1.5 Double and Iterated Integrals

16 The ABC’s of Calculus

case2: y = −x ⇒ x2 − x2 + x2 = 16 or x2 = 16.

x = ±4 But y = −x ⇒ y = ±4 too!

So we get four critical points:

x =4√3

, y =4√3

x = − 4√3

, y = − 4√3

x = 4 , y = −4x = −4 , y = 4

x2 + y2√

x2 + y2

323

√323

323

√323

32√

3232

√32

check critical points for max/min.

Note that maximum is at (4,−4) and (−4, 4) and the value at this point is√

32

Note that minimum is at (± 4√3,± 4√

3) and the value here is

√323

1.5 Double and Iterated Integrals

Let domain f = R and write z = f(x, y) continuous over a finite region R ofthe x, y plane. We divide R into n subregions area ∆A1, ∆A2, ......,∆An in anyfashion what so ever.ro-

xy-

Figure 4

Next we pick a point (x1, y1), (x2, y2), ......, (xn, yn) in each one of these subre-gions and form the sum:

n∑i=1

f(xi, yi)∆Ai = f(x1, y1)∆A1 + ...... + f(xn, yn)∆An

Think about f(xi, yi)∆Ai : This is the signed volume of a parallelepipeds ofbase area ∆Ai and height f(xi, yi).

Functions and their properties 17

If this sum approaches a limit as n → ∞ and at the same time every subregionshrinks to a point (∆Ai → 0) the limit (if it is unique) is called the doubleintegral of f over R and denoted by:

∫ ∫R

fdA or∫R

fdA or∫ ∫

Rf(x, y)dA

by definition:

= limn→+∞; ∆Ai→0

n∑i=1

f(xi, yi)∆Ai

Remark:

1) Let f(x, y) ≥ 0 all (x, y) in R⇒ ∫ ∫

RfdA = volume under that point of s whose projection is R2) If f(x, y) = 1 for (x, y) in R then:∫ ∫

RfdA =∫ ∫

R dA = volume under the part of the plane z = 1 whose projection is in R.

So we claim that∫ ∫

RdA = area of R (numerically)

The double integral (2D) has same properties as (1D) integral, namely:

1) ∫ ∫R

(c · f)dA = c

∫ ∫R

fdA c is constant

2) ∫ ∫R

(f ± g)dA = (∫ ∫

RfdA) ± (

∫ ∫R

gdA)

3) IfR = R1 ∪R2 ∪ ....... ∪Rn where Ri ∩ Rj = 0

then ∫ ∫R

fdA =∫ ∫

R1

fdA +∫ ∫

R2

fdA + .......... +∫ ∫

Rn

fdA

In order to evaluate a double integral we need to set up an ”iterated integral”in cartesian (rectangular) coordinates, such an integral takes the form:

∫ ∫R1

f(x, y)dxdy or∫ ∫

R2

f(x, y)dydx

whereR1 = {(x, y) : f1(y) ≤ x ≤ f2(y), a ≤ y ≤ b}


and

R2 = {(x, y) : g1(x) ≤ y ≤ g2(x), c ≤ x ≤ d}

R1&R2 are simply descriptions of the given region R in rectangular coordinates.

Figure 5

Describing Regions in R2

Example 15 Describe the circle of radius 4 centered at (0, 0) in two differentways = {x2 + y2 = 4}

1) Vertical slices:

Figure 6

= {(x, y) : −2 ≤ x ≤ 2, −√

4 − x2 ≤ y ≤ +√

4 − x2

2) Horizontal slices:

= {(x, y) : −√

4 − y2 ≤ x ≤ +√

4 − y2, −2 ≤ y ≤ 2}

Example 16 Figure 7

R = {(x, y) : 0 ≤ y ≤ (1 − x), 0 ≤ x ≤ 1}

Example 17 Figure 8

R = {(x, y) : x2 ≤ y ≤ x, 0 ≤ x ≤ 1}

NOTE: R1 and R2 are really the same region R described in two differentways.

The iterated integral takes the form:

∫ ∫R1

f(x, y)dxdy =∫ ∫

RfdA (theorem......)

∫ ∫f(x, y)dxdy

means=∫ b

a

∫ f2(y)

f1(y)

f(x, y)dxdy


a, b, f1(y), f2(y):come from the description of R1 by horizontal slices.

∫ ∫R

fdA =∫ b

a

∫ f2(y)

f1(y)

f(x, y)dx

︸︷︷︸integrate this first,keep ”y” fixed.

dy

Then integrate the resulting function of y given inside the box with respect to”y”between a&b.

The other technique for evaluation involves taking vertical slices.

∫ ∫R

fdA =∫ d

c

∫ g2(x)

g1(x)

f(x, y)dy

︸︷︷︸integrate this first, keep ”x” fixed.

dx

The double integral of (piece wise) continuous function f defined on R ⊂ Rn

exists.

We say f is integrable over R if∫ ∫

R | f | dA is finite. If f is integrable overR ⇒ Fubini’s theorem says that:

∫ ∫R1


R2

f(x, y)dydx =∫ ∫

RfdA

Where R1,R2 are the two (different) representations of region R. We canreverse/interchange the order of integration.

Example 18 ∫ 1

0

∫ 2

1

dxdy =∫ ∫

RfdA.

∫ 2

1dx = (2 − 1) = 1 next

∫ 1

0(∫ 2

1dx)dy =

∫ 1

01 · dy = y |10= 1

But this iterated integral by definition:

∫ 1

0

∫ 2

1

dxdy =∫ ∫

RdA

= area of the region R= area of square

Example 19

I =∫ 1

0

∫ x

x2xy2dydx


Figure 9

f(x, y) = xy2

R = {(x, y) : x2 ≤ y ≤ x, 0 ≤ x ≤ 1}

I =∫ 1

0

(x∫ x

x2y2dy)dx

=∫ 1

0

(x[y3

3|y=xy=x2 ])dx

=∫ 1

0

(x[x3

3− x6

3])dx

=∫ 1

0

(x4

3− x7

3)dx

= [x5

15− x8

24]10

=115

− 124

=13· (1

5− 1

8)

= 140

Example 20 HORIZONTAL SLICES:

Redescribe:R = {(x, y) : 0 ≤ y ≤ 1, y ≤ x ≤ √

y}

Figure 10

I =∫ 1

0

∫ √y

y

(xy2)dxdy

=∫ 1

0

(y2

∫ √y

y

xdx)dy

=∫ 1

0

(y2[x2

2]x=

√y

x=y )dy

=∫ 1

0

y2(y

2− y2

2)dy

=12

∫ 1

0

(y3 − y4)dy

=12[y4

4− y5

5]10


=12[14− 1

5]

=12(

120

)

= 140

Example 21 Find the volume cut from 9x2 + 4y2 + 36z = 36 by plane z = 0Figure 11

1) Ensure surface is above xy-plane.

2) Give an idea of shape of region.

3) Set up an integral for volume.

4) Describe the ”projection” involved.

5) Use an appropriate coordinate system.

6) Use Fubini??

7) Write down iterated integrals

8) Evaluate

SOLUTION

1) Yes the surface lies above the xy-plane.

2) Done

3)∫ ∫

R fdA = volume under s.

solve for z then 36z = 36 − 9x2 − 4y2 ⇒ z = 1 − x2

4 − y2

9

The projection of s on the xy-plane (ie z = 0) is given by the use ofgeometry z = 0 1 ≥ x2

4 + y2

9

4) volume =∫ ∫

R(1 − x2

4− y2

9︸︷︷︸=z

)dA

5) R is an ellipse,use elliptical coordinates.

x = ar cosΘy = br sinΘ

(Jacobian :abr)

How to choose a,b? x2

a2 + y2

b2 = r2

x2

4 + y2

9 = 1 ⇒ a = 2 , b = 3


6) R = {(x, y) : x2

4 + y2

9 ≤ 1} in rectangular coordinates.R = {(r, Θ); 0 ≤ Θ < 2π, 0 ≤ r ≤ 1} in elliptical coordinates.

Volume =∫ ∫

R(1 − x2

4− y2

9︸︷︷︸=r2

)dA

=∫ 2π

0

∫ 1

0

(1 − r2)

a·b·r︷︸︸︷6r︸︷︷︸

2·3·rdrdΘ

= 6 ·∫ 2π

0

(∫ 1

0

(r − r3)dr)dΘ

= 12π ·∫ 1

0

(r − r3)dr

= 12π[r2

2− r4

4]10

= 12π[12− 1

4]

= 3π

Example 22 Find the volume common to the cylinders x2 + y2 = 16 and x2 +z2 = 16.

Idea: look at cylinder x2 + y2 = 16 and its pojection onto the xy-plane namely:

D = {(x, y) : 0 < x2 + y2 ≤ 16} =

{(x, y) : −√

16 − x2 ≤ y ≤√

16 − x2, −4 ≤ x ≤ 4}cartesian coordinates

The surface above D is given by x2 +z2 = 16 or z2 = 16−x2 ⇒ z = ±√16 − x2

but z ≥ 0 ⇒ z =√

16 − x2

1.6 Double and Iterated Integrals

Let domain f = R where f : R2 → R and let z = f(x, y) denote that part ofthe surface S which lies above the dom f = R.

Figure 13

We divide R into m subregions of area ∆A1, ∆A2, . . . , ∆Am in any fashion.

Figure 14


Next we pick a point (x1, y1), (x2, y2), . . . , (xm, ym) in each one of these subre-gions and form the sum:

m∑i=1

f(xi, yi)∆Ai = f(x1, y1)∆A1 + f(x2, y2)δA2 + . . . + f(xm, ym)∆Am

If this sum approaches a limit as the number m → ∞ and at the same timeevery subregion shinks to a point (ie ∆Ai → 0), the limit (if it is unique) iscalled the double integral of f over R and is writeen as:

∫ ∫R

fdA =︸︷︷︸or

∫R

fdA =︸︷︷︸or

∫ ∫R

f(x, y)dA

= limm→∞,∆Ai→0

m∑i=1

f(xi, yi)∆Ai

If f(x, y) ≥ 0 then:∫ ∫

RfdA = Volume under surface S lying directly above R

If f = 1, then∫ ∫

R dA = area of R

Figure 15

The double integral(2 variables) has the following properties:

1) ∫ ∫R

(c · f)dA = c

∫ ∫R

fdA c is any constant

2) ∫ ∫R

(f ± g)dA = (∫ ∫

RfdA) ± (

∫ ∫R

gdA)

3) If

R = R1∪R2∪. . .∪Rnwhere Ri ∩ Rj = φ (ie. these are pairwise disjoint)

then ∫ ∫R

fdA =∫ ∫

R1

fdA +∫ ∫

R2

fdA + . . . +∫ ∫

Rn

fdA

=m∑

i=1

∫Ri

fdA


In order to evaluate a double integral we set up a so-called iterated integral .

In cartesian coordinates, such an integral is of the form:∫ ∫R1

f(x, y)dxdy or∫ ∫

R2

f(x, y)dydx

where

R1 = {(x, y) : f1(y) ≤ x ≤ f2(y), a ≤ y ≤ b}andR2 = {(x, y) : c ≤ x ≤ d, g1(x) ≤ y ≤ g2(x)}

Now each one of R1,R2 are a description of R in cartesian coordinates:

Figure 16

So, R1 and R2 are really the same region described in two different ways.

The iterated integral then becomes:

∫ ∫R1

f(x, y)dxdy =∫ b

a

∫ f2(y)

f1(y)︸︷︷︸Limits come from the description of R above

f(x, y)dxdy

=∫ b

a

{∫ f2(y)

f1(y)

f(x, y)dx

︸︷︷︸integrate w.r.t. x and hold y constant

}dy

︸︷︷︸Then integrate w.r.t. y to get a number

So this id the inverse proceedure to partial differentiation or we integrate withrespect to x holding y constant in the inner-most integral, and then integratewith respect to y.

The double integral of a continuous function f defined on R ⊂ R2 exists. Wesay f is integrable over R.

Indeed if |f(x, y)| is integrable over R, the iterated integrals of f over R existand are equal (Fubini), ie∫ ∫

R1


R2

f(x, y)dydx

or, we can reverse the order of integration.


Example 23 ∫ 1

0

∫ 2

1

dxdy = 1

Figure 17

Here R = {(x, y) : 1 ≤ x ≤ 2, 0 ≤ y ≤ 1}, so R is a square. The iteratedintegral represents the area of this square, R.

Now ∫ 1

0

∫ 2

1

dxdy =∫ 1

0

(∫ 2

1

dx)dy =∫ 1

0

(2 − 1)dy =∫ 1

0

dy = 1

If we reverse the order (we need to change the limits!) of integration:∫ 1

0

∫ 2

1

dxdy =∫ 2

1

∫ 1

0

dydx =∫ 2

1

(∫ 1

0

dy)dx =∫ 2

1

(1 − 0)dx = 1

Example 24

I =∫ 1

0

∫ x

x2xy2dydx =

140

I =∫ 1

0

x[x3

3− x6

3]dx

=∫ 1

0

x[y3

3]y=xy=x2dx

=∫ 1

0

x[y3

3− x6

3]dx

=13

∫ 1

0

(x4 − x7)dx

=13(15− 1

8)

=140

Here R = {(x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x} is shown below:

Figure 18

To reverse the order of integration we need to describe R in the ”other” order,ie. by taking a horizontal fibre.

For a given y we need to find the coordinates of the ”ends” of the fibre.

This gives: ( y︸︷︷︸y=x

, y) and (√

y︸︷︷︸y=x2⇒x=

√y

, y)


Therefore:R = {(x, y) : y ≤ x leq

√y, 0 ≤ y ≤ 1}

So

I =∫ 1

0

∫ √y

y

xy2dxdy

=∫ 1

0

y2{∫ √

y

y

xdx}dy

=∫ 1

0

y2(y

2− y2

2)dy

=12

∫ 1

0

(y3 − y4)dy

=12(14− 1

5)

=140

Example 25 ∫ 2π

0

∫ 1−cosΘ

0

ρ3 cos2 ΘdρdΘ =49π

32Here:

R = {(ρ, Θ) : 0 ≤ ρ ≤ 1 − cosΘ, 0 ≤ Θ < 2π}R can also be described as:

R = {(ρ, Θ) : 0 ≤ Θ ≤ cos−1(1 − ρ), 0 ≤ ρ ≤ 2}

Figure 19

So: ∫ 2

0

∫ cos−1(1−ρ)

0

ρ3 cos2 ΘdΘdρ =∫ 2π

0

∫ 1−cosΘ

0

ρ3 cos2 ΘdρdΘ

=∫ 2π

0

cos2 ΘdΘ · [∫ 1−cosΘ

0

ρ3dρ]

=∫ 2π

0

(1 − cosΘ)4

4cos2 ΘdΘ

=14

∫ 2π

0

(1 − cosΘ)4 cos2 ΘdΘ

Example 26 Evaluate using iterated integrals (and interchange) f(x, y) = xover the region R bounded by x2 and x3.

Note: x3 < x2 if 0 < x < 1 Describe R in two different ways:

Figure 20


A horizontal fibre has the end- coordinates (√

y, y), ( 3√

y, y), therefore:

R = {(x, y) :√

y ≤ x ≤ 3√

y, 0 ≤ y ≤ 1}Therefore:

∫ ∫R

xdA =∫ 1

0

∫ 3√y

√y

xdxdy

=∫ 1

0

12(y

23 − y)dy

=12[35y

53 − y2

2]y=1y=0

= 120

Furthermore, a vertical fibre has end coordinates (x, x3) and (x, x2), So:

R = {(x, y) : 0 ≤ x ≤ 1, x3 ≤ y ≤ x2}Therefore∫

RfdA =

∫ ∫R

xdA

=∫ 1

0

∫ x2

x3xdydx

=∫ 1

0

x(∫ x2

x3dy)dx

=∫ 1

0

x(x2 − x3)dx

=∫ 1

0

(x3 − x4)dx

=14− 1

5

=120

=∫ 1

0

∫ 3√y

√y

xdxdy as required by Fubini’s theorem

Example 27 Evaluate∫ ∫

R 1dA where R is the first quadrant region boundedby 2y = x2, y = 3x and x + y = 4

1) Sketch the region Find all points of intersection:

y = 3x meets y = 4 − x at:x = 1 (y = 3)

y =x2

2meets y = 3x at

x2

2= 3x ⇒ x = 0, x = 6


y =x2

2meets y = 4 − x when:

x = −4 and x = 2

2) Describe R by vertical or horizontal fibres:

a) vertical Then R = R1 ∪R2 where:

R1 = {(x, y) : 1 ≤ x ≤ 2, 4 − x ≤ y ≤ 3x}

R2 = {(x, y) : 2 ≤ x ≤ 6,x2

2≤ y ≤ 3x}

Figure 21

So: ∫R

1 · dA =∫ ∫

R1∪R2

1 · dA

=∫ ∫

R1

dydx +∫ ∫

R2

dydx

=∫ 2

1

∫ 3x

4−x

dydx +∫ 6

2

∫ 3x

x22

dydx

=∫ 2

1

(3x − (4 − x))dx +∫ 6

2

(3x − x2

2)dx

= 4 · (x − 1)2

2|x=2x=1 + (3

x2

2− x3

6)|x=6

x=2

= 2 + [3 · 18 − 36 − 6 +86]

Area of R =463

b) Horizontal: Here R = R3 ∪R4 where:

R4 = {(x, y) : 3 ≤ y ≤ 18,y

3≤ x ≤

√2y}

andR3 = {(x, y) : 2 ≤ y ≤ 3, 4 − y ≤ x ≤

√2y}

Figure 22

So area of R :∫R3∪R4

dA = (∫R3

+∫R4

)dA

=∫ 3

2

∫ √2y

4−y

dxdy +∫ 18

3

∫ √2y

y3

dxdy

Area of R =∫ 3

2

(√

2y − (4 − y))dy +∫ 18

3

(√

2y − y

3dy

= (√

2 · 23y

32 − 4y +

y2

2)|y=3

y=2 + (√

2 · 23y

32 − y2

6


= [2√

23

· 3 32 − 12 +

92− 2

√2

3(√

2)3 + 8 − 2]

+[2√

23

1832 − 182

6− 2

√2

36

32 + 6]

=463

Applications:

1) Finding the area of planar regions R, (as we have).

2) Evaluating complicated integrals by interchanging order of integration.

3) Centroids of planar regions:

If R has area A =∫ ∫

R dA then the relations:

x =My

A=

∫ ∫R xdA

A, y =

Mx

A=

∫ ∫R ydA

A

define the centroid (x, y) of R4) finding the mass of planar regions having mass density, per unit area given

by δ(x, y). then:

mass R =∫ ∫

Rδ(x, y)dA

Example 28 POLAR:

I =∫ ∞

0

∫ ∞

0

e−x2−y2dxdy =

∫ ∞

0

(e−y2∫ ∞

0

e−x2dx)dy

Figure 23

{R = (x, y) : 0 ≤ x ≤ ∞, 0 ≤ y ≤ ∞}{R = (x, y) : 0 ≤ y ≤ ∞, 0 ≤ y ≤ ∞}

when x2 + y2 appearing in integrand use ”polar”.

Why? Because:

x2 + y2 = (r cosΘ)2 + (r sin Θ)2

= r2(cos2 Θ + sin2 Θ)= r2

Then R can be described easily in polar coordinates:


R = {(r, Θ) : 0 < r < ∞, 0 ≤ Θ ≤ π2 }

But dA polar = rdrdΘ (= dxdy)

So:

∫ ∫R

fdA =∫ ∫

R textpolar

f(x

r cosΘ,y

r sinΘ) dApolar︸︷︷︸rdrdΘ

=∫ π

2

0

∫ ∞

0

e−r2rdrdΘ

=∫ π

2

0

(∫ ∞

0

re−r2dr︸︷︷︸

not dependant on Θ ! like a constant

)dΘ

= (∫ ∞

0

re−r2dr) · (

∫ π2

0

1 · dΘ︸︷︷︸= π

2

)

=π

2·∫ ∞

0

re−r2dr −r2 = u, 2rdr = du

=π

2·∫

eu · (−du

2)

=π

2· (−1

2

∫eudu)

=π

2· [−1

2e−r2

]r=∞r=0

=π

2· [0 − (

−12

)]

= π4

Example 29 Evaluate:

∫ ∞

−∞

∫ ∞

−∞e−(x2+y2)dxdy

By changing to polar coordinates.

Here R = {(x, y) : −∞ < x < ∞, −∞ < y < +∞}. In polar coordinates thisbecomes:

R′ = {(r, φ) : 0 ≤ r < ∞, 0 ≤ φ < 2π}

Furthermore, the elements of area:

dxdy = rdrdφ earlier

Also x2 + y2 = r2, thus:


∫ ∞

−∞

∫ infty

−∞e−(x2+y2)dxdy =

∫0

∫ ∞

0︸︷︷︸R in polar

e−r2rdrdφ

=∫ 2π

0

(∫ ∞

0

re−r2dr)dφ

= 2π

∫ ∞

0

re−r2dr

= 2π limR→∞

∫ R

0

re−r2dr

= 2π limR→∞

(−12

∫ R

0

(−2r)e−r2dr)

= 2π · (12

limR→∞

(e−r2 |r=Rr=0 )

= −π · limR→∞

[e−R2 − 1]

= π

REMARK Now∫ ∞

−∞

∫ ∞

−∞e−(x2+y2)dxdy =

∫ ∞

−∞e−x2

dx ·∫ ∞

−∞e−y2

dy

= (∫ ∞

−∞e−x2

dx)2

= π above

therefore∫ ∞

−∞e−x2

dx =√

π

Plane areas by double integration

Recall the area of a planar region R is given by:∫ ∫R

dA = area of R

where dA is the element of area in the particular coordinate system used.

Example 30 Find the area of the region R bounded by 3x+4y = 24, x = 0, y =0 (by double integration).

1) sketch the region: Find all points of intersection of the curves making upits boundary.

Figure 24

Here the points are: (0, 0), (0, 6) and (8, 0).


2) Choose a description of R -Either by horizontal or vertical fibres, say,horizontal.

Figure 25

3) Find the coordinates of the end-points of the fibre. -Do this by finding ”in-verse functions”

Figure 26

For fixed y, the set of points on a typical fibre is given by:

{(x, y) : 0 ≤ x ≤ 8 − 43y}

4) give a description of R. Thus:

R = {(x, y) : 0 ≤ x ≤ 8 − 43y, 0 ≤ y ≤ 6}

5) Set up an iterated integral for evaluating the double integral. Here:

∫ ∫R

dA =∫ 6

0

∫ 8− 43y

0

dxdy (in rectangular coordinates)

6) Evaluate the iterated integral to find the area

Area of R =∫ 6

0

(8 − 43y)dy

= [(8y − 23y2)]y=6

y=0

= 48 − 23· 36 = 48 − 24

= 24

(As a check: Area of ∆ = 12 (base) × (height) = 1

2 (8) × (6) = 24)

Remark Use of vertical fibres in step(3) gives the iterated integral:

Area of R =∫ 8

0

∫ 6− 34x

0

dydx (= 24)

Example 31 Area within ρ = 2(1 − cosΘ)

A typical ”fibre” here is Θ = constant which gives a ray from O to the boundaryρ = 2(1 − cos θ)

Figure 27


So:R = {(ρ, Θ) : 0 ≤ ρ ≤ 2(1 − cosΘ 0 ≤ Θ < 2π}

So the area:∫ ∫R

dA =∫ 2π

0

∫ 2−2 cosΘ

0

ρdρdΘ︸︷︷︸element of area in polar coordinates

= 2∫ 2π

0

(1 − cosΘ)2dθ

= 2∫ 2π

0

(1 − 2 cosΘ + cos2 Θ)dΘ

= 2 · 2π + 2∫ 2π

0

cos2 ΘdΘ

= 4π + 2(12· 2π)

= 6π

Example 32 f(x, y) = x over R bounded by y = x2 and y = x3. Graph theregion (if possible).

NOTE: x3 < x2 if x < 1.Figure 28

setting x3 = x2 we get either x = 0 or x = 1 so points of intersection ofthese curves are (0, 0), (1, 1). Describe R(in some coordinate system), stick withcartesian coordinates, (x, y)...so:

R = {(x, y) :√

y ≤ x ≤ 3√

y, 0 ≤ y ≤ 1}Figure 29

Therefore using a horizontal slice:∫ ∫R

fdA =∫ ∫

Rx dxdy → notice the order

=∫ 1

0

∫ 3√y

√y

xdxdy

=∫ 1

0

[x2

2|x= 3√y

x=√

y ]dy

=∫ 1

0

(y

23

2− y

2)dy

=12· [y

53

53

− y2

2]10

=12· (3

5− 1

2)

=12· (6 − 5

10)

= 120


Vertical Slice:

R = {(x, y) : 0 ≤ x ≤ 1, x3 ≤ y ≤ x2}

∫ ∫R

rdA =∫ 1

0

∫ x2

x3x dydx → notice the order

=∫ 1

0

x(∫ x2

x3dy)dx

=∫ 1

0

x(x2 − x3)dx

=∫ 1

0

(x3 − x4)dx

= (x4

4− x5

5) |n=1

n=o

=14− 1

5

=5 − 420

=120

Example 33 ∫ ∫R

1︸︷︷︸this is the area of R!

dA

R is first quadrant region bounded by 2y = x2, y = 3x and x + y = 4

1) sketch & find all points of intersection of these curves:

y = 3x meets y = 4 − x at x = 1 or y = 3y = x2

2 meets y = 3x at x = o and x = 6 or y = 0 and y = 18y = x2

2 meets y = 4− x at x = −4︸︷︷︸not in 1st quadrant, so ignore it

and x = 2 or y = 8

and y = 2.

2) Describe REither way we need to divide R into two pieces.

a) vertical slice.

R = R1 ∪R2

Where:

R1 = {(x, y) : 1 ≤ x ≤ 2, 4 − x ≤ y ≤ 3x}R2 = {(x, y) : 2 ≤ x ≤ 6, x2

2 ≤ y ≤ 3x}


∫ ∫R

1 · dA =∫ ∫

R1

1 · dA +∫ ∫

R2

1 · dA

=∫ 2

1

∫ 3x

4−x

dydx +∫ 6

2

∫ 3x

x22

dydx

=∫ 2

1

(4x − 4)dx +∫ 6

2

(3x − x2

2)dx

= 4 ·∫ 2

1

(x − 1)dx + (3x2

2− x3

6) |x=6

x=2

= 4 · (x − 1)2

2|x=2x=1 +[(54 − 46) − (6 − 4

3)]

= 2 + 18 − 6 +43

=463

Similarly we can show:

R = R3 ∪R4

where:

R3 = {(x, y) : (4 − y) ≤ x ≤ √2y, 2 ≤ y ≤ 3}

R4 = {(x, y) : (y3 ) ≤ x ≤ √

2y, 3 ≤ y ≤ 18}

Example 34 Evaluate the integral∫ 1

0

∫ 1√x

√1 + y3dydx by interchaging the or-

der of integration.Figure 32

∫ ∫R

fdA Here f(x, y) =√

1 + y3

R = {(x, y) :√

x ≤ y ≤ 1, 0 ≤ x ≤ 1}Figure 33

R = {(x, y) : 0 ≤ x ≤ y2, 0 ≤ y ≤ 1}

So by Fubini, using horizontal slices:Figure 34

1 + y3 = u3y2dy = duy2dy = du

3y = 0 ⇒ u = 1y = 1 ⇒ u = 2

∫ 1

0

∫ 1

√x

√1 + y3dydx =

∫ 1

0

∫ y2

0

√1 + y3dxdy


=∫ 1

0

√1 + y3(

∫ y2

0

dx)dy

=∫ 1

0

y2√

1 + y3dy

=∫ 2

1

du

3· √u

=13

∫ 2

1

u12 du

=13· u

32

32

|21

=29· u 3

2 |21

=29· (2

√2 − 1)

Example 35

x2 + y2 + 2x − s√

x2 + y2 = 0 equation in polar?

x = r cosΘy = r sin Θ

} ⇒ x2 + y2 = r2

r2 + 2 · r cosΘ − 2r = 0

Divided by r, (r �= 0)

r + 2 cosΘ − 2 = 0 r = 2 · (1 − cosΘ)

Area of R =∫ ∫

R dA in polar =∫ ∫

R rdrdΘ

Describe R in polar:

R = {(r, Θ) : 0 ≤ r ≤ 2 · (1 − cosΘ), 0 ≤ Θ < 2π}

Area =∫ 2π

0

∫ 2·(1−cosΘ)

0

rdrdΘ

=∫ 2π

0

[r2

2]2·(1−cosΘ)0 dΘ

=∫ 2π

0

(4 · (1 − cosΘ)2

2− 0)dΘ

= 2 ·∫ 2π

0

(1 − cosΘ)2dΘ


=∫ 2π

0

(1 − 2 cosΘ + cos2 Θ)dΘ

Example 36 Find the centroid of:

3x + 4y = 24, x = 0, y = 0

Centroid coordinates are (x, y) where (x = My

A ,, and (y = Mx

A . A is area andMy =

∫ ∫R xdA, Mx =

∫ ∫R ydA

Here A = 24 (found earlier) whereas:

My =∫ ∫

RxdA =

∫ 8

0

∫ 6− 34x

0

xdydx =∫ 8

0

x(6 − 34x)dx = 64

and

Mx =∫ ∫

RydA =

∫ 6

0

∫ 8− 43y

0

ydxdy =∫

+06y(8 − 43y)dy = 48

ie(x, y) = (

My

A,Mx

A) = (

6424

,4824

) = (83, 2)

Note Either description of R in the expression for the iterated integral could beused.

Example 37 Find the centroid of the first quadrant area bounded by y2 =4x, x2 = 5 − 2y, x = 0.

Idea: As above

(x, y) = (

∫ ∫R xdA∫ ∫R dA

,

∫ ∫R ydA∫ ∫R dA

)

Figure 36

The points of intersection are: (0, 0), 0, 52 ), and those for which x = y2

4 and

x2 = 5 − 2y ⇒ y4

16 + 2y − 5 = 0 or y = 2 ⇒ x = 1 This gives (1, 2).

Take a vertical fibre (horizontal ones lead to the sum of two or more complicatedintegrals).

Then:

R = {(x, y) : 2√

x ≤ y ≤ 5 − x2

2, 0 ≤ x ≤ 1}

Figure 37


area, A = 7∫ ∫

R dA

=∫ 1

0

∫ 5−x22

2√

x

dydx

=∫ 1

0

{52− x2

2− 2x

12 }dx

= 1

Also

∫ ∫R

xdA =∫ 1

0

∫ 5−x22

2√

x

xdydx

=∫ 1

0

x(52− x2

2− 2x

12 )dx

=∫ 1

0

{52x − x3

2− 2x

32 }dx

= 1324

And

∫ ∫R

yda =∫ 1

0

∫ 5−x22

2√

x

ydydx

=∫ 1

0

{y2

2|y= 5−x2

2y=2

√x}dx

=12

∫ 1

0

{(5 − x2

2)2 − 4x}dx

=12

∫ 1

0

{25 − 10x2 + x4

4− 4x}dx

=18

∫ 1

0

{25 − 16x − 10x2 + x4}dx

=18(20815

)

=2615

Therefore the centroid is:

(x, y) = (1340

,2615

)


1.7 Volumes under a surface by double integra-tion

We recall that if z = f(x, y) and f(x, y) ≥ 0 over its domain D, a subset of thexy-plane (or z = 0), the volume of that part of the surface S which lies directlyabove the domain D is given by:

Volume under S =∫ ∫

D

zdA =∫ ∫

D

f(x, y)dA

where dA is the element of area on D

Example 38 Find the volume cut from 9x2+4y2+36z = 36 by the plane z = 0.

IDEA

1) Find the projection of this surface on one of the coordinate axes,

Figure 38

Here, say, z = 0. This gives the ellipse:

D = {(x, y) : 9x2 + 4y2 = 36}

2) solve for z in terms of x and y

Here:

z = 1 − x2

4− y2

9= f(x, y)

Note that for z on D, z = 0 while for z in d, z ≥ 0

3) So volume under S: ∫ ∫D

(1 − x2

4− y2

9)dA

4) need to describe D: We need to introduce elliptical (plane) coordinates cen-tered at (0, 0)

x = ar cosΘy = br sin Θ

; r2 = (x

a)2 + (

y

b)2

Figure 39

The Jacobian is: abr

Here a = 2, b = 3

Therefore 1 − x2

4 − y2

9 = 1 − r2


5) Volume∫ ∫

D

(1 − x2

4− y2

9)dA =

∫ 2π

0

∫ 1

0

(1 − r2) (1 · b · r)︸︷︷︸Jacobian

drdθ

=∫ 2π

0

∫ 1

0

(1 − r2) · 6rdrdΘ

= 12π

∫ 1

0

(1 − r2)rdr

= 12π · (12− 1

4)

= 3π

Remark: S can be ”reconstructed” via its ”projections” eg.

S = {(x, y, z) : 9x2 + 4y2 + 36z = 36}Projection of S onto:

xy-plane is:(set z = 0) ⇒ 4x2 + 9y2 = 36xz-plane is:(set y = 0) ⇒ 9x2 + 36z = 36yz-plane is:(set x = 0) ⇒ 4y2 + 36z = 36

Figure 40

The surface S is reconstructed (basically) by superimposing the three projections.the more projections, the better the schematic representation of S.

Figure 41

Example 39 Find the volume in the first octant bounded by xy = 4z, y = xand x = 4.

Here z = xy4 = f(x, y) ≥ 0 for (x, y) in D.

Volume =∫ ∫

D zdA because x > 0 over D.

V ol =∫ ∫

D

(xy

4)dA

Stick with cartesian coordinates, because z = (xy4 ) has an ”interesting form”

and because D is a ”triangle” so:

D = {(x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 4}


orD = {(x, y) : y ≤ x ≤ 4, 0 ≤ y ≤ 4}

Using Vertical Slices:

vol =∫ 4

0

∫ x

0

xy

4dydx

=14

∫ 4

0

x · (∫ x

0

ydy)dx

=14

∫ 4

0

x · [x2

2− 0]dx

=18

∫ 4

0

x3dx

=18[x4

4]40

=18(43)

= 8

Example 40 D is found by projection of S on xy-plane, z = 0.

The domain of f(x, y) = xy4 is clearly all (x, y) on the xy-plane.

We need to restrict D to that part of xy-plane which is bounded by:

y = x (projection of plane y = x on z = 0)x = 4 (projection of plane x = 4 on z = 0)y = 0 (part of projection of z = xy on z = 0)

Figure 45

Thus volume under S is: ∫ ∫D

(xy

4)dA

Figure 46

Now:D = {(x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 4}

So the volume under S is:


∫ 4

0

∫ x

0

(xy

4)dydx =

14

∫ 4

0

{∫ x

0

xydy}dx

= . . . . . .

=18

∫ 4

0

x3dx

= 8

or, using horizontal slices,

vol =∫ 4

0

∫ 4

y

xy

4dxdy

=14

∫y(

∫ 4

y

xdx)dy

=18

∫ 4

0

y(16 − y2)dy

=18[128 − 64]

=648

= 8

Example 41 Find the volume common to the cylinders x2 + y2 = 16 and x2 +z2 = 16.

Idea: We look at the cylinder x2+z2 = 16 and consider the projection of x2+y2 =16 on the xy-plane or z = 0.

Figure 47

the projection of x2 +y2 = 16 on z = 0 is the circle x2 +y2 = 16. The projectionof x2 + z2 = 16 on z = 0 is the region bounded by the parallel lines x = ±4. So,the projection of the solid of intersection is simply:

D = {(x, y) : 0 ≤ x2 + y2 ≤ 16}= {(x, y) : −

√16 − x2 ≤ y ≤

√16 − x2, −4 ≤ x ≤ +4}

Hence the volume of the solid:

= 2∫ ∫

D

√16 − x2︸︷︷︸

z

dA

Figure 48


because there is a symmetric part of the solid below z = 0, (as z = ±√16 − x2)

Thus:

2∫ ∫

D

√16 − x2dA = 2

∫ 4

−4

∫ +√

16−x2

−√16−x2

(√

16 − x2︸︷︷︸z

dydx

= 2∫ 4

−4

2(16 − x2)dx

= 4∫ 4

−4

(16 − x2)dx

= 8∫ 4

0

(16 − x2)dx (because integrand is an even function)

= 8(16x− x3

3)|40

= 8 · 1283

=1024

3

Example 42 Find the volume inside ρ = 2 and outside the cone z2 = ρ2.(cylindricalcoordinates)

Figure 49

The projection of the solid on z = 0 is given by the inside of the circle ρ = 2.

There are four identical parts making up the required volume and so it sufficesto find the volume of that part of the solid which lies above the semi-circle Dshown in the figure.

Next,

volume of solid =∫ ∫

D

zdA

= 4∫ ∫

D

ρdA (z2 = ρ2 ⇒ z = ρ above D)

Recall z2 = ρ2 = x2 +y2 (in cylindrical coordinates). since D is on the xy-planez = 0, it is best to pick a description of D in polar coordinates. Therefore:

D = {(ρ, Θ) : 0 ≤ ρ ≤ 2, 0 ≤ Θ ≤ π︸︷︷︸note this!

}

Therfore:

V olume = 4∫ ∫

D

ρ(ρdρdθ)


= 4∫ π

0

(∫ 2

0

ρ2dρ)dΘ

= 4∫ π

0

83dΘ

=32π

3

Example 43 Find the volume common to r2 + z2 = a2 and r = a sinΘ (cylin-drical coordinates!).

Note r2+z2 = a2 is a sphere centered at (0, 0, 0) of radius a (since r2 = x2+y2).

Furthermore, r = a sinΘ︸︷︷︸=a y

r

⇔√

x2 + y2 = a y√x2+y2

(polar coordinates). Or,

x2 + y2 = ay, But x2 + y2 − ay = 0 ⇒ x2 + (y − a2 )2 = a2

4 (completing thesquare).

Therefore:

r = a sinΘ ⇔ (x − 0)2 + (y − a

2)2 =

a2

4And this is a cylinder centered at (0, a

2 of radius a2 .

Figure 50

The resulting volume is shown in the diagram.

Therefore the volume of the solid:

= 4∫ ∫

D

zdA

where D = {(r, Θ) : 0 ≤ r ≤ a sinΘ, 0 ≤ θ ≤ π2 } centered at (0, a

2 ) in polarcoordinates. Note that θ ≤ π

2 and not Θ ≤ 2π (Why?? else r < 0!).

Or

V olume = 4∫ π

2

0

∫ a sin Θ

0

√a2 − r2(rdrdΘ)

(Why? because volume = 4 × (volume of a half-circular region))

= 4∫ π

2

0

{∫ a sin Θ

0

r(a2 − r2)12 dr}dΘ

=4

(−2)

∫ π2

0

{∫ a sin Θ

0

(−2r)(a2 − r2)12 dr︸︷︷︸

{u=a2−r2; du=−2rdr}

}dΘ

= (−2)∫ π

2

0

{∫ a2 cos2 Θ

a2u

12 du}dΘ


r = 0 → u = a2 r = a sinΘ → u = a2 cos2 Θ

= (−2)∫ π

2

0

{23u

32 |a2 cos2 Θ

a2 }dΘ

=4(−1)

3

∫ π2

0

{a3 cos3 Θ − a3}dΘ

=−4a3

3

∫ π2

0

(1 − sin2 Θ) cosΘdΘ

=−4a3

3+

4a3

3

∫ π2

0

sin2 Θ cosΘdΘ + 2a3π

3

=−4a3

3+

4a3

3· [ sin

3 Θ3

|π20 ] +

2a3π

3

=−4a3

3+

4a3

9+

2a3π

3

=2a3

9(3π − 4)

Example 44 Find the volume cut from paraboloid 4x2 + y2 = 4z by the planez − y = 2.

Note that paraboloid is z = x2 + y2

4 and plane z = y + 2. Therefore the curve ofintersection of the two surfaces is (equates z’s):

y + 2 = x2 +y2

4

Collecting terms we obtain (after completing the square):

4x2 + (y − 2)2 − 12 = 0

Or:

ε :x2

(√

3)2+

(y − 2)2

(√

12)2= 1

Figure 51

Which is an ellipse. So, the projection of this solid of intersection on the planez = 0 is given by This ellipse, whose center is at (0, 2).

The paraboloid is basically constructed by its projections:xz-plane (y = 0): z = x2

yz-plane (x = 0): z = y2

4

Figure 52

To compute the volume of this solid we compute the height of a typical verticalfibre which intersects the solid.


In our case (height of fibre) = (y+2)− (x2 + y2

4 ) (=difference in z- coordinates)

So a volume element = ((y + 2) − ((x2 +y2

4))dA

Where dA = element of area on the projection D of the solid on z = 0,(or theinterior of ε). Therefore:

V olume =∫ ∫

D

((y + 2) − ((x2 +y2

4))dA

We introduce elliptical coordinates centered at (0, 2)(Why? because our D iscentered there!).

ε : (x√3)2 + (

y − 2√12

)2 = 1

So, let:

x =√

3r cosΘy = 2 +

√12r sin Θ

and theJacobian =

√3 ·

√12r

Figure 53

(then ( x√3)2 + (y−2√

12)2 = r2 the basic equation of our ε occurs when r = 1)

So in our coordinate system

D = {(r, Θ) : 0 ≤ r ≤ 1, 0 ≤ Θ ≤ 2π}and thus:

V olume− =∫ ∫

D

((y+2)−(x2+y2

4))dA =

∫ 2π

0

∫ 1

0

3(1−r2)·√

3√

12rdrdΘ︸︷︷︸area element in new coord. syst.

(Also y + 2 − x2 − y2

4 = . . . = 3(1 − r2) as defined above).

V olume =∫ 2π

0

∫ 1

0

3(1 − r2) · 6rdrdΘ (√

3√

12 = 6)

= 36π

∫ 1

0

(1 − r2) · rdr

= 36π(r2

2− r4

4)|r=1

r=0

= 36π(12− 1

4)

= 36π · 14

= 9π


Remark: for elliptical coordinates centered at (x0, y0) based on the ellipse (x−x0a )2+

(y−y0b )2 = 1 : ε

Figure 54

we set

x = x0 + ar cosΘy = y0 + br sin Θ

Jacobian = abr (as before) etc.

1.5 Double and Iterated Integrals

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