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Contents
1 Double Integrals over Rectangular Regions 1
2 Double Integrals Over General Regions 7
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Areas of General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Region Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3 Double Integrals in Polar Coordinates 14
4 Triple Integrals 20
4.1 Rectangular Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.2 General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5 Triple Integrals in Cylindrical Coordinates 26
6 Triple Integrals in Spherical Coordinates 30
1 Double Integrals over Rectangular Regions
Wenowbeginour study of multi-variable integration. In single variable calculus, we learned
that the definite integral Z
b
f(x) dx
a
represents the area between the curve defined by f(x) and the x-axis, between x = a and
x = b. This can be understood geometrically by thinking of f(x) dx as the area of a rectangle
with an infinitesimally small base dx and a height of f(x). The operation of integration is
to add up all of these areas to get the total area. So how does this idea extend to functions
of two variables? A function of the form z = f(x,y) represents a surface in R3. Let R be
some rectangular region in the xy plane. Evaluating f over every point of R creates a solid
“pillar” with a rectangular base and a curved top, where the shape of the top of the pillar
is determined by the shape of the surface, as shown in the figure below. The volume of this
solid can be calculated with multi-variable integration.
Let R be the rectangular region in the xy plane such that a ≤ x ≤ b and c ≤ y ≤ d. We say
that this region is the Cartesian product of the intervals [a,b] and [c,d] and we denote this
as
R=[a,b]×[c,d] = {(x,y) | a ≤ x ≤ b and c ≤ y ≤ d}.
Note that this notation uses the × symbol, but it is not at all related to the cross product
of vectors. Let (x,y) be some point inside the rectangle R and let z = f(x,y) be a surface
in R3. Suppose we draw a small rectangle D with dimensions ∆x and ∆y that is centered
around (x,y). The area of this rectangle is ∆A = ∆x∆y. Then the volume ∆V of the pillar
that lies over the rectangle D and under the surface z = f(x,y) is approximately the area of
Dtimes the height of the surface at (x,y):
∆V ≈f(x,y)∆A=f(x,y)∆x∆y.
Note that this is an approximation because the pillar is actually curved at the top since the
function can change within the rectangle D. However, in the limit as ∆x, ∆y → 0, i.e., each
dimension of D becomes infinitesimal, the above approximation becomes exact:
dV =f(x,y) dA = f(x,y) dxdy,
Figure 1: Illustration of a solid under a surface and over the region [1,2] × [2,3].
which is the volume of the infinitesimal pillar centered at (x,y), with a dx-by-dy rectangular
base. We obtain the total volume under the surface and over the region R = [a,b]×[c,d] by
adding up all of the differential volumes for the pillars centered at every point in R. In other
words, we integrate f(x,y) over the region R. The total volume under the surface and over
the region R is written symbolically as a double integral:
V =ZZ f(x,y) dA.
R
Since dA = dx dy and R = [a,b]×[c,d], we can write the above integral strictly in terms of
x and y, which gives us an iterated integral:
ZZ f(x,y) dA = Z dZ bf(x,y) dx dy .
R c a
Notice that the double integral has an inner integral (with respect to x), and an outer integral
(with respect to y). We interpret the double integral as follows: for a fixed value of y, we
add up all of the pillar volumes along the x axis, which tells us the volume the “slab” made
of all the pillars at that y value. This slab would have a volume
W(y)=Z bf(x,y) dx
a
where the integration is done with respect to x, treating y as a constant. Then we add up all
of the slab volumes for every point along the y axis to obtain the total volume of the solid:
V =Z dW(y) dy =Z dZ bf(x,y) dx dy.
c c a
In other words, we do the inner integral first, while treating the outer variable as constant.
This results in a function of the outer variable only, and reduces the double integral to a
single integral. Note that because the region we are integrating over is perfectly rectangular,
we could have instead done this process with y as the inner variable and x as the outer
variable. Therefore the integrals can be written in either order for this problem. This is
known as Fubini’s Theorem:
Z dZ bf(x,y) dx dy = Z bZ df(x,y) dy dx.
c a a c
However, as we will see in the next section, this is not the case when integrating over non-
rectangular regions.
Many of the common properties and intuitions from single integrals carry over to double
integrals. For example, surfaces that lie below the xy plane contribute negative volume, just
as curves that lie below the x-axis contribute negative area in single integrals. The linearity
properties of single integrals apply to double integrals as well:
ZZ c1f(x,y)+c2g(x,y) dA = c1ZZ f(x,y) dA+c2ZZ g(x,y) dA
D D D
for any constants c1 and c2 and any two-variable functions f and g. Another useful property
to know is that if f(x,y) can be factored into a function of x only times a function of y only,
i.e., f(x,y) = g(x)h(y), then
Z bZ df(x,y) dy dx = Z bZ dg(x)h(y) dy dx = Z bg(x) dx Z dh(y) dy .
a c a c a c
In other words, we can do the single integrals for g(x) and h(y) separately, then multiply
them to the get the answer for the double integral. Functions that can be factored this way
are said to be separable.
Example 1.1. Find the volume under the surface of f(x,y) = x2y−y+1 over the rectangle
R=[0,1]×[−1,2].
Solution: We need to do the double integral
ZZ f(x,y) dA = Z 2 Z 1x2y −y +1 dx dy = Z 2 Z 1x2y −y +1 dx dy.
R −1 0 −1 0
Wefirst do the inner integral with respect to x, treating y as constant and get
Z 1 x=1
2 1 3 1 2
x y −y+1 dx= x y−xy+x = y−y+1=− y+1.
0 3 x=0 3 3
Now plugging this into the outer integral we get
Z 2 y=2
2 1 2 4 1 4 1
− y+1dy=− y +y = − +2 − − −1 =− +2+ +1=2.
−1 3 3 y=−1 3 3 3 3
Now let us do the integral in the other order, to verify that it is the same:
Z 1 Z 2 x2y −y +1 dy dx.
0 −1
The inner integral is
Z 2 y=2
2 1 2 2 1 2 2 1 2 1 3 2 3
x y −y+1 dy = x y − y +y =(2x −2+2)−( x − −1)= x + .
−1 2 2 y=−1 2 2 2 2
Plugging this into the outer integral, we get
Z 1 x=1
3 2 3 1 3 3 3 1
x + dx= x + x = + =2.
0 2 2 2 2 x=0 2 2 y
Example 1.2. Find the volume under the surface of f(x,y) = xe over the rectangle
R=[2,3]×[0,1].
Solution: Since this function can be written as f(x,y) = g(x)h(y) with g(x) = x and
y
h(y) = e , the double integral is the product of the two single integrals:
ZZ Z 3Z 1 y Z 3 Z 1 y
f(x,y) dA = xe dy dx = x dx e dy .
R 2 0 2 0
The x integral is Z
3 1 x=3 9 5
x dx = x2 = −2= .
The y integral is 2 2 x=2 2 2
Z 1 y=1
y y
e dy = e =e−1.
Therefore the double integral is 0 y=0
ZZ f(x,y) dA = 5(e−1).
R 2
Example 1.3. Calculate the iterated integral
Z 2Z 1 −x
x+e dxdy.
1 0 y
Solution: This function doesn’t appear to be separable but if we split this into two integrals
as Z Z Z Z
2 1 2 1 −x
x dx dy + e dx dy
1 0 1 0 y
then each of these integrals has a separable integrand and can be reduced to the product of
two single integrals:
Z 1 Z 2 Z 1 −x Z 2 −1
x dx dy + e dx y dy
0 1 0 1
" 1#" 2# " 1#" 2#
1 2 −x
= x y + −e ln|y|
1 2 0 0 −1 1 0 1 1 −1
=2(1−0)(2−1)+(e −e )(ln(2)−ln(1))= 2 +(1−e )ln(2).
Example 1.4. Calculate the iterated integral
Z 1Z 1√x+y dx dy.
0 0
Solution: We first do the inner integral
Z 1√x+y dx.
0
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