(iii) ∆xi = xi−xi−1, ∆yj = yj−yj−1and kP k = max{ q ∆x2i + ∆yj2 | 1 ≤ i ≤ n , 1 ≤ j ≤ m} is defined to be the norm of the partition P

Definitions

(a) Let D be a bounded, closed and connected region in the xy−plane, let ρ : D → R be a continuous function on D and, for each p = (x, y) ∈ D, let ρ(p) be the density (in units of mass per unit area) at the point p. The total mass m of D is defined to be

m = Z Z

D

ρ(x, y) dA = lim

kP k→0 m

X

j=1 n

X

i=1

ρ(x^∗_ij, y_ij^∗) ∆x_i∆y_j,

where

(i) R = [a, b] × [c, d] is a rectangle containing D.

(ii) P = {Rij = [xi−1, xi] × [yj−1, yj] | 1 ≤ i ≤ n , 1 ≤ j ≤ m} is a partition of R, i.e. {xi}ⁿ_i=0 and {y_j}^m_j=0 are partitions of [a, b] and [c, d], respectively.

(iii) ∆x_i = x_i−x_i−1, ∆y_j = y_j−y_j−1and kP k = max{

q

∆x²_i + ∆y_j² | 1 ≤ i ≤ n , 1 ≤ j ≤ m}

is defined to be the norm of the partition P.

(iv) (x^∗_ij, y_ij^∗) ∈ [x_i−1, x_i] × [y_j−1, y_j] ∩ D for each 1 ≤ i ≤ n and 1 ≤ j ≤ m.

(b) The moment of D about the x−axis, denoted M_x, is defined by M_x =

Z Z

D

y ρ(x, y) dA = lim

kP k→0 m

X

j=1 n

X

i=1

y_ij^∗ ρ(x^∗_ij, y_ij^∗) ∆x_i∆y_j.

Note that |y| equals the distance between the point p = (x, y) and the x−axis.

(c) The moment of D about the y−axis, denoted M_y, is defined by

M_y = Z Z

D

x ρ(x, y) dA = lim

kP k→0 m

X

j=1 n

X

i=1

x^∗_ijρ(x^∗_ij, y_ij^∗) ∆x_i∆y_j.

Note that |x| equals the distance between the point p = (x, y) and the y−axis.

(d) The coordinate (¯x, ¯y) of the center of mass of D is defined by

where the total mass m of D is given by m =

Z Z

D

ρ(x, y) dA.

(e) The moment of inertia of D about the x−axis, denoted I_x, is defined by

I_x = Z Z

D

y²ρ(x, y) dA = lim

kP k→0 m

X

j=1 n

X

i=1

(y_ij^∗)²ρ(x^∗_ij, y_ij^∗) ∆x_i∆y_j.

Note that |y| equals the distance between the point p = (x, y) and the x−axis.

(f) The moment inertia of D about the y−axis, denoted I_y, is defined by

I_y = Z Z

D

x²ρ(x, y) dA = lim

kP k→0 m

X

j=1 n

X

i=1

(x^∗_ij)²ρ(x^∗_ij, y_ij^∗) ∆x_i∆y_j.

Note that |x| equals the distance between the point p = (x, y) and the y−axis.

(g) A function f is called a probability density function (pdf) of a continuous random variable x if

(i) f (x) ≥ 0 for all x.

(ii) Z ∞

−∞

f (x) dx = 1.

(iii) the probability that x lies between a and b is given by P (a ≤ x ≤ b) =

Z b a

f (x) dx.

(h) A function f is called a (joint) probability density function (pdf) of continuous random variables x and y if

(i) f (x, y) ≥ 0 for all x, y.

(ii) Z ∞

−∞

Z ∞

−∞

f (x, y) dxdy = 1.

(iii) the probability that (x, y) lies in D is given by P ((x, y) ∈ D) =

Z Z

D

f (x, y) dA.

(i) Let f be the probability density function (pdf) of the random variable x. Thenthe expected value µ of x is defined to be

µ = Z ∞

−∞

xf (x) dx = R∞

−∞ xf (x) dx R∞

−∞ f (x) dx =the mean value of x.

(j) Let f be the (joint) probability density function (pdf) of random variables x and y. Then the expected value µx of x and the expected value µy of y, also called the x−mean and y−mean respectively, are defined by

µ_x= Z Z

R²

xf (x, y) dA and µ_y = Z Z

R²

yf (x, y) dA.

Remarks

(a) A single random variable isnormally distributed if its probability density function is of the form

f (x) = 1 σ√

2πe^{−(x−µ)2/(2σ2)},

where µ is the mean or expectation of the distribution (and also its median and mode), σ is the standard deviation and σ² is the variance.

(b) Let I = Z ∞

−∞

e^−x2dx. Since

I² =

Z ∞

−∞

e^−x2dx
Z ∞

−∞

e^−y2dy

= Z ∞

−∞

Z ∞

−∞

e^−x2−y2dx dy

= Z 2π

0

Z ∞ 0

e^−r2r dr dθ where we have used x = r cos θ, y = r sin θ and dx dy = rdr dθ

= Z 2π

0

−e^−r2 2 |^∞₀ dθ

= π,

we have shown that

Z ∞

−∞

e^−x2dx =√

π. (1)

(c) Setting

y = x − µ σ√

2 =⇒ dx = σ√ 2dy, we find

1 σ√

2π Z ∞

−∞

e^{−(x−µ)2/(2σ2)}dx = 1

√π Z ∞

−∞

e^−y2dy =

√π

√π = 1, (2)

1 Z ∞

1 Z ∞ √

which implies that

1 σ√

2π Z ∞

−∞

(x − µ) e^{−(x−µ)2/(2σ2)}dx = 0, (4) and

1 σ√

2π Z ∞

−∞

(x − µ)²e^{−(x−µ)2/(2σ2)}dx

= − σ

√2π Z ∞

−∞

(x − µ) · −2(x − µ)

2σ² · e^{−(x−µ)2/(2σ2)}

 dx

= − σ

√2π Z ∞

−∞

(x − µ) · d

dxe^{−(x−µ)2/(2σ2)}dx

= − σ

√2π(x − µ) · e^{−(x−µ)2/(2σ2)}|_−∞^∞ + σ² σ√

2π Z ∞

−∞

e^{−(x−µ)2/(2σ2)}dx

= σ². (5)

By combining (3), (4) and (5), we get 1

σ√ 2π

Z ∞

−∞

x²e^{−(x−µ)2/(2σ2)}dx

= 1

σ√ 2π

Z ∞

−∞

(x − µ) + µ²e^{−(x−µ)2/(2σ2)}dx

= 1

σ√ 2π

Z ∞

−∞

(x − µ)²+ 2µ(x − µ) + µ² e^{−(x−µ)2/(2σ2)}dx

= σ²+ µ². (6)

(d) Let f : R × R+→ R+ be defined by f (x, t) = 1

√4πte^−x2/4t for all x ∈ R and t > 0.

Then (i) 1

√4πt Z ∞

−∞

e^−x2/4tdx = 1 for all t > 0.

(ii) f_xx− f_t= 1

√4πte^−x2/4t



−2

4t + 4x² 16t² + 1

2t − x² 4t²



= 0 for all x ∈ R and t > 0, i.e. f is a solution of the heat equation in R × R+.

Definition Let U and E be smooth bounded, closed regions in Rⁿ. A map T : U → E is said to be an onto C¹ transformation with nonzero Jacobian if

(a) for each x = (x₁, x₂, . . . , x_n) ∈ E, there exists a u = (u₁, u₂, . . . , u_n) ∈ U such that (x₁, x₂, . . . , x_n) = x = T (u) = (T₁(u), T₂(u), . . . , T_n(u)),

(b) for each 1 ≤ i ≤ n, the function T_i : U → R is continuously differentiable on U,

(c) for each u = (u₁, u₂, . . . , u_n) ∈ U, the Jacobian J_T(u) of T at u is nonzero, i.e.

J_T(u) = ∂(x₁, x₂, . . . , x_n)

∂(u₁, u₂, . . . , u_n) =             

∂x1

∂u₁

∂x1

∂u₂ · · · ∂x1

∂u_n

∂x₂

∂u1

∂x₂

∂u2

· · · ∂x₂

∂un

· · · ·

∂x_n

∂u₁

∂x_n

∂u₂ · · · ∂x_n

∂u_n             

= det



















∂x1

∂u₁

∂x1

∂u₂ · · · ∂x1

∂u_n

∂x₂

∂u1

∂x₂

∂u2

· · · ∂x₂

∂un

· · · ·

∂x_n

∂u₁

∂x_n

∂u₂ · · · ∂x_n

∂u_n

















 6= 0

Change of Variables in Multiple Integrals Let U and E be smooth bounded, closed regions in Rⁿ, let T : U → E be an onto C¹ transformation with nonzero Jacobian. Suppose that f : E → R is a continuous function on E and T is 1 − 1 except perhaps on the boundary of U.

Then

Z

E

f (x₁, x₂, . . . , x_n) dV = Z

U

f ◦ T (u₁, u₂, . . . , u_n) |J_T(u)| du₁du₂· · · du_n

= Z

U

f (x₁(u), x₂(u), . . . , x_n(u))    

∂(x₁, x₂, . . . , x_n)

∂(u₁, u₂, . . . , u_n)    

du₁du₂· · · du_n.

Remark If n = 2 and T : S → R is an 1−1 onto transformation from a region S in the uv−plane onto R in the xy−plane.

We can approximate the image region R = T (S) by a parallelogram determined by the secant vectors

a = r(u0 + ∆u, v0) − r(u0, v0) b = r(u0, v0+ ∆v) − r(u0, v0) shown in the following

But

ru = lim

∆u→0

r(u₀+ ∆u, v₀) − r(u₀, v₀)

∆u .

and so

r(u₀+ ∆u, v₀) − r(u₀, v₀) ≈ ∆u r_u. Similarly

r(u₀, v₀+ ∆v) − r(u₀, v₀) ≈ ∆v r_v.

This means that we can approximate R by a parallelogram determined by the vectors ∆u r_u and

∆v r_v shown in the following

|(∆u r_u) × (∆v r_v)| = |r_u × r_v| ∆u ∆v Computing the cross product, we obtain

r_u× r_v =         

i j k

∂x

∂u

∂x

∂v 0

∂y

∂u

∂y

∂v 0         

=       

∂x

∂u

∂y

∂x ∂u

∂v

∂y

∂v       

k =       

∂x

∂u

∂x

∂y ∂v

∂u

∂y

∂v       

k

The determinant that arises in this calculation is called the Jacobian of the transformation and is given a special notation.

Definition Let T : S → R be a transformation defined by

(x, y) = T (u, v) = (g(u, v), h(u, v)) ∀ (u, v) ∈ S.

Then the Jacobian of the transformation T is

∂(x, y)

∂(u, v) =       

∂x

∂u

∂x

∂y ∂v

∂u

∂y

∂v       

= ∂x

∂u

∂y

∂v − ∂x

∂v

∂y

∂u

Examples (a) Evaluate

Z Z

E

e

x−y

x+y dA, where E = {(x, y) | x ≥ 0, y ≥ 0, x + y ≤ 1}.

y

E x

1 1

0

x + y = 1

x = ¹₂(u + v) y = ¹₂(v − u)

v

1 u

−1

1

u = v u = −v

0 S

Solution Let u = x − y, v = x + y. Then

(x, y) ∈ E ⇐⇒ u + v ≥ 0, v − u ≥ 0 and 0 ≤ v ≤ 1.

Let

S = {(u, v) | u + v ≥ 0, v − u ≥ 0, 0 ≤ v ≤ 1} = {(u, v) | u ≥ −v, v ≥ u, 0 ≤ v ≤ 1}

and T : S → E be defined by

(x, y) = T (u, v) = 1

2(u + v) , 1

2(v − u)

∀ (u, v) ∈ S.

Then

J_T(u, v) =    

∂x

∂u

∂x

∂y ∂v

∂u

∂y

∂v

=    

1 2

1 2

−¹₂ ¹₂    

= 1

2 =⇒ |J_T(u, v)| = 1 2, and

Z Z

E

e

x−y

x+y dA = Z Z

S

e

x(u,v)−y(u,v)

x(u,v)+y(u,v) |J_T(u, v)| dA

= Z 1

0

Z v

−v

e^u/v1 2du dv

= Z 1

0

v 2e^u/v

|^u=v_u=−vdv

= Z 1

0

v 2

 e − 1

e

 dv

= v² 4

 e − 1

e



|¹₀

= 1 4

 e − 1

e

 .

(b) Find the volume V inside the paraboloid z = x² + y² for 0 ≤ z ≤ 1.

Solution Using vertical slices, we see that V =

Z Z

E

(1 − z) dA = Z Z

E

1 − (x²+ y²)
dA, where E = {(x, y) | 0 ≤ x²+ y² ≤ 1}.

Let S = {(r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π} and T : S → E be defined by (x, y) = T (r, θ) = r cos θ , r sin θ
∀ (r, θ) ∈ S.

Then

J_T(r, θ) =    

∂x

∂r

∂x

∂y ∂θ

∂r

∂y

∂θ

=    

cos θ −r sin θ sin θ r cos θ    

= r =⇒ |J_T(r, θ)| = r, and

V =

Z 2π 0

Z 1 0

(1 − r²) |J_T(r, θ)| dr dθ

= Z 2π

0

Z 1 0

(r − r³) dr dθ

= Z 2π

0

 r² 2 − r⁴

4



|¹₀dθ

= Z 2π

0

1 4dθ

= π

2. (c) Find the volume V inside the cone z =p

x²+ y² for 0 ≤ z ≤ 1.

Solution Using vertical slices, we see that V =

Z Z

E

(1 − z) dA = Z Z

E

1 −p

x²+ y²)big) dA, where E = {(x, y) | 0 ≤ x²+ y² ≤ 1}.

Let S = {(r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π} and T : S → E be defined by (x, y) = T (r, θ) = r cos θ , r sin θ
∀ (r, θ) ∈ S.

Then

V =

Z 2π 0

Z 1 0

(1 − r) |J_T(r, θ)| dr dθ

= Z 2π

0

Z 1 0

(r − r²) dr dθ

= Z 2π

0

 r² 2 − r³

3



|¹₀dθ

= Z 2π

0

1 6dθ

= π

3.

(d) For a > 0, find the volume V inside the sphere x² + y² + z² = a².

Solution Let S = {(ρ, φ, θ) | 0 ≤ ρ ≤ a, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π}, E = {(x, y, z) | 0 ≤ x²+ y²+ z² ≤ a²} and T : S → E be defined by

(x, y, z) = T (ρ, φ, θ) = ρ sin φ cos θ , ρ sin φ sin θ , ρ cos φ
∀ (ρ, φ, θ) ∈ S.

Then

J_T(ρ, φ, θ) =       

∂x

∂ρ

∂x

∂φ

∂x

∂y ∂θ

∂ρ

∂y

∂φ

∂y

∂θ

∂z

∂ρ

∂z

∂φ

∂z

∂θ

=      

sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ

cos φ −ρ sin φ 0

= cos φρ²sin φ cos φ(cos²θ + sin²θ) + ρ sin φρ sin²φ(cos²θ + sin²θ)

= ρ²sin φ(cos²φ + sin²φ)

= ρ²sin φ,

and

V =

Z Z Z

S

dV

= Z 2π

0

Z π 0

Z a 0

|J_T(ρ, φ, θ)| dρ dφ dθ

= Z 2π

0

Z π 0

Z a 0

ρ²sin φ dρ dφ dθ

= Z 2π

0

Z π 0

 ρ³ 3



|^a₀ sin φ dφ dθ

= Z 2π

0

a³

3 (− cos φ) |^π₀ dθ

= Z 2π

0

2a³ 3 dθ

= 2a³θ 3 |^2π₀

= 4πa³ 3 .