Mathematics. Multivariable Calculus

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Mathematics. Multivariable Calculus

Artur Siemaszko

Faculty of Mathematics and Computer Science University of Warmia and Mazury in Olsztyn

April 17, 2013

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The Definite Integral

Example

Estimate population = 18 000 foxes

Lower estimate = 4 000; Upper estimate = 35 000

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Example

Estimate population = 18 000 foxes

Lower estimate = 4 000; Upper estimate = 35 000

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∆x_i =x_i− x_i−1, ∆y_j =y_j− y_j−1

L_ij =inf{f (x , y ) : x ∈ [x_i−1,x_i),y ∈ [y_j−1,y_j)}

M_ij =sup{f (x , y ) : x ∈ [x_i−1,x_i),y ∈ [y_j−1,y_j)}

Definition

Suppose the function f is continuous on R = [a, b] × [c, d ]. The definite integral (double integral) of f over R is the quantity

Z

R

f dA = lim

∆x →0, ∆y →0

X

i,j

f (u_ij,v_ij)∆x_i∆y_j = Z

R

f (x , y ) dxdy .

Definition

Suppose the function f is bounded on R. Thedefinite integral (double integral) of f over R is the quantity

Z

R

f dA = sup

P

X

i,j

L_ij∆x_i∆y_j =inf

P

X

i,j

M_ij∆x_i∆y_j

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∆x_i =x_i− x_i−1, ∆y_j =y_j− y_j−1

Definition

Z

R

f dA = lim

∆x →0, ∆y →0

X

i,j

f (u_ij,v_ij)∆x_i∆y_j

=

R

f (x , y ) dxdy .

Definition

Z

R

f dA = sup

P

X

i,j

P

X

i,j

M_ij∆x_i∆y_j

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∆x_i =x_i− x_i−1, ∆y_j =y_j− y_j−1

Definition

Z

R

f dA = lim

∆x →0, ∆y →0

X

i,j

R

f (x , y ) dxdy .

Definition

Z

R

f dA = sup

P

X

i,j

P

X

i,j

M_ij∆x_i∆y_j

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∆x_i =x_i− x_i−1, ∆y_j =y_j− y_j−1

Definition

Z

R

f dA = lim

∆x →0, ∆y →0

X

i,j

R

f (x , y ) dxdy .

Definition

Z

R

f dA = sup

P

X

i,j

P

X

i,j

M_ij∆x_i∆y_j

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Interpretation if the Double Integral as Volume

Fact

If x , y , z, represent length and f is non-negative, then Z

R

f dA = volume under graph of f above region R.

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Interpretation if the Double Integral as Volume

Fact

If x , y , z, represent length and f is non-negative, then Z

R

f dA = volume under graph of f above region R.

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Interpretation if the Double Integral as Average Value

Fact 1 Area of R

Z

R

f dA = Average value of f on the region R.

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Interpretation if the Double Integral as Average Value

Fact 1 Area of R

Z

R

f dA = Average value of f on the region R.

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Exercise 7.1

Decide (without calculation) whether the integrals are positive, negative, or zero. Let D be the region inside the unite circle centered at the origin, let R be the right half of D and let B be the bottom half of D.

1. R

D dA;

2. R

B dA;

3. R

R5x dA;

4. R

B5x dA;

5. R

D5x dA;

6. R

D(y³+y⁵)dA;

7. R

B(y³+y⁵)dA;

8. R

R(y³+y⁵)dA;

9. R

B(y − y³)dA;

10. R

D(y − y³)dA;

11. R

Dsin y dA;

12. R

Dcos y dA;

13. R

De^xdA;

14. R

Dxe^xdA;

15. R

Dxy²dA;

16. R

Bx cos y dA.

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Exercise 7.2

Let f (x , y ) be a function of x and y which is independent of y , that is, f (x , y ) = g(x ) for some one variable function g.

(a) What does the graph of f look like?

(b) Let R be the rectangle a ≤ x ≤ b, c ≤ y ≤ d . By interpreting the integral as a volume, and using your answer to part (a), expressR

Rf dA in terms of a one-variable integral.

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Exercise 7.3

It is known that

|a + b| ≤ |a| + |b|, for any numbers a and b.

Use this to explain why Z

R

f dA

≤ Z

R

|f | dA.

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Exercise 7.3

It is known that

|a + b| ≤ |a| + |b|, for any numbers a and b.

Use this to explain why Z

R

f dA

≤ Z

R

|f | dA.

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Iterated Integral

Fact

If R is the rectangle a ≤ x ≤ b, c ≤ y ≤ d and f is a continuous function on R, then the double integral of f over R is equal to theiterated integral

Z

R

f dA = Z d

c

Z b a

f (x , y )dx

! dy

= Z b

a

Z d c

f (x , y )dy

! dx .

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Iterated Integral

Fact

If R is the rectangle a ≤ x ≤ b, c ≤ y ≤ d and f is a continuous function on R, then the double integral of f over R is equal to theiterated integral

Z

R

f dA = Z d

c

Z b a

f (x , y )dx

! dy =

Z b a

Z d c

f (x , y )dy

! dx .

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Exercise 7.4

A building is 8 miters wide and 16 meters long. It has a flat roof that is 12 meters high at one corner, and 10 meters high at each of the adjacent corners. What is the volume of the building?

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Exercise 7.4

A building is 8 miters wide and 16 meters long. It has a flat roof that is 12 meters high at one corner, and 10 meters high at each of the adjacent corners. What is the volume of the building?

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx

dy =R16

0 (88 − y )dy = 1280 V =R8

0

R16

0 12 −^x₄−^y₈ dy

dx =R8

0(176 − 4x )dx = 1280

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx dy

0

V =R8 0

R16

0 12 −^x₄−^y₈ dy

dx =R8

0(176 − 4x )dx = 1280

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx

dy =R16

0 (88 − y )dy

=1280 V =R8

0

R16

0 12 −^x₄−^y₈ dy

dx =R8

0(176 − 4x )dx = 1280

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx

dy =R16

0 (88 − y )dy = 1280

0 0 4 8 0

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx

dy =R16

0 (88 − y )dy = 1280 V =R8

0

R16

0 12 −^x₄−^y₈ dy dx

=R8

0(176 − 4x )dx = 1280

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx

dy =R16

0 (88 − y )dy = 1280 V =R8

0

R16

0 12 −^x₄−^y₈ dy

dx =R8

0(176 − 4x )dx

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Solution of Exercise 7.4

f (x , y ) = 12 −^x₄−^y₈

V =R16 0

R8

0 12 −^x₄−^y₈ dx

dy =R16

0 (88 − y )dy = 1280 V =R8

0

R16

0 12 −^x₄−^y₈ dy

dx =R8

0(176 − 4x )dx = 1280

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Iterated Integrals Over Non-Rectangular Regions

The density at the point (x , y ) of a triangular metal plate, as shown below, is δ(x , y ).

Express its mass as an iterated integral.

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Iterated Integrals Over Non-Rectangular Regions

The density at the point (x , y ) of a triangular metal plate, as shown below, is δ(x , y ).

Express its mass as an iterated integral.

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Mass = Z 1

0

Z 2−2x 0

δ(x , y )dydx

=

0 0

δ(x , y )dxdy .

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Mass = Z 1

0

Z 2−2x 0

δ(x , y )dydx = Z 2

0

Z 1−y /2 0

δ(x , y )dxdy .

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Exercise 7.5

Find the mass M of a metal plate R bounded by y = x and y = x², with density given by δ(x , y ) = 1 + xy kg/meter². Solution:

M =

0 x²

(1 + xy )dydx = 14kg; M =

Z 1 0

Z ^√y y

(1 + xy )dxdy = 5 14kg;

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Exercise 7.5

M = Z 1

0

Z x x²

(1 + xy )dydx

= 5 14kg; M =

Z 1 0

Z ^√y y

(1 + xy )dxdy = 5 14kg;

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Exercise 7.5

M = Z 1

0

Z x x²

(1 + xy )dydx = 5 14kg;

M =

0 y

(1 + xy )dxdy = 5 14kg;

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Exercise 7.5

M = Z 1

0

Z x x²

(1 + xy )dydx = 5 14kg;

M = Z 1

0

Z ^√y y

(1 + xy )dxdy

= 5 14kg;

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Exercise 7.5

M = Z 1

0

Z x x²

(1 + xy )dydx = 5 14kg;

M = Z 1

0

Z ^√y y

(1 + xy )dxdy = 5 14kg;

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Exercise 7.6

A city occupies a semicircular region of radius 3 km bordering on the ocean. Find the average distance d from points in the city to the ocean.

Solution:

d = 1

Area(R) Z

R

ydA. Z

R

ydA = Z 3

−3

Z

√

9−x² 0

ydy

!

dx = 18, d = 4/π

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Exercise 7.6

Solution:

d = 1

Area(R) Z

R

ydA.

Z

R

ydA = Z 3

−3

Z 9−x 0

ydy dx = 18, d = 4/π

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Exercise 7.6

Solution:

d = 1

Area(R) Z

R

ydA.

Z

R

ydA = Z 3

−3

Z

√

9−x² 0

ydy

! dx

=18, d = 4/π

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Exercise 7.6

Solution:

d = 1

Area(R) Z

R

ydA.

Z

R

ydA = Z 3

−3

Z

√

9−x² 0

ydy

!

dx = 18,

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Exercise 7.6

Solution:

d = 1

Area(R) Z

R

ydA.

Z

R

ydA = Z 3

−3

Z

√

9−x² 0

ydy

!

dx = 18, d = 4/π

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Polar Coordinates

When computing integrals in polar coordinates use







x = r cos θ y = r sin θ

,

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Polar Coordinates







, x²+y²=r²

, dA = rdrd θ.

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Polar Coordinates







, x²+y²=r², dA = rdrd θ.

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Exercise 7.7

Compute the integral of f (x , y ) = 1/(x²+y²)^3/2over the region R shown below.

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2=1/(r²)^3/2=1/r³. Z

R

fdA = Z π/4

0

Z 2 1

1

r³rdrd θ = Z π/4

0

Z 2 1

1

r²drd θ = π 8.

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Exercise 7.7

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2

Z

R

fdA = Z π/4

0

Z 2 1

1

r³rdrd θ = Z π/4

0

Z 2 1

1

r²drd θ = π 8.

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Exercise 7.7

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2=1/(r²)^3/2

=1/r³. Z

R

fdA = Z π/4

0

Z 2 1

1

r³rdrd θ = Z π/4

0

Z 2 1

1

r²drd θ = π 8.

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Exercise 7.7

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2=1/(r²)^3/2=1/r³.

R

fdA =

0 1 r³rdrd θ =

0 1 r²drd θ = 8.

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Exercise 7.7

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2=1/(r²)^3/2=1/r³. Z

R

fdA = Z π/4

0

Z 2 1

1 r³rdrd θ

= Z π/4

0

Z 2 1

1

r²drd θ = π 8.

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Exercise 7.7

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2=1/(r²)^3/2=1/r³. Z

R

fdA = Z π/4

0

Z 2 1

1

r³rdrd θ = Z π/4

0

Z 2 1

1 r²drd θ

= 8.

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Exercise 7.7

Solution:

f (x (r , θ), y (r , θ)) = 1/(x²+y²)^3/2=1/(r²)^3/2=1/r³. Z

R

fdA = Z π/4

0

Z 2 1

1

r³rdrd θ = Z π/4

0

Z 2 1

1

r²drd θ = π 8.

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Cylindrical Coordinates

When computing integrals in cylindrical coordinates use











x = r cos θ y = r sin θ z = z

,

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Cylindrical Coordinates











, x²+y²=r²

, dV = rdrd θdz.

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Cylindrical Coordinates











, x²+y²=r², dV = rdrd θdz.

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Spherical coordinates











x = ρ sin φ cos θ y = ρ sin φ cos θ z = ρcos φ

, x²+y²+z²= ρ².

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Spherical coordinates











,

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Spherical coordinates











, x²+y²+z²= ρ².

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Volume element in spherical coordinates

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Volume element in spherical coordinates

dV = ρ²sin φd ρd φd θ