all

2.  The integral of an integral
 Another Method for finding
› Volume
› Mass density
› Centers of mass
› Joint probability
› Expected value
 Fibuni’s Theorem states that if f is continuous on a plane
region R

3. The two intervals determine the region of integration R on the iterated integral

4. Integrate with respect to x.
Treat y as a constant
Integrate with respect to y
NOTE: similar to partial derivatives
Concerning treatment of variables as a constant
**Do Inner Integral first!

5. ∫ ∫ (x2-2y2+1) dx dy
∫ [(x3)/3-2y2x+x] | dy
∫ [((64/3)-8y2+4)-(0 -0 +0)] dy
[(64y)/3- (8y3)/3+4y] |
[(64(2))/3-(8(23)/3+4(2)]-[(64(1))/3-(8(13))/3+4(1)]
(128-64)/3+(-64+8)/3 +(8-4)
64/3-56/3+4
8/3+4
20/3
2
1
4
0
2
1
4
0
2
1
2
1
Integrate with respect to x. Treat y as constant
Integrate with respect to y

6.  In mathematics, a planar lamina is a closed
surface of mass m and surface density such
that:, over
the closed surface.
 Planar laminas can be used to compute
mass, electric charge, moments of
inertia, or center of mass.

7.  Suppose the lamina occupies a region D of the xy-plane and its density at a
point (x,y) in D is given by ρ(x,y) where ρ is a continuous function on D. This
means that:
 Ρ(x,y)=lim
 where ∆m and ∆A are the mass and area of a small rectangle that
contains (x,y) and the limit is taken as the dimensions of the rectangle
approach 0.

 Therefore we arrive at the definition of total mass in the lamina. All one has to
do is find the double integral of the density function.

m=∬ρ(x,y)dA
∆m
___
∆A

8.  The center of mass of a lamina with density function ρ(x,y) that occupies a region D. To
find the center of mass we first have to find the moment of a particle about an
axis, which is defined as the product of its mass and its directed distance from the axis.
 The moment of the entire lamina about the x-axis:

Mx=∬yρ(x,y)dA

 Similarly, the moment about the y-axis:

My=∬xρ(x,y)dA

 You can define the center of mass (α,ŷ) so that mα=My and mŷ=Mx The physical
significance is that the lamina behaves as if its entire mass is concentrated at its center
of mass. Thus, the lamina balances horizontally when supported at its center of mass.
 The coordinates (α,ŷ) of the center of mass of a lamina occupying the region D and
having density function ρ(x,y) are:

α= = ∬xρ(x,y)dA ŷ= = ∬yρ(x,y)dA
My
__
m m
1 __
__
m
My
m
__ 1

9.  The moment of inertia of a particle of mass m about an axis is defined to be mr^2,
where r is the distance from the particle to the axis. We extend this concept to a lamina
with density function ρ(x,y) and occupying a region D by proceeding as we did for
ordinary moments: we use the double integral:

 The moment of inertia of the lamina about the x-axis:

Ix =y^2ρ(x,y)dA

 Similarly the moment about the y-axis is:

Iy=x^2ρ(x,y)dA

 It is also of interest to consider the moment of inertia about the origin, also called the
polar moment of inertia:

I0=∬(x^2+y^2)ρ(x,y)dA

 Also notice the following:
 I0=Ix+Iy