COALITION MATH 152, SPRING 1998

              (S. A. Fulling, assisted by Vera Rice)

                             Day 21.T


    1.  The usual paper shuffling:  Old homework, CAPA, revised 
    new homework.  The next lab (centroid of Texas) is in the 
    book; no handout.  Reminders:

        A.  Area of Texas lab may be in the stack outside my 
        office.

        B.  Rework of last test is due Thursday; see Web for 
        instructions.

    2.  Call on teams to report results on the 4 flavors of the 
    final exercise of last class.

    3.  Admonition to start reading the Web material on centers 
    of mass and moments of inertia.  Apology that TeX and HTML 
    are not well integrated yet, and graphics nonexistent 
    (volunteers?).  Will try to distribute graphics in hard copy 
    after the lectures.
        
    4.  A bridge between volumes and moments/centroids:  Total 
    mass, with an introduction to integration in polar and 
    cylindrical cordinates.  (I use free transparencies from 
    Larson & Hostetler to make some of these points.)

        A.  A homogeneous material has a DENSITY \rho (with mks 
        units kg/m^3).  The total mass of a body of that material 
        is its volume times its density.

        B.  To find the total mass of a body built up from finite 
        disks (like a barbell), we would add the mass of the 
        disks (which might be made of different materials):

        \Sum_i \rho_i V_i = \Sum_i \pi r_i^2 \rho_i \Delta u_i.

        C.  In the limit of thin disks (continuously varying 
        density), this becomes an integral: 

        \int \pi r(u)^2 \rho(u) du.

        Here u is the coordinate along the axis.  This formula 
        applies when the body is a solid of revolution (without 
        washers, for simplicity) AND the density depends on u 
        only (each cross section is homogeneous).

        D.  Example:   The body has the shape formed by rotating 
        about the y axis the region bounded by

        y = \sqrt{1 - x^2},  x = 0,  y = 0  
 
        (coordinates in meters).   The density is given by the 
        formula 

        \rho = 3 + 2y  kg/m^3

        Find the total mass of this body.

        M = \int_0^1 \pi (1 - y^2) (3 + 2y) dy

        E.  Now consider a cylinder made of shells of different 
        materials.  The total mass is the sum of the masses of 
        the cylinders:

        \Sum_i \rho_i V_i 
            = \Sum_i 2\pi r_i h_i \rho_i \Delta r_i. 

        F.  In the limit of thin cylinders, this becomes

        \int 2\pi r h(r) \rho(r) dr.

        (In particular, for a "squarely chopped off" cylindrical 
        body, h is a constant, the length of the cylinder.)  This 
        formula applies when the body is a solid of revolution 
        AND the density depends on r only.

        G.  Example:  A beam with circular cross section has a 
        metal core of radius 0.1 m and a plaster exterior whose 
        outer layers have compressed due to weathering, so that 
        the density function is

        \rho = { 10      for 0 < r < 0.1,
    
               { 1 + 3r  for 0.1 < r < 0.5 . 

        What is the total mass of the beam per unit length?
        (TEAM EXERCISE)

THERE WAS SUFFICIENT TIME TO TALK OFF-THE-CUFF BRIEFLY ABOUT 
POLAR COORDINATES (POINT 5 OF NEXT TIME).