MAT3100: Multivariable Calculus - Course Notes, Lecture notes of Mathematics

MAT 3100 – Multivariable Calculus Rationale Multivariable functions arise in many real world situations, where physical quantities often depend on two or more variables. This course takes calculus from the two dimensional world of single variable functions into the three dimensional world of multivariable functions which are required to understand and manipulate planes and surfaces, curves in two or three dimensions and scalar-valued and vector-valued functions of several variables.

Typology: Lecture notes

2021/2022

Uploaded on 01/13/2022

josphat-nyirongo
josphat-nyirongo 🇿🇲

4

(2)

5 documents

1 / 14

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MAT3100:MULTIVARIABLECALCULUS
MWALEDavid
2020
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

Partial preview of the text

Download MAT3100: Multivariable Calculus - Course Notes and more Lecture notes Mathematics in PDF only on Docsity!

MAT3100: MULTIVARIABLE CALCULUS

MWALE David

Contents

  • 1 FUNCTIONS OF SEVERAL VARIABLES
  • 2 Further Differential Calculus
  • 3 Surfaces
  • 4 Multiple Integration
    • 4.1 Double Integrals
    • 4.2 Triple Integrals
    • 4.3 Application of multiple integration
      • 4.3.1 Area rectangular coordinates
      • 4.3.4 Volume in Rectangular Coordinates
  • Bibliography

Chapter 2

Further Differential Calculus

Chapter 3

Surfaces

∫ (^) d c

{∫ (^) g 2 (y) g 1 (y)^ f(x; y)^ dx

dy: Remember that:

  1. We work from the inside integral.
  2. We integrate with respect to x if the limits are of x; (treating y as a constant).
  3. We integrate with respect to y if the limits are of y;(treating x as a con- stant).

Example 4.1.1. Evaluate

I =

1

2 (x^ + 2y)^ dx dy: Solution

I =

1

2 (x^ + 2y)^ dx

dy

1

[x 2 2 + 2yx

] 4

2 dy

1 (6 + 4y)dy = [(6y + 2y^2 )]^21 = 12:

Theorem 4.1.2. (Fubini’s Theorem for double integrals) If is continuous on the rectangle R = { (x; y) : a x b; c y d } ; then ∫ (^) b a

∫ (^) d c^ f(x; y)^ dA^ =

∫ (^) b a

∫ (^) d c^ f(x; y)^ dy dx^ =

∫ (^) d c

∫ (^) b a^ f(x; y)^ dx dy:

More generally, this is true if we assume that f is bounded on R; f is discon-

tinuous only on a finite number of smooth curves, and the iterated integrals exist.

Example 4.1.3. Evaluate

I =

1

∫ (^) π 0 (3 + sin^ θ)^ dθ dr: Solution

I =

1

∫ (^) π 0 (3 + sin^ θ)^ dθ dr =

1 [(3θ^ ^ cos^ θ)]

π 0 dr

=

1 (3π^ + 2)dr = [(3π + 2)r]^21 = 3π + 2:

Example 4.1.4. Confirm the Theorem above by showing that

I =

∫ (^) π 0

1 y^ sin(xy)^ dx dy^ =

1

∫ (^) π 0 y^ sin(xy)^ dy dx: Solution Home Work

4.2 Triple Integrals

Triple integrals are of the form ∫ (^) z 2 z 1

∫ (^) y 2 y 1

∫ (^) x 2 x 1 f(x; y)^ dx dy dz:

As in the previous section, we work from the inside out. All variables are regarded as constants for the time being, except for the one variable with respect to which the stage of integration is taking place.

In general, a double integral is denoted by ∫ ∫ Ω^ f(x; y)^ dA;

where Ω is the region over which the integral is to be evaluated and dA = dx dy is an area element. If f(x; y) = 1 for all (x; y) Ω; then this integral represents the area of the region Ω: That is (^) ∫ ∫

Ω^ dA^ =

Ω^ dx dy^ =^ area of^ Ω: If f(x; y) 0 for all (x; y) Ω; then the integral ∫ ∫ Ω^ f(x; y)dA^ =

Ω^ f(x; y)dx dy;

represents the volume of the solid bounded above by the surface z = f(x; y) and below by the region Ω: This leads us to the following discussion

4.3 Application of multiple integration

The following example will be useful to appreciate the notion of area as an application of double integration.

4.3.1 Area rectangular coordinates

Example 4.3.2. Find the area of the region bounded by y = 45 x ; the x-axis and the ordinate at x = 5: Solution We begin by sketching the required region. To find the area, we may, take vertical strips and use the usual formula for area;

Area = δy · δx:

But the total area of the strip is given by y ∑=y 1 y=

δy · δx:

Consequently, the sum of all the strips across the region is given by

A ∑^ x= x=

y ∑=y 1 y=

δy · δx

∑^ x= x=

y ∑=y 1 y=

δy · δx

It follows that if δx Ï 0 and δy Ï 0 ; we get

Then, as usual, if δx Ï 0 ; δy Ï 0 and δz Ï 0 :

V =

∫ (^) x 2 x 1

∫ (^) y 2 y 1

∫ (^) z 1 0 dx dy dz:

Example 4.3.5. A solid is enclosed by the plane z = 0; the planes x = 1; x = 4; y = 2; y = 5 and the surface z = x + y: Find the volume of the solid. Solution First of all, it is important to give a sketch of the desired solid. As in the description of volume above. It follows that

volume of element δx · δy · δz volume of column δx · δy

z= ∑x+y z=

δz

volume of slice δx ∑^ y= y=

δy

z= ∑x+y z=

δz

volume of total solid ∑^ y= x=

δx ∑^ y= y=

δy

z= ∑x+y z=

δz

Consequently,

V =

1 dx

2 dy

∫ (^) x+y 0 dz =

1 dx

2 (x^ +^ y)^ dy

1 dx

[

xy + y 2 2

] 5

2

1

3 x +^212

dx =

[ 3 x 2 2 +

21 x 2

] 4

1 = 54^ units

Bibliography

Prescribed Readings

  1. Grossman S.T., (1996) Multivariable Calculus, Linear Algebra and Differential Equations, 3rd ed. HBL Publishers. ISBN: 0 155 64751 2 Recommended Readings
  2. Stewart J., (2009) Calculus 6th ed. Brooks/Cole Cengage Learning, ISBN: 0 495 38362 8
  3. Wylie R.C. and Barret L.C. (1995). Advanced Engineering Mathematics. McGraw-Hill. Company. ISBN: 0 070 72206 4
  4. Kreyzig E. (2011). Advanced Engineering Mathematics. 10th Ed. John Willey and Sons. ISBN: 0 470 45836 5