> 
# Problems from Ostobee Zorn
# 8.2 
# #9  Find the volume of the solid obtained by rotating the 
# region
# trapped by
# y=sqrt(x), y=0,x=1 around the line y=1 
# A solution:   
> plot({sqrt(x),1,[[0,0],[0,1.5]],[[1,0],[1,1.5]]},x=-.1..1.1);
# We can think of the desired solid as being drilled from a 
# cylinder of
# radius 1 with centerline
# y=1
> Volume := Pi*1^2*1 - Int(Pi*(1-sqrt(x))^2,x=0..1);

                                       1
                                       /
                                      |           1/2 2
                      Volume := Pi -  |  Pi (1 - x   )  dx
                                      |
                                     /
                                     0
> value(Volume);

                                     5/6 Pi
# #14  Find a formula for the volume of a pyramid with 
# height h and
# square base l  
> volume :=Int( (l/h*(h-x))^2,x=0..h);

                                      h
                                      /  2        2
                                     |  l  (h - x)
                          volume :=  |  ----------- dx
                                     |        2
                                    /        h
                                    0
> value(volume);

                                         2
                                    1/3 l  h
#  Question: What's the center of mass of such a pyramid, 
# assuming it,s
# homogeneous?
> Totalmoment := Int((x-CM)*(l/h*(h-x))^2,x=0..h);

                                    h
                                    /           2        2
                                   |  (x - CM) l  (h - x)
                   Totalmoment :=  |  -------------------- dx
                                   |            2
                                  /            h
                                  0
> sol := solve(Totalmoment,CM);

                               h              h           h
                               /              /           /
                           2  |              |   2       |   3
                        - h   |  x dx + 2 h  |  x  dx -  |  x  dx
                              |              |           |
                             /              /           /
                             0              0           0
               sol := - -----------------------------------------
                               h              h          h
                               /              /          /
                           2  |              |          |   2
                          h   |  1 dx - 2 h  |  x dx +  |  x  dx
                              |              |          |
                             /              /          /
                             0              0          0
> sol := value(sol);

                                  sol := 1/4 h
# Whoever guessed 1/3h in class was wrong, but not by much. 
#  It's 1/4 h.
# 
# #20   A cylindrical gasoline tank with radius 4 feet and 
# length 25
# feet is buried
# on its side underneath a service station (i.e., its flat 
# ends are
# perpendicular to the surface
# above).  If the gasoline in the tank is 6 feet deep, what 
# is the
# volume of the gas in the
# tank.
# 
# Solution:  Lets change 6 to a here.  That way we can 
# solve it no
# matter how you interpret
# the last statement.
#  
# If  we slice the tank with planes parallel to the ends of 
# the tanks,
# then each cross-section is the same.   Its area is
> area := int(sqrt(4^2-x^2)-(-sqrt(4^2-x^2)),x=a..4);

                                      2 1/2
                area := 8 Pi - (16 - a )    a - 16 arcsin(1/4 a)
# So the volume will be 
> volume := 25*area;

                                         2 1/2
            volume := 200 Pi - 25 (16 - a )    a - 400 arcsin(1/4 a)
# Now if the gasoline is really 6 feet deep in the tank, a 
# will be -2.
> subs(a=-2,volume);

                                    1/2
                      200 Pi + 50 12    - 400 arcsin(-1/2)
> evalf(");

                                   1010.963122
# The volume is 1011 cubic feet.
# 
# If we slice the tank with plane parallel to the surface 
# of the
# gasoline, we see the
# cross-sections are all rectangles  25 feet long and of 
# varying widths.
> vol2 :=Int(25*(sqrt(4^2-x^2)-(-sqrt(4^2-x^2))),x=a..4);

                                   4
                                   /
                                  |            2 1/2
                         vol2 :=  |  50 (16 - x )    dx
                                  |
                                 /
                                 a
> value(vol2);

                                    2 1/2
                 200 Pi - 25 (16 - a )    a - 400 arcsin(1/4 a)
> subs(a=-2,value(vol2));

                                    1/2
                      200 Pi + 50 12    - 400 arcsin(-1/2)
> 
# Same answer, as it should be.
# 
# Further problem: How deep would the gasoline be when
# there is 900 cubic feet of gas in the tank?
# 
# Solution:  solve the equation volume = 900 for a 
> solve(volume=900,a);
# Solve returns the question:  so use plot  and fsolve
# Actually, we don't need plot here because we can estimate
# that a will greater than -2 (volume = 1011) and less than 
#   4 (volume=0) 
> fsolve(volume=900,a,-2..4);

                                  -1.386707838
# So, the gas will be about 5.387 feet deep in the tank 
# when there is 900 cubic feet of gas in the tank.