Steps to set up iterated integrals
Double integral, rectangular coordinate (x, y)
• Sketch the shape of the region R by sketching bounding curves and using common sense (You need
to know how to (i) find intercepts of lines with x, y axis, (ii) find center and radius of sphere, (ii)
find vertex and direction of parabola from their equations), determine intersection points of bounding
curves if necessary (by solving simultaneous equations, e.g. y = x2 and y = x appear as bounding
curve, solve x2 = x to get x = 0 or 1)
• Determine whether R is vertically or horizontally simple (e.g. vertically simple means contained in a
vertical strip), or both, or none. In case it is neither vertically or horizontally simple, cut it up into
pieces that are, call them R1 , R2 , . . .. Example:
• For R (or some Ri ) horizontally simple,
(note that the region R can be bounded by any number of bounding curves, but typically the number of
bounding curves for a vertically/horizontally simple region is either 4, or 3, or 2 where two of them will be the
upper and lower bounds for y integral) Find the bounds a, b and g1 (x), g2 (x) where y = gi (x) is usually
obtained from equations of bounding curves by solving for y.
• Now the integral reads:
Z
a
b
Z
g2 (x)
f (x, y)dydx
g1 (x)
and setting up for horizontally simple region is entirely similar, just replace ‘upper’ and ‘lower’ limits
by ‘right’ and ‘left’ limits given by some function x = h1 (y) and x = h2 (y). Finally if you have cut the
region, sum integral of each subregion.
Double integral, polar coordinate
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• Sketch the region R by sketching bounding curves which may occure in polar form (You need to be
very familiar with equations of horizontal/vertical lines and tangent circles in polar coordinates e.g.
and note a can also be negative. For unfamiliar curves r = r(θ), try a few special angles 0, π, 2π, π/2, 3π/2
or even π/4, 3π/4, . . . to sketch points and link them smoothly)
• If any bounding curve is not already in polar form, turn them into polar form by x = r cos θ, y = r sin θ.
Determine points of intersection of these curves if necessary. (e.g. intersection of r = 1 + cos θ and
r = 1/2 is cos θ = −1/2 and you’ll need to be able to know that this means θ = 2π/3, 4π/3)
• Most regions R you will meet in polar coordinate are ‘radially’ simple:
(Note r1 (θ) is often 0, also, the bounding curves can be circles, lines or other complicated shaps) Find the
bounds θ1 , θ2 (from polar coordinate of intersection points determined above), and r1 (θ), r2 (θ) possibly
by solving equation of bounding curves (in polar form) for r.
• Now the integral is
Z
θ2
θ1
Z
r2 (θ)
f (r cos θ, r sin θ)rdrdθ
r1 (θ)
Triple integral, rectangular coordinate
• (Note here we assume that the only type of solid region D is z-vertically simple, other solids may or
may not appear, I wish you are lucky)
• List all equations of bounding surfaces/planes (pay attention to phrases like ‘first octant’, they amounts
to more bounding planes!). Locate the ones with z’s, solve for z to get upper limit z = g1 (x, y) and
lower limit z = g2 (x, y), and solve for intersection curve g1 (x, y) = g2 (x, y).
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• Sketch the region R on XY-PLANE bounded by the equations involving only x and y, and the intersection curve
• Express the integral in terms of a nested integral:
ZZ Z
R
g1 (x,y)
f (x, y, z)dzdA
g2 (x,y)
• Now use steps in ‘Double integral, rectangular coordinate’ to express
RR
R
as iterated integral
Triple integral, cylindrical coordinates
• List equations of bounding surfaces, look for the ones involving z, solve for z to find upper limit
z = g1 (x, y) or z = h1 (r, θ) and lower limit z = g2 (x, y) or z = h2 (r, θ). Solve for intersection curve
h1 (r, θ) = h2 (r, θ) if necessary. (Note: you need to know for example first-octant amounts to three
more bounding surface given by θ = 0, θ = π/2, and z = 0)
• Sketch the region R on XY-PLANE bounded by the equations involving only x, y or r, θ, and the
intersection curve derived above. Transform all remaining x, y in bounding equations to r, θ
• Express the integral in terms of a nested integral:
ZZ Z
R
h1 (r,θ)
f (. . .)dz rdrdθ
h2 (r,θ)
and use (the remaining) steps in ‘Double integral, polar coordinate’ to set up the iterated integral
Triple integral, spherical coordinates
• (Note spherical coordinate is fundamentally different from cylindrical, and we can no longer ‘project’
to any plane to get a planar region. We assume the solids to use spherical coordinates are ‘radially’
simple, and essentially we project to the unit sphere)
• List equation of all bounding surfaces, translate them into spherical coordinates (It helps to be familiar
with the level surfaces of coordinate functions, e.g level surfaces with constant ρ, φ, θ are cocentric
spheres, conesp
and half-planes in general) Example: above the upper nappe of√z 2 = (x2 + y 2 )/3, means
1
above z = √3 x2 + y 2 , this translates into r cos φ = √13 r sin φ thus tan φ = 3 considering the range
0 ≤ φ ≤ π we have φ = π/3. Above this cone means between this cone and positive z-axis φ = 0, thus
the range of φ is 0 ∼ π/3.
• Locate equations containing ρ after simplifying, identify the ‘outer’ and ‘inner’ limits ρ = ρ1 (φ, θ) and
ρ = ρ2 (φ, θ). Note very often the ‘inner’ bound ρ2 (φ, θ) is just zero!
• Find the bounds on φ and θ (For this note that first octant means 0 ≤ φ, θ ≤ π/2, also note φ = 0, π/2, π
are positive z-axis, xy-plane, negative z-axis respectively.)
• Now we write our iterated integral. Usually the order of integration is as follows
Z
θ2
θ1
Z
φ2
φ1
Z
ρ1 (θ,φ)
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ2 sin φ dρdφdθ
ρ2 (θ,φ)
Example To illustrate the use of spherical coordinate, we evaluate the volume of a cylinder in a ‘stupid’
way: Find volume of the solid region bounded above by z = 1, below by xy-plane (z = 0), on the side by
x2 + y 2 = 1:
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First let’s translate the bounds into spherical coordinates:
1
cos φ
z = 0 ⇔ ρ cos φ = 0 ⇔ cos φ = 0 ⇔ φ = π/2
z = 1 ⇔ ρ cos φ = 1 ⇔ ρ =
x2 + y 2 = 1 ⇔ ρ2 sin2 φ = 1 ⇔ ρ sin φ = 1 ⇔ ρ =
1
sin φ
the last equation we used the fact that sin φ ≥ 0 since 0 ≤ φ ≤ π. Now only two equation still involve ρ but
turns out they both give ‘outer’ limit, and inner limit is 0. To see where the two different ‘outer’ limit meets,
we compute the intersection curve by setting r = cos1 φ = sin1 φ we get sin φ = cos φ so tan φ = 1 so φ = π/4
√
and intersection curve is given by r = 1/ 2 and φ = π/4 but only the last one is useful to us. Now the
bounds on φ are clear and there are no bound on θ, we can set up the iterated integral (and I’m evaluating
them to assure you that this gives the correct volume)
V =
Z
0
2π
Z
0
π/4
Z
1
cos φ
2
ρ sin φdρdφdθ +
0
= π/3 + 2π/3 = π
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Z
2π
0
Z
π/2
π/4
Z
1
sin φ
0
ρ2 sin φdρdφdθ