1. Double Integrals - Çankaya Üniversitesi

MULTIPLE INTEGRALS

1. Double Integrals

Let R be a simple region defined by a ≤ x ≤ b and g1(x) ≤ y ≤ g2(x), where g1(x) and g2(x)

are continuous functions on [a, b] and let f (x, y) be a function defined on R. Then

ˆ b ˆ g2(x)

IR = f (x, y)dy dx

a g1(x)

is called an iterated integral of f over R.

Similarly, if R is a simple region defined by c ≤ y ≤ d and h1(y) ≤ x ≤ h2(y), where h1(y) and

h2(y) are continuous functions on [c, d] and let f (x, y) be a function defined on R. Then

ˆ d ˆ h2(y)

JR = f (x, y)dx dy

c h1(y)

is called an iterated integral of f over R.

Notation. Sometimes IR may be written as

ˆ b ˆ g2(x) ˆ b ˆ g2(x) ˆ b ˆ g2(x)

IR = f (x, y)dy dx = f (x, y) dy dx = dx f (x, y) dy,

a g1(x) a g1(x) a g1(x)

and JR may be written as

ˆ d ˆ h2(y) ˆ d ˆ h2(y) ˆ d ˆ h2(y)

JR = f (x, y)dx dy = f (x, y) dx dy = dy f (x, y) dx.

c h1(y) c h1(y) c h1(y)

Example 1.1. Calculate both iterated integrals IR and JR of the function f (x, y) = 3xy2−2x2y
over R where R is bounded by x = 0, x = 2, y = 1 and y = 2.

1

2 MULTIPLE INTEGRALS

Theorem 1.1. Suppose that f (x, y) is continuous on R, where R is defined by

a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)

or
c ≤ y ≤ d, h1(y)¨≤ x ≤ h2(y).

Then the double integral of f over R, denoted by f (x, y)dA, is

R

¨ ˆ b ˆ g2(x) ˆ d ˆ h2(y)

f (x, y) dA = f (x, y) dy dx = f (x, y) dx dy.

a g1(x) c h1(y)
R

Properties. Let f (x, y) and g(x, y) be two integrable functions over R, a and b be two arbitrary

constan¨ts, and R = R1 ∪ R2. Then ¨ ¨

(i) af (x, y) + bg(x, y) dA = a f (x, y) dA + b g(x, y) dA.

¨R ¨ R¨ R

(i) f (x, y) dA = f (x, y) dA + f (x, y) dA.

R R1 R1 ¨

Example 1.2. (1) If R = {(x, y)| 0 ≤ x ≤ 2 and 1 ≤ y ≤ 4}, evaluate (6x2 + 4xy3) dA.

¨R
(2) Let R be the region bounded by y = x, y = 0 and x = 4, write f (x, y) dA as an

R

iterated in¨tegral.

(3) Evaluate x dA, where R is the portion of disk x2 + y2 ≤ 16 in the second quadrant.

¨R
(4) Write f (x, y) dA as an iterated integral, where R is the region bounded by x = y2

R

and x = 2 − y.

(5) Let R be the region bounded by the graphs of y = x, y = 0 and x = 4. Evaluate
¨
(4ex2 − 5 sin y)dA.

R

√
(6) ¨Let R be the region bounded by the graphs of y = x, x = 0 and y = 3. Evaluate

(2xy2 + 2y cos x) dA.

R¨
(7) Write f (x, y) dA as an iterated integral, where R is the region bounded by x2+y2 = 1

R

and x2 + 2y2 = 1.

Example 1.3. Reverse the order of the integration

MULTIPLE INTEGRALS 3

ˆ 2 ˆ 2x

(1) f (x, y) dy dx.

ˆ1 2 ˆx √
4−y2

(2) √ f (x, y) dx dy.
ˆ−√ 4−y2
ˆ0 1 2x−x2

(3) f (x, y) dy dx.

ˆ0 2 ˆx2x

(4) f (x, y) dy dx.
ˆ0 e ˆx2ln y

(5) f (x, y) dx dy.

ˆ1 3 ˆ0 x ˆ 6 ˆ 6−x

(6) f (x, y) dy dx + f (x, y) dy dx.

00 30

Example 1.4. Evaluate the integral.
ˆ 1ˆ 1

(1) ex2dxdy.

ˆ0 1 ˆy 1
(2) √ ey3 dy dx.
ˆ0 9 ˆ x

3 ex2−2x
(3) dx dy.
¨1 √y x + 1

(4) x(x − 1)exy dA where R is bounded by the lines x = 0, y = 0 and x + y = 2.

¨R
(5) (x + y) dA where R is bounded by the lines y = 2x, x = 0 and y = 4.

ˆR 8 ˆ ln x
(6) ey dy dx.

ˆ2 π ˆ0 1+cos y
(7) x2 sin y dx dy.

ˆ0 1 ˆ 0 y2
1

(8) dy dx.
x2 + 1
ˆ0 1 ˆ0 1 x

(9) (1 + x2 + y2) 3 dx dy.
y2 2
ˆ0 2 ˆ0 x

(10) dy dx.
x1 2
1
x

2. Area ¨

The area of a bounded region in the real plane can be computed by dA = Area of R.

R

4 MULTIPLE INTEGRALS

For example, if R has the shape

ˆ b ˆ g2(x)

then, A(R) = dy dx. If R has the shape

a g1(x)

ˆ d ˆ h2(y)

then, A(R) = dx dy.

c h1(y)

Example 2.1. Find the area of the region R where R is bounded by

(1) y = x and y = x2 in the first quadrant.
(2) y = x2 and y = x + 2.

(3) xy = 1, y = x, x = e.
(4) y = 1 x2 − 1 and y = 2 − x.

4
(5) x = y2, y − x = 3, y = −3, and y = 2.

Example 2.2. Let R1 = {(x, y)| x2 + 2y2 ≤ 1}, R2 = {(x, y)| 2x2 + y2 ≤ 1} and R = R1 ∩ R2.
Find the area of R.

¨ 3. Volume
The double integral f (x, y)dA of a positive function f (x, y) can be interpreted as the volume

D

under the surface z = f (x, y) over the region D.

MULTIPLE INTEGRALS 5

¨
If f (x, y) ≥ g(x, y) for all (x, y) in D, then the double integral (f (x, y) − g(x, y))dA is the

D

volume between the surface z = f (x, y) and the surface z = g(x, y).

Example 3.1. For example, the volume of the tetrahedron bounded by the plane 2x+y +z = 2
and the three coordinate planes (See the figure below)

is given by ˆ 1 ˆ 2−2x
V = (2 − 2x − y) dy dx

00

Example 3.2. Find the volume of the solid lying in the first octant and bounded by the graphs
of z = 4 − x2, x + y = 2, x = 0, y = 0 and z = 0.

Example 3.3. (1) Find the volume of the solid bounded by the graphs of z = 2, z = x2 +1,
y = 0 and x + y = 2.

(2) A solid T is bounded by the coordinate planes, the plane z = x + 2y + 1 and the planes
x = 1 and y = 2. Find the volume of T .

(3) A solid T lies in the first octant and is bounded by the coordinate planes, and the planes
x + 2y = 2 and x + 4y + 2z = 8. Find the volume of T .

6 MULTIPLE INTEGRALS

(4) A solid T is bounded by the elliptic paraboloid z = 1 x2 + 1 y2, the coordinate planes
49

and the planes x = −2 and y = 3. Find the volume of T .
(5) A solid T lies in the first octant and is bounded by the elliptic cylinder 4x2 + z2 = 1,

the plane y = x, and the planes y = 0 and z = 0.
(6) Find the volume enclosed between the two surfaces z = x2 + 3y2 and z = 8 − x2 − y2.
(7) A solid T is bounded above by z = 2 − x2 − y2 and below by z = x2 + y2. Find the

volume of T .
(8) A solid T is bounded by the sphere x2 +y2 +z2 = 4 and the paraboloid x2 +y2 = 4(1−z).

Find the volume of T .
(9) Find the volume V of the solid region T where T is bounded by the sphere x2+y2+z2 = 4

and the paraboloid x2 + y2 = 4(1 − z).
(10) A solid T is bounded by the paraboloids x2 + y2 − z = 0 and x2 + y2 + 2z = 1. Find

the volume of T .
(11) Find the volume of the solid in the first octant bounded by the circular paraboloid

z = x2 + y2, the cylinder x2 + y2 = 4 and the coordinate planes. Find the volume of T .
(12) A solid T is bounded by the cone z2 = xy and the plane x + y = 4. Find the volume of

T.
(13) Use double integration to find the volume of the tetrahedron bounded by the coordinate

planes and the plane 3x + 6y + 4z − 12 = 0.
(14) Find the volume of the solid lying in the first octant and bounded by z = 4 − x2,

x + y = 2, x = 0 and z = 0.
(15) Find the volume enclosed between the two surfaces z = x2 + y2 and z = 8 − x2 − y2.

4. Change of Variables in Double Integrals
A transformation T from the uv-plane to the xy-plane is a function that maps points in the
uv-plane to points in the xy-plane, so that

T (u, v) = (x, y),

where x = g(u, v) and y = h(u, v), for some functions g and h. We consider changes of variables
in double integrals as defined by a transformation T from a region S in the uv-plane onto a
region R in the xy-plane

Example 4.1. Let R be the region bounded by the straight lines y = 2x + 3, y = 2x + 1,
y = 5 − x and y = 2 − x. Find a transformation T mapping a region S in the uv-plane onto R,
where S is a rectangular region, with sides parallel to the u- and v-axes.

MULTIPLE INTEGRALS 7

The determinant ∂x ∂x
∂u ∂v

∂y ∂y

∂u ∂v

is referred to as the Jacobian of the transformation T and is written using the notation ∂(x, y) .

∂(u, v)

Theorem 4.1 (Change of Variables in Double Integrals). Suppose that the region S in the uv-

plane is mapped onto the region R in the xy-plane by the one-to-one transformation T defined

by x = g(u, v) and y = h(u, v), where g and h have continuous first partial derivatives on S. If

f is continuous on R and the Jacobian ∂(x, y) is nonzero on S, then

¨¨ ∂(u, v)

f (x, y)dA = f (g(u, v), h(u, v)) ∂(x, y) dudv.

∂(u, v)

RS

Remark. Note that ∂(x, y) stands for the absolute value of the Jacobian.

∂(u, v)

Remark. Sometimes, it is useful to use the formula

∂(u, v) 1
=.
∂(x, y) ∂(x, y)

Example 4.2. ∂(u, v)
¨

(1) Evaluate (x2 + 2xy) dx dy where R is the region bounded by the lines

R 5¨− −

y = 2x + 3, y = 2x + 1, y = x, and y = 2 x.
ex−y
(2) Evaluate the double integral dA, where R is the rectangle bounded by the lines
x+y

R

y = x, y = x + 5, y = 2 − x and y = 4 − x.

(3) Find the area of the region in the first quadrant bounded by the curves xy = 1, xy = 4,

y = x, and y = 2x.

8 MULTIPLE INTEGRALS

¨ x2 dx dy where R is the region bounded by xy = 2, xy = 4, y2 = x, and
(4) Evaluate y4

R

y2 = 3x. ¨

(5) Evaluate xy dx dy where R is the region bounded by the lines x+3y = −1, x+3y = 3,

R

x − y = 1¨, and x − y = 2.
(6) Evaluate (x2 +y2) dA where R is the region bounded by the lines x+y = 1, x+y = 2,

R

3x + 4y = 5, and 3x + 4y = 6.
(7) Find the a¨rea of the region bounded by the curves y = x2, y = 2x2, x = 3y2, and x = y2.

(8) Evaluate (x2 + y2) dA where R is the region in the first quadrant bounded by y = 0,

R

y = x, xy = 1, x2 − y2 = 1.

5. Double Integrals in Polar Coordinates

Let

T: x = r cos θ
y = r sin θ

be a transformation from rθ-plane to xy-plane. Which transforms regions expressed in polar

coordinates to regions in Cartesian coordinates.

Example 5.1 (A Transformation Involving Polar Coordinates). Let R be the region inside the
circle x2 + y2 = 9 and outside the circle x2 + y2 = 4 and lying in the first quadrant between the
lines y = 0 and y = x. Find a transformation T from a rectangular region S in the rθ-plane to
the region R.

Example 5.2. Derive the evaluation formula for polar coordinates (r > 0):
¨¨
f (x, y)dA = f (r cos θ, r sin θ)rdrdθ.

RS

Theorem 5.1. Suppose that f (r, θ) is continuous on the region

R = {(r, θ)|α ≤ θ ≤ β and g1(θ) ≤ r ≤ g2(θ)},

where 0 ≤ g1(θ) ≤ g2(θ) for all θ in [α, β]. Then,
¨ ˆ β ˆ g2(θ)

f (r, θ)dA = f (r, θ)rdrdθ.

R α g1(θ)

¨

Example 5.3. Evaluate x2 + y2 dx dy where R is the disk x2 + y2 ≤ 9.

R
¨
dx dy
Example 5.4. Evaluate x2 + y2 + 1 where R is the half disk x2 + y2 ≤ 1, y ≥ 0.

R

MULTIPLE INTEGRALS 9

¨ dx dy where R is the annulus 1 ≤ x2 + y2 ≤ 4.
Example 5.5. Evaluate x2 + y2

R

Example 5.6. Find the area inside the curve defined by r = 2 − 2 sin θ.

¨
Example 5.7. Evaluate (x2 + y2 + 3)dA, where R is the circle of radius 2 centered at the

R

origin.

Example 5.8. Find the volume inside the paraboloid z = 9 − x2 − y2, outside the cylinder
x2 + y2 = 4 and above the xy-plane.

ˆ 1 ˆ √
1−x2

Example 5.9. Evaluate the iterated integral x2(x2 + y2)2dydx.

−1 0

Example 5.10. Find the volume of the solid bounded by z = 8 − x2 − y2 and z = x2 + y2.

Example 5.11. Find the volume cut out of the sphere x2 + y2 + z2 = 4 by the cylinder
x2 + y2 = 2y.

10 MULTIPLE INTEGRALS

Example 5.12. Find the volume of the solid in the first octant bounded by the circular
paraboloid z = x2 + y2, the cylinder x2 + y2 = 4 and the coordinate planes.

Example 5.13. Find the volume of the solid under the surface z = x2 + y2 above the xy-plane
and inside the cylinder x2 + y2 = 2y.

Example 5.14. Find the volume V of the solid region T where T is bounded by xy-plane, the
cylinder x2 + y2 = x and the cone z = x2 + y2.

Example 5.15. Find the volume V of the solid region T where T is bounded above by the
plane z = 1 and below by the cone z = x2 + y2.

ˆ 1ˆ √
1−x2

Example 5.16. Evaluate x2 x2 + y2 dy dx.

00

6. Triple Integrals
Suppose that f (x, y, z) is continuous on the box Q defined by

Q = {(x, y, z)| a ≤ x ≤ b, c ≤ y ≤ d and r ≤ z ≤ s}.

Then, we can write the triple integral over Q as a triple iterated integral:

˚ ˆ sˆ dˆ b

f (x, y, z)dV = f (x, y, z)dxdydz.

rc a
Q

˚

Example 6.1. Evaluate the triple integral 2xey sin zdV , where Q is the rectangle defined

Q

by

Q = {(x, y, z)| 1 ≤ x ≤ 2, 0 ≤ y ≤ 1 and 0 ≤ z ≤ π}.

In general, if F (x, y, z) is a function defined on a closed bounded region Q in space, we evaluate
the triple integral by applying a three dimensional version of iterated integration.
For instance, if the region Q can be written in the form

Q = {(x, y, z)|(x, y) ∈ R and g1(x, y) ≤ z ≤ g2(x, y)},

MULTIPLE INTEGRALS 11

where R is some region in the xy-plane and where g1(x, y) ≤ g2(x, y) for all (x, y) in R, then

˚ ¨ ˆ g2(x,y)

f (x, y, z)dV = f (x, y, z)dzdA.

g1(x,y)
QR

˚

Example 6.2. Evaluate 6xydV , where Q is the tetrahedron bounded by the planes x = 0,

Q

y = 0, z = 0 and 2x + y + z = 4.

˚
Example 6.3. Evaluate 6xydV , where Q is the tetrahedron bounded by the planes x = 0,

Q

y = 0, z = 0 and 2x + y + z = 4 as in the previous example, but this time, integrate first with
respect to x.

Example 6.4. Evaluate

ˆ 4ˆ 4ˆ y 6

0 x 0 1 + 48z − z3 dzdydx.

¨

Recall that for double integrals, we had mentioned that 1dA gives the area of the region R.

R

Similarly for solid region in space ˚
V = 1dV

Q

where V is the volume of the solid Q.

Example 6.5. Find the volume of the solid bounded by the graphs of z = 4 − y2, x + z = 4,
x = 0 and z = 0.

12 MULTIPLE INTEGRALS

Example 6.6. Find the volume of the solid region bounded by the planes x = 0, x = 3, z = 1
and z = y2.

Steps to evaluate a triple by integration over a solid region Q first with respect to
z, then with respect to y, and finally with respect to x.

• Step 1. Sketch: Sketch the region Q along with its “shadow” R (vertical projection)
in the xy-plane. Label the upper and lower bounding surfaces of Q and the upper and
lower bounding curves of R.

• Step 2. Find the z- limits of integration: Draw a line M passing through a typical
point (x, y) in R parallel to the z- axis. As z increases, M enters Q at z = f1(x, y) and
leaves at z = f2(x, y). These are the z- limits of integration.

• Step 3. Find the y- limits of integration: Draw a line L through (x, y) parallel to
the y- axis. As y increases, L enters R at y = g1(x) and leaves at y = g2(x). These are
the y- limits of integration.

• Step 4. Find the x- limits of integration: Choose x- limits that include all lines
through R parallel to the y- axis. (x = a, x = b).

Then the integral is

˚ ˆ x=b ˆ y=g2(x) ˆ z=f2(x,y)

F (x, y, z)dV = F (x, y, z) dz dy dx.

x=a y=g1(x) z=f1(x,y)
Q

Follow a similar procedure if you change the order of integration.

Change of Variables in Triple Integrals. For a transformation T from a region S of uvw-

space onto a region R in xyz-space, defined by x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w),

the Jacobian of the transformation is the determinant ∂(x, y, z) defined by

∂(u, v, w)

∂x ∂x ∂x

∂(x, y, z) ∂u ∂v ∂w
= ∂y ∂y ∂z

∂(u, v, w) ∂u ∂v ∂w
∂z ∂z ∂z

∂u ∂v ∂w

Theorem 6.1. Suppose that the region S in uvw-space is mapped onto the region R in xyz-space
by the one-to-one transformation T defined by x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w),
where g, h and l have continuous first partial derivatives in S. If f is continuous in R and the

MULTIPLE INTEGRALS 13

Jacobian ∂(x, y, z) is nonzero in S, then

∂(u, v, w)

˚˚ ∂(x, y, z)

f (x, y, z)dV = f (g(u, v, w), h(u, v, w), l(u, v, w)) ∂(u, v, w) dudvdw.

RS

7. Cylindrical Coordinates

Example 7.1. Derive the evaluation formula for triple integrals in spherical coordinates:
˚˚
f (x, y, z)dV = f (r cos θ, r sin θ, z)r dz dr dθ.

RS

˚
Example 7.2. Evaluate ex2+y2dV , where Q is the solid bounded by the cylinder x2+y2 = 9,

Q

the xy-plane and the plane z = 5.

˚
Example 7.3. Write f (r, θ, z)dV as a triple iterated integral in cylindrical coordinates if

Q

Q = {(x, y, z)| x2 + y2 ≤ z ≤ 18 − x2 − y2}.

14 MULTIPLE INTEGRALS

Example 7.4. Evaluate the triple iterated integral

ˆ 1 ˆ √ ˆ 2−x2−y2
1−x2
3
√ (x2 + y2) 2 dzdydz.

−1 − 1−x2 x2+y2

Example 7.5. Use a triple integral to find the volume of the solid Q bounded by the graph of
y = 4 − x2 − z2 and the xz-plane.

Example 7.6. Find the volume V of D where D is the solid region bounded by z = x2 + y2

and z = x2 + y2.

ˆ 2ˆ √ ˆ 3
2x−x2

Example 7.7. Evaluate z x2 + y2 dz dy dx using cylindrical coordinates.

˚0 0 0

Example 7.8. Evaluate (x2 + y2 + 1) dz dy dx where D is the solid region bounded by the

D

cylinder x2 + y2 = 2y and the panes z = 0 and z = 2.

Example 7.9. Find the volume V of the solid region bounded by the plane x + z = 1 and the
paraboloid z = 1 + x2 + y2.

˚

Example 7.10. Evaluate zy dV where D is the solid region bounded above by the plane

D

z = 1 and below by the cone z = x2 + y2.
˚

Example 7.11. Evaluate xy dV where D is the solid region in the first octant bounded

D 4 − x2 − y2 and on the sides and the bottom by the

above by the the hemisphere z =
coordinate planes.

MULTIPLE INTEGRALS 15

˚

Example 7.12. Evaluate y2z dx dy dz where D is the solid region bounded above by the

D

the sphere x2 + y2 + z2 = 4 and below by the cone z = x2 + y2.
˚

Example 7.13. Evaluate (x2 + y2) dV where D is the solid region between the cylinders

D

x2 + y2 = 1 and x2 + y2 = 4 and the planes z = 1 and z = 2 in the first octant.

ˆ 1ˆ √ ˆ 1
1−x2

Example 7.14. Evaluate √ x dz dy dx.

0 − 1−x2 x2+y2

8. Spherical Coordinates

Example ˚8.1. Derive the eva˚luation formula for triple integrals in spherical coordinates:
f (x, y, z)dV = f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρ dφ dθ.

RS

Example 8.2. Find the volume of the sphere of radius a.

Example 8.3. Evaluate the triple integral ˝ cos(x2 + y2 + z 2 ) 3 dV , where Q is the unit ball:
2

Q

x2 + y2 + z2 ≤ 1.

Example 8.4. Find the volume lying inside the sphere x2 + y2 + z2 = 2z and inside the cone
z2 = x2 + y2.

ˆ 2 ˆ √ ˆ √
4−x2 4−x2−y2

Example 8.5. Evaluate the triple iterated integral (x2 + y2 + z2)dzdydx.

−2 0 0

16 MULTIPLE INTEGRALS

˚ x2 + y2 + z2 dx dy dz, where D is the ball x2 + y2 + z2 ≤ a2.

Example 8.6. Evaluate

D

˚
Example 8.7. Evaluate xz dx dy dz, where D is the part of the spherical shell 1 ≤ x2 +

D

y2 + z2 ≤ 4 in the first octant.