Merced College; Don Power

 

MULTIVARIATE CALCULUS - CH 15, LECTURE

 

MULTIPLE INTEGRALS

 

15.1  Double Integrals

 

Classic volume problem:  Given a surface z=f(x,y), that is continuous and non-negative on a region R in the xy-plane, find the volume of the solid located between the surface and the region.

 

For lesson 14.1:  rectangular regions, a£x£b, c£y£d

 

Meaning of double integral:

      ∫∫ f(x,y) dxdy means

lim(as Δx and Δy approach 0) of ΣΣf(xi,yj)ΔxΔy, which is a compact way of writing

limit of Σf(x,y1)ΔxΔy + Σf(x,y2)ΔxΔy + ...+ Σf(x,yn)ΔxΔy  (each sum is for x1 + x2 + ... + xm)

 

Table example:  col totals, then row totals from rectangular table.

      Δx = 1, Δy = 2; row totals 2+5+14+29=50 etc.; table totals 50,66,82,98 (bottom to top);

V is approximately 296 * ΔxΔy  = 296*2 = 592

     

f(x,y)=3x^2+2y

 

 

Note:  this actually estimates the volume over the region
0 £ x £ 4, 1 £ y £ 9 (why 4 and 9?)
 

  y     \  x

0

1

2

3

7

14

17

26

41

5

10

13

22

37

3

6

9

18

33

1

2

5

14

29

 

Iterated Integrals:  Meaning of  Int(1,4)Int(2,5)x3y dydx

      What does region represent?

      How to calculate

 

Fubini:  dA = dxdy = dydx; condition: continuity or finite number of discontinuities)

                  Ex setup prev as dA, then in reverse order

                 

     

15.2  Double Integrals over Nonrectangular Regions

 

Double Int over general regions

      Type 1:  inner:  y between functions of x, outer:  x between constants;  cf Type II regions

      Tasks:  set up integral & eval;

                  describe region given by an integral

      Crucial question for "region btwn 2 curves" ...where do they intersect

 

      Tasks: 

Set up integral & eval;   X11a

                  Describe region given by an integral       X6

                  Reverse order of integr       X50

 

Volume vs area (for area, integrand = 1)

 

Special case that makes simplification easy:  If integrand can be factored into separate functions purely of x and purely of y, and if the limits of integration are constant, then you can separate the integral into two single integrals: 

 

 

15.3  Double Integrals in Polar Coordinates

 

Description of a "simple polar region"

 

Increment of area: dy dx is replaced by dr ds, where s is arc length,

but since s=rq, (hence ds = r dq), we use dA = r dr dq

 

Thm 3 gives setup for double integral over a simple polar region

      Toughest task:  determine the θ-limits of integration

 

Examples:

X4       basic area computation:  What is the region?  Note that we get single integral of 1/2 r2dθ along the way (technique from last semester for polar areas)

X8       area inside rose curve

X10     inside circle of radius 2 and to right of a line x=1

X14     set up the integral -- with a picture

X24     with an integrand

X37     setup, without a picture.

 

15.4  Parametric Surfaces; Surface Area

 

For a review of parametric equations, see under MATH-04B, Lecture Notes for Chapter 11

 

Fact:     A vector function of one variable is a line or curve in space

            A vector function of two variables is a surface

      Ex:  If z = f(x,y), let x=u, y=v, then z=f(u,v)

            a.  Find a parametric rep of z=x2+y2

            b.  Sketch the parametric surface x=u2+v2, y=u,z=v

Cylinders:  For right circular cylinders, try polar representation

      X5b

Surfaces of revolution parametrically

      If about x-axis, x=u, y=f(u)cos(v), z=f(u) sin(v)

      X8 about both x-axis and y-axis

Parameterizations using

      Cyl coords:  Ex 2

      Sph coords:  Ex 3, X15

 

S = doubleint(||ru ´ rv||)dA  for parameterized surface r(u,v)=<x(u,v),y(u,v),z(u,v)>

 

S = doubleint(sqrt(1+zx2+zy2)dA for function z=f(x,y)

 

 

 

Ex:  Find SA for f(x,y)=xy, R = {(x,y): x^2+y^2£16}:  describe region, set up in rect, then polar; eval.

      int(  int(  sqrt(1+y^2+x^2),  y,-sqrt(16-x^2),sqrt(16-x^2)),  x,-4,4)

      int(  int(  sqrt(1+r^2)*r,  r,0,4),  q,0,2p)

 

Ex:  Find SA for f(x,y) = 2x+y^2,  R = triangle with vertices (0,0), (2,0), (2,2)

      int(  int(  sqrt(5+4x^2),  y,0,x)  x,0,2)

 

15.5  Triple Integrals

 

Over rectangular regions

      X4

      Same properties as for double integrals

      For rectangular regions, easy to change order of integration

More general regions

      See text's def of "simple xy-solid" bounded above and below by two functions z=gi(x,y)

      Key is then to determine the region in the xy-plane

            X10

 

15.6  Centroid, Center of Gravity, Theorem of Pappus

 

Basic principle:  total moment = sum(mass*distance);

 center of mass:  point for which total mass = 0, or

                              total moment / total mass

 

1.  Density: delta(x,y) or delta (x,y,z)-- variable within a region, which is

"lamina" when working with 2 variables x,y

solid when working with 3 variables x,y,z

 

2.   Mass:

m = double int, where integrand = delta(x,y).

m = triple int, where integrand = delta(x,y,z).

 

3.   Moments and centers of mass:  Motivation:  moment = mass * distance (teeter-totter or lever)

      Moments are Mx and My [and Mz], with center of mass at xbar = Mx/m etc.

Ex:  Region in the 1st quadrant bounded by y=x2 and y=1, delta=xy.  Sol: m=1/6, xbar=4/7, ybar=3/4.

 

4.   "density is proportional to..."  use constant of variation 'k' "  except we can let k=1 for centers of mass (it divides out)

For centroid (center of the area), let delta = 1

 

5.   Moment of inertia (second moments)  Ix contains y2, (whereas Mx just had y)  etc

      Physical explanation pg 1079, before X43

 

4.  Theorem of Pappus:  Volume of a solid of revolution = area of plane region being revolved * distance traveled by its centroid.

 

5.  Probability:   pdf, joint density func -- not in Anton

 

6.  Expected values -- not in Anton

 

 

15.7  Triple Integrals in Cylindrical and Spherical Coordinates

 

recall:

      x = r cos q = ρ sin φ cos q

      y = r sin q = ρ sin φ sin q

      z = ρ cos φ

 

Increment of volume for cylindrical coordinates (same adjustment as for polar coords in 2D)

dy dx is replaced by dr ds, where s is arc length,

but since s=rq, (hence ds = r dq), we use dA = r dr dq

 

Increment of volume for spherical coordinates:

      In the xy-plane, we retain the ds = r dq, except that we replace r = ρ sinφ

      In the "rz"-plane, we have ds = ρ dφ

      Increment of distance from the origin is dρ

      Thus, dV = ρ2sinφ dρ dφ dq

 

More formal treatment of the change of variables later (in lesson on Jacobians)

 

Examples

X13  translation to cyl coords

X15 transl to sph coords

X26 centroid (cyl); use symmetry and geom formulas where possible

X27 centroid (sph)

 

15.8  Change of Variables in Multiple Integrals; Jacobians

 

Conversion from a coord system in terms of u,v,w to one in terms of x,y,z, where x = x(u,v,w), y = y(u,v,w), z = z(u,v,w): 

 

J(u,v,w) =

 

Applications of Jacobians:

 

1.  Calculate integrating factors for conversions from rectangular to polar, cylindrical, spherical coordinates.

 

2.  Deal with problem integrands
Ex:  (#33)  Doubleint ( sin(x-y) / cos(x+y) ) dA over region bounded by y=0, y=x, x+y=p/4.
With u=x-y, v=x+y, solve for x and y:  x = 1/2 (u+v), y = 1/2 (-u+v)
Jacobian is 1/2

      Boundaries of region are v=p/4, u=0, u=v
Doubleint (1/2 (sin u / cos v), u=0..v, v=0..p/4) Note: the other order of integration won't work.

      Result: 1/2 [ ln(1 + sqrt 2) - p/4]

 

3.  Deal with problem regions
Ex: (#23 -- this actually has both problems)

      Doubleint (sin(4x2+9y2)dA over elliptical region 4x2+9y2=1 in the first quadrant

      With u=2x, v=3y, we get x=1/2 u, y=1/3 v and a region of u2+v2=1

      Jacobian is 1/6

      Convert to polar coords to get doubleint (1/6 sin(r2)*r, r=0..1, q=0..p/2)

      Result: p/24 [1-cos(1)]

 

 

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