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18.02 Multivariable Calculus
Fall 2007
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18.02 Lecture 16. – Thu, Oct 18, 2007
Handouts: PS6 solutions, PS7.
Double integrals.
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Recall integral in 1-variable calculus: a f (x) dx = area below graph y = f (x) over [a, b].
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Now: double integral R f (x, y) dA = volume below graph z = f (x, y) over plane region R.
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Cut��R into small pieces ΔA ⇒ the volume is approximately f (xi , yi ) ΔAi . Limit as ΔA → 0
gives R f (x, y) dA. (picture shown)
How to compute the integral? By taking slices: S(x) = area of the slice by a plane parallel to
yz-plane (picture shown): then
� xmax �
volume = S(x) dx, and for given x, S(x) = f (x, y) dy.
xmin
In the inner integral, x is a fixed parameter, y is the integration variable. We get an iterated
integral.
Example 1: z = 1 − x2 − y 2 , region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (picture shown):
� 1� 1
(1 − x2 − y 2 ) dy dx.
0 0
(note: dA = dy dx, limit of ΔA = Δy Δx for small rectangles).
How to evaluate:
1 3 1
� 1 � �
2 2 2 1 2
1) inner integral (x is constant): (1 − x − y ) dy = (1 − x )y − y = (1 − x2 ) − = − x2 .
0 3 0 3 3
� 1 � �1
2 2 1 2 1 1
2) outer integral: ( − x2 ) dx = x − x3 = − = .
0 3 3 3 0 3 3 3
Example 2: same function over the quarter disc R : x2 + y 2 < 1 in the first quadrant.
How to find the bounds of √ integration? Fix x constant: what is a slice parallel to y-axis? bounds
for y = from y = 0 to y = 1 − x2 in the inner integral. For the outer integral: first slice is x = 0,
last slice is x = 1. So we get:
� 1 � √1−x2
(1 − x2 − y 2 ) dy dx.
0 0
(note the inner bounds depend on the outer variable x; the outer bounds are constants!)
2 3
�√1−x2 2
= (1 − x2 )3/2 .
�
Inner: (1 − x )y − y /3 0
3
� 1
2 π
Outer: (1 − x2 )3/2 dx = · · · = .
0 3 8
(. . . = trig. substitution x = sin θ, dx = cos θ dθ, (1 − x2 )3/2 = cos3 θ. Then use double angle
formulas... complicated! I carried out part of the calculation to show how it would be done but
then stopped before the end to save time; students may be confused about what happened exactly.)
Exchanging order of integration.
�1�2 �2�1
0 0 dx dy = 0 0 dy dx, since region is a rectangle (shown). In general, more complicated!
1
2
√
1� x
ey
�
Example 3: dy dx: inner integral has no formula. To exchange order:
0 x y
√
1) draw the region (here: x < y < x for 0 ≤ x ≤ 1 – picture drawn on blackboard).
2) figure out bounds in other direction: fixing a value of y, what are the bounds for x? here: left
border is x = y 2 , right is x = y; first slice is y = 0, last slice is y = 1, so we get
� 1� y y � 1 y � 1
e e
dx dy = 2
(y − y ) dy = ey − yey dy = [−yey + 2ey ]10 = e − 2.
0 y2 y 0 y 0
(the last integration can be done either by parts, or by starting from the guess −yey and adjusting;).
18.02 Lecture 17. – Fri, Oct 19, 2007
Integration in polar coordinates. (x = r cos θ, y = r sin θ): useful if either integrand or
region have a simpler expression in polar coordinates.
Area element: ΔA � (rΔθ) Δr (picture drawn of a small element with sides Δr and rΔθ).
Taking Δθ, Δr → 0, we get dA = r dr dθ.
�� � π/2 � 1
Example (same as last time): (1 − x2 − y 2 ) dx dy = (1 − r2 ) r dr dθ.
x2 +y 2 ≤1, x≥0, y≥0 0 0
� �1 � π/2
1 2 1 4 1 1 π1 π
Inner: r − r
= . Outer: dθ = = .
2 04 4 0 4 24 8
��
In general: when setting up f r dr dθ, find bounds as usual: given a fixed θ, find initial and
final values of r (sweep region by rays).
Applications.
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1) The area of the region R is R 1 dA. Also, the total mass of a planar object with density
δ = lim Δm/ΔA (mass per unit area, δ = δ(x, y) – if uniform material, constant) is given by:
ΔA=0
��
M= δ dA.
R
��
1
2) recall the average value of f over R is f¯ = f dA. The center of mass, or centroid,
Area R
of a plate with density δ is given by weighted average
�� ��
1 1
x̄ = x δ dA, ȳ = y δ dA
mass R mass R
3) moment of inertia: physical equivalent of mass for rotational motion. (mass = how hard
it is to impart translation motion; moment of inertia about some axis = same for rotation motion
around that axis)
Idea: kinetic energy for a single mass m at distance r rotating at angular speed ω = dθ/dt (so
velocity v = rω) is 12 mv 2 = 12 mr2 ω 2 ; I0 = mr2 is the moment of inertia.
��
For a solid with density δ, I0 = r2 δ dA (moment of inertia / origin). (the rotational energy
R
is 12 I0 ω 2 ).
3
��
Moment of inertia about an axis: I = (distance to axis)2 δ dA. E.g. about x-axis, distance
R
is |y|, so ��
Ix = y 2 δ dA.
R
Examples: 1) disk of radius a around its center (δ = 1):
� 2π � a � 4 �a
2 r πa4
I0 = r r dr dθ = 2π = .
0 0 4 0 2
2) same disk, about a point on the circumference?
Setup: place origin at point so integrand is easier; diameter along x-axis; then polar equation of
circle is r = 2a cos θ (explained on a picture). Thus
� π/2 � 2a cos θ
3
I0 = r2 r dr dθ = ... = πa4 .
−π/2 0 2
