Application of Integration: Probability (summary review)
A continuous random variable X has a probability densityfunction f (x), such that
P(a ≤ X ≤ b) =∫ ba f (x)dx .
In particular: f (x) ≥ 0,∫∞−∞ f (x)dx = 1
• the mean of f is µ =∫∞−∞ x f (x) dx
• the median of f is
the number m such that∫∞m f (x) dx = 1
2
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Application of Integration: Centre of Mass
Goal: compute the centre of mass of a lamina (thin, flat plate)
i.e. the point on which it balances horizontally.
For example:
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Centre of mass of 1D objects
First: what is the centre of mass of several point masses on a line?If mass mk sits at position xk :
x =∑n
k=1 mkxk∑nk=1 mk
= Mm =
moment (about x = 0)
total mass
Next: what is the centre of mass of a continuous 1D object (wire,rod) a ≤ x ≤ b with given linear density (mass/unit length) ρ(x)?
x =∫ ba xρ(x)dx∫ ba ρ(x)dx
= Mm =
moment (about x = 0)
total mass
Example: Find the centre of mass of a wire 0 ≤ x ≤ L with (linear)density ρ(x) = k x :
m =∫ R0 ρ(x)dx = k
∫ R0 xdx = k x2
2 |L0 = k
2L2
M =∫ R0 xρ(x)dx = k
∫ R0 x2dx = k x3
3 |L0 = k
3L3
x = Mm =
k3L23k2L2
= 23L
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Centre of mass of 2D laminaFirst: what is the centre of mass of several point masses in a plane?If mass mk sits at position (xk , yk): (x , y), where
x =∑n
k=1 mkxk∑nk=1 mk
=My
m , y =∑n
k=1 mkyk∑nk=1 mk
= Mxm
Next: what is the centre of mass (x , y) of a 2D lamina of constantdensity whose shape is the region below y = f (x), a ≤ x ≤ b?
x =My
m =ρ∫ ba xf (x)dx
ρ∫ ba f (x)dx
=∫ ba xf (x)dx∫ ba f (x)dx
= 1A
∫ ba xf (x)dx
y = Mxm =
ρ∫ ba
12[f (x)]2dx
ρ∫ ba f (x)dx
=∫ ba
12[f (x)]2dx∫ b
a f (x)dx= 1
A
∫ ba
12 [f (x)]2dx
The constant density cancels. We also call (x , y) the centroid.
Example: Find the centroid of {0 ≤ y ≤ 1− x2, 0 ≤ x ≤ 1}:M = A =
∫ 10 (1− x2)dx = (x − 1
3x3)|10 = 1− 1
3 = 23
My =∫ 10 x(1− x2)dx = (12x
2− 14x
4)|10 = 12 −
14 = 1
4 , x = 14/
23 = 3
8
Mx =∫ 10
12(1−x2)2dx = 1
2
∫ 10 (1−2x2+x4)dx = 4
15 , y = 4/152/3 = 2
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A little more on centroidsFor a region between two graphs, {g(x) ≤ y ≤ f (x), a ≤ x ≤ b}:
x = 1A
∫ ba x(f (x)− g(x))dx , y = 1
A
∫ ba
12([f (x)]2 − [g(x)]2)dx
Pappus’s Theorem: if a region of area A in the plane is rotatedabout a line L not intersecting it, the resulting volume isV = 2πrA , r = the distance from the region’s centroid to L.
Proof, for L = y−axis, and region {f (x) ≤ y ≤ g(x), a ≤ x ≤ b}:by “shells”, V = 2π
∫ ba x(g(x)− f (x))dx = 2πxA.
Example: Use Pappus to find the volume of a doughnut (torus).
A doughnut is obtained by rotating a disk of radius r about a linea distance R > r away from its centre. Pappus says:
V = 2πR(πr2) = 2π2Rr2 . (Fun: do this using “shells”.)
Challenge: Find the centre of mass of the unit upper half-disk if its(area) density (mass/unit area) is proportional to (a) the distancefrom the x-axis; (b) the distance from the y -axis.
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