·

16

3=

4

3

21

Example: 3.10c

Derive the formula for the volume of revolution. for the following sphere:

x2 + y2 + z2 = a2.

x

y

z

x + y = a

x=0

y=0

region

surface z = f(x,y)

a

a

a

2 2 2

xy

Consider the +’ve octant, i.e. the 1/8 ’th of the sphere.

In polar coordinates we have, z = f(r, θ), where

z =√a2 − r2

Substituting back into the Cartesian formulation for volume we get

V =∫ a0

∫ √a2−x2

0

√a2 − x2 − y2dydx

This is very messy to work with and leads us over to polar coordinates.

A column at an arbitrary (r, θ) has a base of r dr dθ and a height of z =√a2 − r2. If we then

integrate over r to build up a wedge slice with θ fixed

slice volume =(∫ ar=0

√a2 − r2rdr

)θ fixed

dθ

Now integrate this over θ to add up the slices

V =∫ π/20

(∫ ar=0

√a2 − r2rdr

)dθ

22

The volume of the sphere is then given by

V =∫ ∫Rf(r, θ)︸ ︷︷ ︸

z

rdrdθ︸ ︷︷ ︸area

From Schaums 17.11.9

V =∫ π/20

(∫ ar=0

√a2 − r2 r dr

)dθ

Evaluation gives (Schaum’s 17.11.9)

V =∫ π/2θ=0−

(a2 − r2)3/2

3

∣∣∣∣∣∣a

r=0

dθ

=∫ π/2θ=0

a3

3︸︷︷︸wedge volume

dθ

=a3

3θ

∣∣∣∣∣π/2

0

=πa3

6

The total volume of the sphere is 8 times this value

Vtotal = 8 ·πa3

6=

4

3πa3

Aside:

let u = a2 − r2 and du = −2rdr. Therefore we can write rdr = −du/2.

The limits of integration are r = 0→ u = a2 and r = a→ u = 0.

The integral then becomes

−∫ 0

a2

u1/2

2du = −

[2

3·

1

3u3/2

]0a2

= −1

3[0− a3] =

1

3a3

23

Example: 3.11

Find the volume of the paraboloid, z = x2 + y2 for 0 ≤ z ≤ 4. Consider only the +’veoctant, i.e. 1/4 of the volume.

x

y

z

x sum

2

2

4z = x + y surface2 2

2 2

projection onto (x,y) plane

x + y = 4

Start with a differential volume

dV = dxdydz

at an arbitrary (x, y, z) in dV .

Build up a column - sum over z

(∫ 4

z=x2+y2dz)dxdy

↪→lower bound is the surface of the paraboloidupper limit is the circle at a radius of 4

Build up a slice, and sum over x with y fixed.

∫ √4−y2

x=0

∫ 4

z=x2+y2dzdx

dySum over y

V =∫ 2

y=0

∫ √4−y2

x=0

∫ 4

z=x2+y2dzdxdy

24

Evaluate

V =∫ 2

0

∫ √4−y2

0(4− x2 − y2)dxdy

=∫ 2

0(4− y2)x

∣∣∣∣√

4−y2

0−x3

3

∣∣∣∣∣√

4−y2

0

dy

=∫ 2

0(4− y2)3/2dy −

1

3

∫ 2

0(4− y2)3/2dy

=2

3

∫ 2

0(4− y2)3/2dy

Using the integral tables or through substitution

V =2

3· 3π = 2π

The total volume is

V = 4(2π) = 8π

We could check this value using a volume ofrevolution

dV = πy2dz = πzdz

and

V =∫ 4

z=0πzdz = π

z2

2

∣∣∣∣∣4

0

= 8π

z

4

y

z=y2

o

Area = y2

dV= y dz = z dz2

25

Example: 3.12

Find the volume bounded by a cylinder,

x2 + y2 = a2

and a paraboloid,

z = x2 + y2

z

y

x

z = x + y2 2

x + y = a2 2 2

aa

Cartesian Coordinates

Consider the positive octant, start with

V = dx dy dz

We will sum over z(x, y fixed) to make a column.

vol =

(∫ x2+y2

z=0dz

)dxdy

Now sum the columns over y (x fixed) to make a slice.

vol =

∫ √a2−x2

y=0

(∫ x2+y2

z=0dz

)dy

dxFinally, sum the slices over x to get the total volume

V =∫ ax=0

∫ √a2−x2

y=0

∫ x2+y2

z=0dz dy dx

=∫ ax=0

∫ √a2−x2

y=0(x2 + y2)dy dx

=∫ ax=0

(x2y +

y3

3

)∣∣∣∣∣√a2−x2

y=0

dx

26

=∫ ax=0

(x2√a2 − x2 +

1

3

(a2 − x2

)3/2)dx

= −x

4

(a2 − x2

)3/2+a2

8

[x√a2 − x2 + a2 sin−1 x

a

]∣∣∣∣∣a

x=0

=πa4

8

The total for all 4 octants is

Total V = 4πa4

8=πa4

2

Cylindrical Coordinates (r, θ, z)

Start with

dV = rdr dθ dz

We will sum over z to make a column with (r, θ) fixed.

vol =

(∫ r2z=0dz

)rdr dθ

Now sum the columns over r with θ fixed to make a wedge slice.

vol =

∫ ar=0

(∫ r2z=0dz

)rdrdθ︸ ︷︷ ︸

planar area

Sum the wedges over θ to get

V =∫ 2π

θ=0

∫ ar=0

∫ r2z=0dzrdr dθ

27

Finally, evaluate from inner to outer

V =∫ 2π

θ=0

∫ ar=0r2rdr dθ

=∫ 2π

θ=0

r4

4

∣∣∣∣∣a

0

dθ

=a4

4

∫ 2π

θ=0dθ

= 2πa4

4=πa4

2

Note how much easier the integrations are when an axi-symmetric problem is solved using cylin-drical coordinates.

28

Example: 3.13

Derive a formula for the volume of a sphere with radius, a

x2 + y2 + z2 = a2

In Cartesian coordinates, we will consider the positive octant where

z =√a2 − x2 − y2

As in the previous example, we will build up from dA = dx dy, then sum over z to obtain acolumn, the sum the columns over y to obtain a slice, and finally sum the slices.

V = 8∫ ax=0

∫ √a2−x2

y=0

∫ √a2−x2−y2

z=0dzdydx =

4

3πa3

Note: although not shown, the integrations are messy.

Using spherical coordinates, (R, θ, φ). The sphere surface is justR = a.

Start with

dV = R2 sinφ dR dθ dφ

Sum over R to make a column, (0 ≤ R ≤ a). Sum the columns over φ to make a wedge(0 ≤ φ ≤ π). The sum the wedges over θ (0 ≤ θ ≤ 2π) to get the total volume

V =∫ 2π

θ=0

∫ πφ=0

∫ aR=0

R2 sinφdR dφ dθ

=a3

3

∫ 2π

θ=0(− cosφ)|πφ=0 dθ

=2a3

3

∫ 2π

θ=0dθ =

4

3πa3

29

Example: 3.14

Find the centroid, center of mass and the 1st moment of mass for a quarter circle of radiusa with an inner circle of radius a/2 made of lead with a density of ρ1 = 11, 000 kg/m3

and an outer circle of radius amade aluminum with a density of ρ2 = 2, 500 kg/m3. Thethickness is uniform throughout at t = 10mm.

x

y

a

a

a/2

a/2

x + y = a2 2 2

1

2 ρ∗1 = 11 g/cm2 = 110 kg/m2

ρ∗2 = 2.5 g/cm2 = 25 kg/m2

Centroid:

The total area is given by

A =∫ ∫Rdxdy =

1

4πa2

The 1st moment of area about the y− axis is

Fy =∫ ∫RxdA

x

y2 2 2y = a - x

dA = dx dy

x

In Cartesian coordinates

Fy =∫ ax=0

∫ √a2−x2

y=0dy

xdx=

∫ ax=0

x√a2 − x2dx

= −(a2 − x2)3/2

3

∣∣∣∣∣a

0

=a3

3

If we use polar coordinates with x = r cos θ and dA = rdrdθ

30

x

y2 2 2y = a - x

dA = r dr d

x

Fy =∫ π/2θ=0

∫ ar=0

(r cos θrdr)dθ

=a3

3

∫ π/2θ=0

cos θdθ

=a3

3sin θ|π/20 =

a3

3

Therefore

x =FyA

=a3/3

(πa2)/4=

4a

3π≈ 0.424a

y =FxA

=

∫ ∫R ydxdy

A=

4a

3π≈ 0.424a (by symmetry)

Center of Mass

If the thickness, t and the density, ρ are constant, then the center of mass is equivalent to thecentroid. But in this example we have more mass concentrated toward the origin, so we can expectthe center of mass, xc, yc < x, y.

The total mass is

M =∫ ∫Rρ(x, y)tdA

= ρ1t∫ ∫R1

dA+ ρ2t∫ ∫R2

dA

= ρ1t

[1

4π(a

2

)2]

+ ρ2t1

4π

[a2 −

(a2

)2]

= ρ1tπa2

16+ ρ2tπ

a2

4− ρ2tπ

a2

16

= πta2

(ρ1 − ρ2

16+ρ2

4

)

=πta2(ρ1 + 3ρ2)

16

31

The 1st moment of mass about the y− axis

∫ ∫Rxdm =

∫ ∫R1

xρtdA+∫ ∫R2

xρtdA

ClearlyR1 andR2 are easiest to define in polar coordinates

ρ1t∫ π/2θ=0

∫ a/2r=0

r cos θrdrdθ + ρ2t∫ π/2θ=0

∫ ar=a/2

r cos θrdrdθ

= ρ1t∫ π/2θ=0

1

3

(a2

)3

cos θdθ + ρ2t∫ π/2θ=0

1

3

[a3 −

(a2

)3]

cos θdθ

=ρ1ta

3

24

∫ π/20

cos θdθ︸ ︷︷ ︸=1

+ρ2ta

3

3

(1−

1

8

) ∫ π/20

cos θdθ︸ ︷︷ ︸=1

=ρ1ta

3

24+

7

24ρ2ta

3 =ta3(ρ1 + 7ρ2)

24⇐ 1st moment of mass

Therefore

xc =

∫ ∫R xdM∫ ∫R dM

=

ta3

24(ρ1 + 7ρ2)

πta2

16(ρ1 + 3ρ2)

=2a

3π

(ρ∗1 + 7ρ∗2ρ∗1 + 3ρ∗2

)=

2a

3π

(110 + 7× 25

110 + 3× 25

)= 0.327a

By symmetry yc = xc

center of mass = 0.327a

center of area = 0.424a

x

y

a

a

a/2

a/2

centroid

center of

mass

32

Example: 3.15

Find the area of the paraboloid z = x2 + y2 below the plane z = 1

The surface area S projects into the interior of the circle

x2 + y2 = 1

in the xy− plane. This region is denoted in the section above asG. Here

z = f(x, y) = x2 + y2

so that

∂f

∂x= 2x,

∂f

∂y= 2y

and

S =∫ ∫

x2+y2≤1

√4x2 + 4y2 + 1dy dx

The double integral above is easier to integrate in polar coordinates, since the combination ofx2 + y2 may be replaced by r2. Taking the element area to be

dA = r dr dθ

in place of dy dx, we obtain

S =∫ ∫

r≤1

√4r2 + 1 r dr dθ

=∫ 2π

0

∫ 1

0

√4r2 + 1 r dr dθ

=π

6(5√

5− 1)

33

Example: 3.16

Find the moment of inertia about the y− axis of the area enclosed by the cardioidr = a(1− cos θ)

It is customary to take the density as unity when working with a geometrical area. Thus we have

Iy =∫ ∫

Ax2dA

with

x = r cos θ, dA = rdrdθ

If we integrate first with respect to r, then for any θ between 0 and 2π, r may vary from 0 toa(1− cos θ)

Iy =∫ 2π

0

∫ a(1−cos θ)

0r3 cos2 θdrdθ

=∫ 2π

0

a4

4cos2 θ(1− cos θ)4dθ

The integral above can be solved using the reduction formula

∫ 2π

0cosn θdθ =

cosn−1 θ sin θ

n

∣∣∣∣∣2π

0

+n− 1

n

∫ 2π

0cosn−2 θdθ

to give

Iy =a4

4

∫ 2π

0(cos2 θ − 4 cos3 θ + 6 cos4 θ − 4 cos5 θ + cos6 θ) dθ

=a4

4

[1 +

18

4+

5

8

]π

=49πa4

32

34

Example: 3.17

Find the center of gravity of a homogeneous solid hemisphere of radius a

We may choose the origin at the center of a sphere and consider the hemisphere that lies above thexy− plane. The equation of the hemispherical surface is

z =√a2 − x2 − y2

or in terms of cylindrical coordinates

z =√a2 − r2

By symmetry we have

x = y = 0

We calculate z:

z =

∫ ∫ ∫zdV∫ ∫ ∫dV

=

∫ 2π

0

∫ a0

∫ √a2−r2

0

2

3πa3

=3a

8

35

Example 4.1

Show for the 3D case f(x, y, z) that curl grad f = 0

∇f = i∂f

∂x+ j

∂f

∂y+ k

∂f

∂z

holds for any function f(x, y, z), for instance

f(x, y, z) = x2 + y2 + y sinx+ z2

∇f = i∂f

∂x+ j

∂f

∂y+ k

∂f

∂z

= i(2x+ y cosx) + j(2y + sinx) + k(2z)

∇× (∇f) =

∣∣∣∣∣∣∣i j k

∂/∂x ∂/∂y ∂/∂z2x+ y cosx 2y + sinx 2z

∣∣∣∣∣∣∣= i(0− 0)− j(0− 0) + k(cosx− cosx)

= 0

1

Example 4.2

Given a 3D temperature field

T = f(x, y, z) = 8x+ 6xy + 30z

find the average temperature, T along a line from (0, 0, 0) to (1, 1, 1).

x

y

z

curve CParametrize so t increases in the direction of travel.

C x = t, y = t, z = t, 0 ≤ t ≤ 1

The length of C is given by

L =∫ 1

t=0

√√√√(dxdt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt =√

3

The line integral is given by

∫CfdS =

∫ 1

t=0(8t+ 6t2 + 30t)︸ ︷︷ ︸

f value on C

√1 + 1 + 1dt︸ ︷︷ ︸dS on C

=√

3∫ 1

0(38t+ 6t2)dt

=√

3

38t2

2+ 6

t3

3

∣∣∣∣∣1

0

= 21√

3

The average temperature is

T =21√

3√

3= 21 ◦C

2

Example: 4.3a

Suppose the temperature near the floor of a room (say at z = 1) is described by

T = f(x, y) = 20−x2 + y2

3

where

−5 ≤ x ≤ 5

−4 ≤ y ≤ 4

What is the average temperature along the straight line path fromA(0, 0) toB(4, 3).

T =

∫Cf(x, y)dS∫CdS

where C is the curve x(t) = t

y(t) =3

4t

for 0 ≤ t ≤ 4, where t increases in the direction of travel.

The numerator is

∫Cf(x, y)dS =

∫ 4

t=0f [x(t), y(t)]

√√√√(dxdt

)2

+

(dy

dt

)2

dt

=∫ 4

t=0

20−

(t2 +

9

16t2)

3

√

1 +9

16dt = 86.1

The denominator is the arc length L

L =∫ 4

t=0

√√√√(dxdt

)2

+

(dy

dt

)2

dt = 5

The mean temperature is

T =86.1

5= 17.2

3

Example: 4.3b

What is the average room temperature along the walls of the room?

T =

∮Cf(x, y)dS∮CdS

where the closed curve C is defined in 4 sections

C1 y = −4 x = t −5 ≤ t ≤ 5

C2 x = 5 y = t −4 ≤ t ≤ 4

C3 y = 4 x = 5− t 0 ≤ t ≤ 10

C4 x = −5 y = 8− t 0 ≤ t ≤ 8

Note: Set up so that t increases in the direction of travel.

∮C1

fdS =∫C1

fdS +∫C2

fdS +∫C3

fdS +∫C4

fdS

∫C1

fdS =∫ 5

t=−5

{20−

t2 + 16

3

}√12 + 02dt = 118.9

Similarly

∫C2

= 82.7∫C3

= 118.9∫C4

= 82.7

The room perimeter,∮dS, in this instance is 36.

The mean temperature is given by

T =118.9 + 82.7 + 118.9 + 82.7

36= 11.2

4

Example: 4.3c

What is the average temperature around a closed circular path→ x2 + y2 = 9?

T =

∮Cf(x, y)dS∮CdS

where C is a closed circular path

x(t) = 3 cos t

y(t) = 3 sin t

for 0 ≤ t ≤ 2π.

Show that this integration gives

T = 20−9

3= 17

as it should for this f(x, y), i.e. curve C is actually the T = 17 contour.

x = r cos θ

y = r sin θ

let θ = t

5

Example 4.4

Given a force field in 3D:

~F = i(3x2 − 6yx) + j(2y + 3xz) + k(1− 4xyz2)

What is the work done by ~F on a particle (i.e. energy added to the particle) if it moves in astraight line from (0, 0, 0) to (1, 1, 1) through the force field.

W =∫C

~F · d~r →∫Pdx+Qdy +Rdz

=∫C

(3x2 − 6yz)dx+ (2y + 3xz)dy + (1− 4xyz2)dz

the parametric form of C leads to:

x = t dx = dt

y = t dy = dt

z = t dz = dt

for 0 ≤ t ≤ 1. Therefore

W =∫ 1

t=0(3t2 − 6t2)dt+ (2t+ 3t2)dt+ (1− 4t4)dt

=∫ 1

t=0(1 + 2t− 4t4)dt

= (t+ t2 −4

5t2)

∣∣∣∣∣1

0

= 1 + 1−4

5=

6

5Joules

6

Example: 4.5a

The gravitational force on a mass,m, due to mass,M , at the origin is

~F = −GMm~r

|~r|3= −K

~r

|~r|3

whereK = GMm.The vector field is given by:

~F (x, y, z) = iP + jQ+ kR

where

P = −Kx

(x2 + y2 + z2)3/2

Q = −Ky

(x2 + y2 + z2)3/2

R = −Kz

(x2 + y2 + z2)3/2

Compute the work, W , if the mass, m moves from A to B along a semi-circular path inthe (y, z) plane:

y2 + z2 = 16y or z =√

16y − y2 and x = 0

FromA(0, 0.1, 1.261) toB(0, 16, 0)

A B

x

y

M

z

m

curve C in the x=0 plane

(xz plane) - circle, radius 8

center (0,8,0)

W =∫C

~F · d~r =∫C

(iP + jQ+ kR) · (idx+ jdy + kdx)

The integrand can either be rewritten using the parametric equations for curve,C, or can be rewrit-ten in terms of the y variable here since we have an explicit equation for C.

7

On curve C

P = −Kx

(x2 + y2 + z2)3/2= 0

Q = −Ky

(x2 + y2 + z2)3/2= −

Ky

(y2 + 16y − y2)3/2= −

K

64√y

R = −Kz

(x2 + y2 + z2)3/2= −

K√

16y − y2

64y3/2

Also

dx = 0

dz =1

2√

16y − y2(16− 2y)dy

Therefore

W =∫CPdx+Qdy +Rdz

=∫ 16

y=0.1−

K

64√ydy −

K√

16y − y2

64y3/2

16− 2y

2√

16y − y2dy

=∫ 16

y=0.1−

K

64√y

(1 +

8− yy

dy

)

=K

4

1√y

∣∣∣∣∣16

0.1

= −0.728K Joules

Energy of 0.728K Joules must be supplied to move mass m along this line from A to B i.e tocounteract gravity.

8

Example: 4.5b

Find the work to move through the same field, but following a straight line path fromA(0, 0.1, 1.261) toB(0, 16, 0).

A

B

x

y

M

zm curve C

ComputeW , expressing the path C in parametric form. The equation of C is:

z = 1.261−1.261

16− 0.1(y − 0.1) and x = 0

or in parametric form

x = 0 y = t z = 1.269− 0.0793t for 0.1 ≤ t ≤ 16

W =∫CPdx+Qdy +Rdz

P = −Kx

(x2 + y2 + z2)3/2= 0 dx = 0

Q = −Ky

(x2 + y2 + z2)3/2= −

Kt

{t2 + (1.269− 0.0793t)2}3/2dy = dt

R = −Kz

(x2 + y2 + z2)3/2= −

K(1.269− 0.0793t)

{t2 + (1.269− 0.0793t)2}3/2dz = −0.0793dt

Substitution and simplification yields

W = K∫ 16

t=0.1

0.101− 0.994t

(1.0063t2 − 0.201t+ 1.61)3/2dt

= −0.728K Joules

This is an example of a conservative force field - workW is the same for all paths betweenA andB.

9

Example: 4.6

Suppose the temperature variation (same for all (x, y)) in the atmosphere near the groundis

T (z) = 40−z2

5

where T is in ◦C and z is inm. Look at a cylindrical building roof as follows:

x

z

30 m long

10 m high

10 m

What is the air temperature in contact with the roof?

The equation of the roof is

y2 + z2 = 100 0 ≤ x ≤ 30

The temperature equation in space is

T = f(x, y, z) = 40−z2

5

and the roof equation for S is

z = g(x, y) =√

100− y2 0 ≤ x ≤ 30

The average temperature over the roof is

T =

∫ ∫Sf(x, y, z)dS

Area of S

10

x

y

z

30

10projection

is rectangular here

z=g(x,y) = 100 - y2

xy

dx dy

Consider half of the roof - i.e. the positive octant

∫ ∫SfdS

where

f = 40−z2

5

= 40−1

5(100− y2)

= 20 +1

5y2

dS =

√√√√1 +

(∂g

∂x

)2

+

(∂g

∂y

)2

dxdy

=

√√√√1 + 0 +y2

100− y2

=

√√√√ 100

100− y2

11

Therefore

∫ ∫SfdS =

∫ 10

y=0

∫ 30

x=0

(20 +

1

5y2

)√√√√ 100

100− y2dxdy

= 30∫ 10

y=0(200 + 2y2)

1√

100− y2dy

We can either go to the tables or use numerical integration. For instance we could use Simpson’srule to obtain

∫ 10

y=0() dy = 465.6

The surface area is

∫ ∫SdS =

1

4cylinder

=1

42πrh =

π

2(10)(30) = 471.2

T =30(465.6)

471.2= 29.6 ◦C

12

Example: 4.7

Given a velocity field in 3D space

~V = i(2x+ z) + j(x2y) + k(xz)

find

a) the flow rateQ (m3/s) across the surface z = 1 for0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 in the +’ve z direction

b) the average velocity across the surface

z

x

y

1

1

1

xy

dSsurface S

Given surface S, with z = g(x, y) = 1 or G = z − 1 = 0, and dS being a surface elementand n a unit normal vector given as

n = ±∇G|∇G|

→ i∂G

∂x+ j

∂G

∂y+ k

∂G

∂z= k(1)

Choose n = k for flow in the +’ve direction.

Q =∫ ∫

S

~V · n dS

where

dS =

√√√√1 +

(∂g

∂x

)2

+

(∂g

∂y

)2

dxdy = dxdy

13

and

~V · n = ~V · k = xz = x

On surface z = 1

~V = i(2x+ 1) + j(x2y) + kx

n

V

The ~V changes in magnitude and direction over S.

Take ~V · n components for each dS and add up to get total flow.

Q =∫ ∫

S

~V · n dS

=∫ ∫

Rxy

xdxdy

=∫ 1

x=0

∫ 1

y=0xdxdy

=x2

2

∣∣∣∣∣1

0

y|10 =1

2m3/s

b) average velocity across S

Vavg =Qacross S

Area of S=

1

2m/s

1m2=

1

2m/s

14

Example: 4.8

Given

~V = i(1 + x) + j(1 + y2) + k(1 + z3)

verify the divergence theorem for a cube, where 0 ≤ x, y, z ≤ 1 i.e. show that∮ ∮S

~V · ndS =∫ ∫ ∫

V(∇ · ~V )dV

where

S = cube surface (closed)V = interior volume of the cube

For a general surface z = g(x, y) we can use a projection of this surface to obtain dS

dS =

√√√√1 +

(∂g

∂x

)2

+

(∂g

∂y

)2

dxdy

Since all surface are just z = const. we get

∂g

∂x→ 0

∂g

∂y→ 0

Which gives that dS = dxdy.

Similarly there is no need for n =∇G|∇G|

contributions.

1. at the y = 0 face:

n(outward) = −j dS = dxdz

~V · n = −(1 + y2) = −1 since y = 0∫ ∫S

~V · ndS =∫ 1

x=0

∫ 1

z=0− 1dzdx = −1

This means that 1m3/s inflow i.e. -’ve direction

15

2. at the y = 1 face:

n(outward) = j dS = dxdz

~V · n = +(1 + y2) = 2 since y = 1∫ ∫S

~V · ndS =∫ 1

x=0

∫ 1

z=02dzdx = 2

This means that 2m3/s outflow i.e. +’ve direction

3. at the z = 0 face:

n(outward) = −k dS = dxdy

~V · n = −(1 + z3) = −1∫ ∫S

~V · ndS =∫ 1

x=0

∫ 1

y=0− 1dydx = −1

4. at the z = 1 face:

n(outward) = k dS = dxdy

~V · n = +(1 + z3) = 2 since y = 1∫ ∫S

~V · ndS =∫ 1

x=0

∫ 1

y=02dydx = 2

5. at the x = 0 face:

n(outward) = −i dS = dydz

~V · n = −(1 + x) = −1 since x = 0∫ ∫S

~V · ndS =∫ 1

y=0

∫ 1

z=0− 1dzdy = −1

This means that 1m3/s inflow i.e. -’ve direction

6. at the x = 1 face:

n(outward) = i dS = dydz

~V · n = +(1 + x) = 2 since x = 1∫ ∫S

~V · ndS =∫ 1

y=0

∫ 1

z=02dzdy = 2

This means that 2m3/s outflow i.e. +’ve direction

16

Summing up these components, we get

∮ ∮S

~V · ~ndS = −1 + 2− 1 + 2− 1 + 2 = 3m3/s

The right hand side is given as

∫ ∫ ∫V

(∇ · ~V )dV

where

∇ · ~V =∂u

∂x+∂v

∂y+∂w

∂z

= 1 + 2y + 3z2

The RHS integral is solved as

∫ 1

x=0

∫ 1

y=0

∫ 1

z=0(1 + 2y + 3z2)dzdydx =

∫ 1

x=0

∫ 1

y=0

(z + 2yz + z3

)∣∣∣10dydx

=∫ 1

x=0

∫ 1

y=0(2 + 2y) dydx

=∫ 1

x=0

(2y + y2

)∣∣∣1y=0

dx

=∫ 1

x=03 dx = 3x|1x=0 = 3m3/s

17

Example: 4.8

Given: ~F = ix+ j2z + ky (a force field in 3D).

The closed path C is given by the intersection of:

x2 + y2 = 4

z = 4− x− y

The object moves once in a CW direction around C starting at (2, 0, 2).

Verify Stoke’ theorem:∮C

~F · d~r =∫ ∫

S(∇× ~F ) · n dS

Aside:

Check the curl of ~F

∇× ~F =

∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂zx 2z y

∣∣∣∣∣∣∣∣∣= i(1− 2) + j(0) + k(0)

= −i

We see that we cannot make the following assumption

∮~F · d~r = 0

It may or may not be dependent of C.

The LHS is given as

∮CPdx+Qdy +Rdz

18

x = (±)2 cos t = +2 cos t dx = −2 sin tdt

y = (±)2 sin t = −2 sin t dy = −2 cos tdt

z = 4− 2 cos t+ 2 sin t dz = 2 sin tdt+ 2 cos tdt

Note the sign convention due to the CW motion.

P = x = 2 cos t on C

Q = 2z = 8− 4 cos t+ 4 sin t on C

R = y = −2 sin t on C

∮=

∫ 2π

t=0(−4 sin t cos t− 16 cos t+ 8 cos2 t− 8 sin t cos t

= −4 sin2 t− 4 sin t cos t)dt

=∫ 2π

t=0(−16 sin t cos t− 16 cos t− 4 sin2 t+ 8 cos2 t)dt

= −16sin2 t

2

∣∣∣∣∣2π

0

− 16 sin t|2π0 −4

(t

2−

sin 2t

4

)∣∣∣∣∣2π

0

+ 8

(t

2+

sin 2t

4

)∣∣∣∣∣2π

0

= −4 ·2π

2+ 8 ·

2π

2= −4π + 8π = 4π

The RHS is

∫ ∫S(∇× ~F ) · ~ndS

We can choose any S with C as the boundary. Here we will select

z = g(x, y) = 4− x− y

G = z − 4 + x+ y = 0

19

∇× ~F =

∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂zx 2z y

∣∣∣∣∣∣∣∣∣ = i(1− 2) + j(0) + k(0) = −i

n = (±)∇G|∇G|

= (±)+i+ j + k√

1 + 1 + 1

= −i√

3−

j√

3−

k√

3

Therefore

(∇× ~F ) · n = +1√

3

dS =

√√√√1 +

(∂g

∂x

)2

+

(∂g

∂y

)2

dxdy =√

3 dxdy

Therefore

∫ ∫S

=∫ ∫Rxy

(+

1√

3

)√3dxdy =

∫ ∫Rxy

dxdy︸ ︷︷ ︸area of Rxy

=∫ ∫Rxy

rdrdθ

=∫ 2

r=0

∫ 2π

θ=0rdrdθ

=r2

2

∣∣∣∣∣2

0

· θ|2π0

=4

2· 2π

= 4π

LHS = RHS, both are positive. Work is done by the ~F field on the object.

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Example: 4.9

Given a 2D force field, ~F (x, y) = i(xy3) + j(x2y) and a path C in the field:

Verify Green’s theorem

∮CPdx+Qdy =

∫ ∫R

(∂Q

∂x−∂P

∂y

)dxdy

with

P = xy3

Q = x2y

LHS: For the general curveC, we need the parametric form to compute∮CPdx+Qdy. However,

because of the shape ofC in this example, a parametric representation of the curve is not required.

1. on C1:

y = 1 dy = 0 P = x Q = x2

W =∫C1

Pdx+Qdy =∫ 0

x=1xdx =

x2

2

∣∣∣∣∣0

1

= −1

2

2. on C2:

x = 0 dx = 0 P = 0 Q = 0

∫C2

Pdx+Qdy = 0

3. on C3:

y = −1 dy = 0 P = −x Q = −x2

∫C3

Pdx+Qdy =∫ 1

x=0− xdx = −

x2

2

∣∣∣∣∣1

0

= −1

2

4. on C4:

x = 1 dx = 0 P = y3 Q = y

∫C4

Pdx+Qdy =∫ 1

y=−1ydy =

y2

2

∣∣∣∣∣1

−1

= 0

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Therefore

W =∮CPdx+Qdy = −

1

2−

1

2= −1 Joules

A negative value ofW implies that work has to be supplied by the object.

RHS:

∫ ∫R

(∂Q

∂x−∂P

∂y

)dxdy =

∫ ∫R

(2xy − 3xy2

)dxdy

In this particular case the calculation of the double integral is made easier becauseR limits are allconstants.

∫ 1

x=0

[∫ 1

y=−1(2xy − 3xy2)dy

]dx =

∫ 1

x=0

[(xy2 − xy3)

∣∣∣1y=−1

]dx

=∫ 1

x=0(−2x)dx = −x2

∣∣∣10

= −1 Joules

Therefore LHS = RHS.

We can use Green’s theorem to change a line integral computation ofW to a∫ ∫R instead.

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