Lecture 37

1 Voume integrals and change of varibles

1.1 Triple (Volume) integrals
Let f (x, y, z) be a real valued function defined over a closed and bounded region of space IR3 .
For example the solid ball or rectangular box. Now we want to define the definite integral of
f (x, y, z) over such regions.
We partition the region by small planes parallel to the coordinate axes. Then we obtain small
rectangular cubes over which the function will be approximated by f (xk yk , zk ). We form the
Riemann sum
X
Sn = f (xk , yk , zk )|Ωk |,
k

where |Ωk | is the volume of the small rectangle. Now by our understanding of Riemann sums
we choose refinement of partitions in such way that maxk |Ωk | → 0. Then we obtain the
definite integral as ZZZ
f (x, y, z)dV = lim Sn .
Ω n→∞

Evaluation of integrals in three dimensions is done again using Fubini’s theorem. In this
case again Fubini’s theorem states

Theorem 1.1.1 Suppose f (x, y, z) is integrable over Ω ⊂ IR3 , then

ZZZ Z Z Z Z Z Z
f (x, y, z)dV = f (x, y, z) dz dy dx = f (x, y, z) dx dz dy
Ω x y z x z y
Z Z Z Z Z Z
f (x, y, z) dy dx dz = f (x, y, z) dx dy dz
z x y z y x
Z Z Z Z Z Z
= f (x, y, z) dz dx dy = f (x, y, z) dy dz dx
y x z y z x

To evaluate the triple integrals we follow the following steps:

1. Draw a line parallel to z axis that passes through the point (x, y) of R where R is the
projection of Ω onto IR2 .

2. Identify the upper surface and lower surface through which this line passes at most once.

3. Identify the upper curve and lower curve of the projection R and limits of integration.

It is easy to see from the definition, the volume of Ω is

X X ZZZ
V = lim |Ωk | = lim 1 |Ωk | = 1 dV
k→∞ k→∞ Ω
k k

1
 Figure 1: Volume bounded by the surfaces

Problem 1.1.1 Find the volume of the region bounded by the surfaces z = x2 + 3y 2 and
z = 8 − x2 − y 2 .
RRR
Solution: The volume is V = Ω dzdydx, where Ω is bounded above by the surface z =
8 − x − y and below by the surface z = x2 + 3y 2 . Therefore, the limits of z are from
2 2

z = x2 + 3y 2 to z = 8 − x2 − y 2 .
The Projection of Ω on xy-plane is the solution of

8 − x2 − y 2 = x2 + 3y 2 =⇒ x2 + 2y 2 = 4.

Therefore the limits of x and y are to be determined by R : x2 + 2y 2 = 4. Hence

ZZ Z 8−x2 −y 2
V = dzdA
R y=x2 +3y 2
Z 2 Z √(4−x2 )/2
= √ (8 − 2x2 − 4y 2 )dydx
−2 − (4−x2 )/2

2 
Z √(4−x2 )/2
4
= (8 − x2 )y − y 3 √
−2 3 y=− (4−x2 )/2
√ Z 2
4 2 √
= (4 − x2 )3/2 dx = 8π 2.
3 −2

Problem 1.1.2 Find the volume of the region bounded by x + z = 1, y + 2z = 2 in the first
quadrant.

2
Solution: Draw line parallel to z-axis and note that the upper surfaces are: 2z + y = 2
over triangle bounded by x = 0, y = 1y = 2x and z = 1 − x over the triangle bounded by
y = 0, x = 1, y = 2x. Therefore,
Z 2 Z y/2 Z 2−y Z 1 Z 2x Z 1−x
2
V = dz dx dy + dz dy dx
y=0 x=0 z=0 x=0 y=0 z=0

On the other hand, by first drawing the line parallel to x-axis, we get
Z 1 Z 2−2z Z 1−z
V = dx dy dz
z=0 y=0 x=0

Taking the line parallel to y-axis we get

Z 1 Z 1−x Z 2−2z
V = dy dz dx
x=0 z=0 y=0

4 1 2
2 cos(x2 )
Z Z Z
Example 1.1.1 (Order of integration) Evaluate √ dx dy dz.
z=0 y=0 x=2y z

Note that the projection of Ω onto xyplane is the triangle bounded by y = 0, x = 2 and
x = 2y. So changing the order of integration in x and y, we get
Z 4 Z 2 Z x/2
2 cos(x2 )
I= √ dy dx dz.
z=0 x=0 y=0 z
Z 4 Z 2
x cos(x2 )
= √ dx dz = 2 sin 4.
z=0 x=0 z

1.2 Substitutions in multiple integrals

Suppose a domain G in uv-plane is transformed onto a domain Ω of xy-plane by a transfor-
mation x = g(u, v), y = h(u, v). Then any function of x, y may be written as a function of
u, v. Then the relation between the double integral over G and Ω is
ZZ ZZ
f (x, y)dx dy = f (g(u, v), h(u, v))|J(u, v)|du dv
Ω G

where J is the Jacobian given by

∂(x, y) xu xv
J= =
∂(u, v) yu yv

The main idea of the proof is as follows. Let (u, v), (u + ∆u, v), (u + ∆u, v + ∆v) and
(u, v + ∆v) be the vertices of the rectangle in the uv−plane. Let ∆Ak be its area element.
Under the transformation this points are mapped to (x1 , y1 ) = (g(u, v), h(u, v)), (x2 , y2 ) =
(g(u+∆u, v), h(u+∆u, v)), (x3 , y3 ) = (g(u+∆u, v +∆v) and (x4 , y4 ) = (g(u, v +∆v), h(u, v +
∆v)). Then by Taylor’s theorem
∂g
g(u + ∆u, v) = g(u, v) + ∆u + o((∆u)2 )
∂u

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 ∂g ∂g
g(u + ∆u, v + ∆v) = g(u, v) + ∆u + ∆v + o((∆u)2 ) + o((∆v)2 )
∂u ∂v
Then the area of the ”rectangle” in xy-plane ∆A˜k is

∆A˜k ≈|(x3 − x1 )(y3 − y1 ) − (x3 − x2 )(y3 − y2 )|

≈|J|∆u∆v + o((∆u)2 ) + o((∆v)2 )

Taking this as the area in the Riemann sum of f (x, y) we get the required formula.

4 Z 1+ y2
2x − y
Z
Example 1.2.1 Evaluate the integral I = dxdy.
0 y/2 2

Solution: The domain of integration is a parallelogram with vertices (0, 0), (1, 0), (3, 4) and
(2, 4). One has to divide the domain into 3 domains. Instead we can take the transformation
u = 2x−y y
2 , v = 2 . Then the inverse transformation is x = u + v, y = 2v. Then

1 1
J= = 2.
0 2

Under this transformation, the parallelogram is transformed into cube with vertices (0, 0), (1, 0)(1, 2)
and (0, 2). Now by change of variable formula
ZZ Z 2Z 1
I= f (u + v, 2v)2dudv = 2udu dv = 2.
0 0
Z 1 Z 1−x √
Example 1.2.2 Evaluate the integral I = x + y(y − 2x)2 dA.
0 0

Solution: The given domain is the triangle bounded by x = 0, y = 0 and x + y = 1. In this

case the integrand is complicated....so we can take transformation u = x + y and v = y − 2x.
Under this transformation, the given triangle will be transformed into triangle bounded by
v = u, v = −2u and u = 1. The inverse of this transformation is x = u−v 3 and y =
2u+v
3 .
Hence the Jacobian
1/3 −1/3
J= = 1/3.
2/3 1/3
Hence Z 1Z u √
I= uv 2 dv du
0 v=−2u

dA
over R : x2 + y 2 ≤ 1.
RR
Example 1.2.3 Evaluate the integral I = R (2−x2 −y 2 )2

Solution: Taking the transformation x = r cos θ, y = r sin θ, we get

cos θ −r sin θ
J= = r.
sin θ r cos θ

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By substitution formula,
Z 2π Z 1 Z 2
r dr dθ dt
I= = 2π = π/2.
0 r=0 (2 − r2 )2 1 2t2

Substitution formula for triple integrals

As discussed above suppose a three dimensional domain G is transformed onto a domain D
with a transformation x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), then
ZZZ ZZZ
f (x, y, z)dV = F (u, v, w)|J(u, v, w)|dV
D G

where
xu xv xw
∂(x, y, z)
J= = yu yv yw .
∂(u, v, w)
z u zv zw
ZZZ
Example 1.2.4 Evaluate (x2 y + 3xyz)dV where R = {(x, y, z) : 1 ≤ x ≤ 2, 0 ≤ xy ≤
Ω
2, 0 ≤ z ≤ 1}.

solution: We take the transformation u = x, v = xy and w = z. Then the planes x = 1, 2

transforms to u = 1, 2. The plane y = 0 transforms to v = 0. The surface xy = 2 transforms
to v = 2. Then the Jacobian J is

1 0 0
1
= y x 0 = x = u.
J
0 0 1

Now by substitution formula,

Z 2 Z 2 Z 1
1
I= (uv + 3vw) dw dv du
u=1 v=0 w=0 u
Z 2Z 2
3v
= (v + )dv du
1 0 2u
Z 2
3
= (2 + )du = 2 + 3 ln 2.
1 u