SPHERE
BRIEF ABSTRACT
Sphere is designed as a big homework problem using spherical coordinates (among other things!)  The problem revolves -- no pun intended -- around issues of painting a hemispherical dome.
GENERAL INFORMATION
FileName:    SPHERE
Full title:   Painting a Hemispherical Dome
Last Update: 6/1/96
Developers:
Aaron D. Klebanoff, Department of Mathematics,  Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA.
Sandra K. Dawson, Glenbrook South High School, Glenview IL 60025 USA.
Dave Horn, Roger High School, Michigan City IN 46360 USA.
Brian J. Winkel, Department of Mathematics,  Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA.  
Contact:  
Aaron Klebanoff, Department of Mathematics, 
Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. 
Phone:  812-877-8151.  Email:  Klebanoff@rose-hulman.edu.  
FAX:  812-877-3198.
Support:  
The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849:  Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. 
STATEMENT OF PROBLEM 
NOTE: Throughout this problem, we make use of a standard spherical coordinate system where the usual point from rectangular coordinates P:(x, y, z) is now represented as the point P:(rho, phi, theta) defined as follows: the first coordinate, rho, is the distance from the origin O: (0, 0, 0), to the point P; the second coordinate, phi, called the vertical angle, is the angle (from 0 and Pi radians) from the positive z-axis (vertical) to the point P; and the third coordinate, theta, called the polar angle, is the angle (from 0 and 2 Pi radians) between the x-axis (polar axis) and the projection of the ray OP onto the xy-plane.  See, e.g., p. 744 of Calculus with Analytic Geometry, 4th ed., by Edwards and Penney or a section on spherical coordinates in most other calculus books.
The roof of a hemispherical dome of radius 50 feet at an amusement park is to be painted red, white, and blue in 36 10-degree vertical swaths.  Each painter works in one of the 10-degree vertical swaths from bottom to top with a color, and then paints the neighboring strip in a different color.
(1) What is the roof's surface area?  How about the surface area of a 10-degree swath?
(2) Suppose a painter covers one of the 36 10-degree strips moving from bottom to top.  Determine the painted area as a function of the vertical angle.
(3) Suppose the painter covers a 10-degree strip (polar angle) at a constant angular rate.  Determine the rate at which the painted area changes.
(4) As the painter covers the strip, he slows because of the increasing proportion of edge work near the narrow top of each swath.  As a result, he rises up the roof at a rate of r(t) = Sqrt[1.6 t] radians per hour.  Find the area painted as a function of time as well as the rate at which the area covered changes.
(5) Suppose the painter can cover 300 square feet of roof per hour.  If the painter covers a 10 degree swath, determine his angular position as a function of time.
(6) The painter actually starts at the bottom of a swath painting at a rate of 400 square feet per hour and slows at a rate proportional to his angular position to 100 square feet per hour by the time he reaches the top of a swath due to the intricate work involved at the narrow top of the swath.  Determine his angular position as a function of time.
KEYWORDS  
Integration, sphere, spherical coordinates, surface area.
TEACHER NOTES
ISSUES RELATED TO THE PROBLEM 
We were unable to find any text book that talked about computing surface area in spherical coordinates.  Formulas for computing surface area in rectangular coordinates typically require knowledge of the projection of the surface onto the plane over which the integration is done.  Thus, the students must have at least a basic understanding of the derivation of the differential volume of a cube in spherical coordinates in order to realize that they need only look, then, at a differential rectangle on the surface of the sphere.
Prerequisites
Multiple integrals, spherical coordinates, derivative.
Time allotment - time management
If assigning the entire problem, 1-2 weeks out of class should suffice.  The last two problems require the most ingenuity on the part of the students, and it is there that the instructor should leave time for questions.
Expectations 
Some students may quickly arrive at the problem integration formulas and have little problem at the first 4 questions in this exercise.  However, expect many students to have a lot of trouble with this.  Students have a difficult time visualizing in 3D, and adding to that spherical coordinates (not a very natural coordinate system for the average calculus student) may be a substantial stumbling block.  It is partially for these reasons that we strongly advise this to be assigned as a group project.  The last two problems are a bit challenging, and the instructor may wish to offer these as extra credit problems.
Future payoffs
This problem gives the students a better grasp of spherical coordinates and integrating in spherical coordinates.  It also reinforces how to apply the derivative.
Extensions
In (6), have the rate slow (or speed up) at a different rate, e.g. at a rate proportional to time.

Don't require that the swaths be 10 degrees across, but instead an arbitrary angle.
           
References and Sources
POSSIBLE SOLUTIONS
(1) What is the roof's surface area?  How about the surface area of a 10-degree swath?
The radius of the roof.
In[2]:=
R = 50;
The surface area of the roof.
In[3]:=
area = (1/2) 4 Pi R^2;
The surface area of a 10-degree section of roof.
In[4]:=
section = area/36//N;
(2) Suppose a painter covers one of the 36 10-degree strips moving from bottom to top.  Determine the painted area as a function of the vertical angle.
A[th1, th2, phi1, phi2] is the surface area of a spherical rectangle with polar angles satisfying 0 <= th1 < th2 <= 2 Pi and vertical angles satisfying 0 <= phi1 < phi2 <= Pi/2.
In[5]:=
A[th1_,th2_,phi1_,phi2_] = Integrate[R^2 Sin[phi], 
			{phi,phi1,phi2},{th, th1, th2}];
The area covered from the bottom in a 10-degree swath as a function of the vertical angle phi along with a plot of this.
In[6]:=
coverage[phi_] = A[0, Pi/18, Pi/2 - phi, Pi/2];
In[7]:=
Plot[coverage[phi], {phi, 0, Pi/2},
   AxesLabel -> {"radians", "ft^2"},
   AxesOrigin -> Automatic]

Out[7]=
-Graphics-
(3) Suppose the painter covers a 10-degree strip (polar angle) at a constant angular rate.  Determine the rate at which the painted area changes.
The rate at which area is covered from the bottom in a 10-degree swath as a function of the vertical angle phi along with a plot of this.
In[8]:=
rate[phi_] = coverage'[phi];
In[9]:=
Plot[rate[phi], {phi, 0, Pi/2},
   AxesLabel -> {"radians", "ft^2/radian"},
   AxesOrigin -> Automatic]

Out[9]=
-Graphics-
(4) As the painter covers the strip, he slows because of the increasing proportion of edge work near the narrow top of each swath.  As a result, he rises up the roof at a rate of r(t) = Sqrt[1.6 t] radians per hour.  Find the area painted as a function of time as well as the rate at which the area covered changes.
The rate at which the vertical angle (measured in radians) increases with respect to time (measured in hours).
In[10]:=
r[t_] = Sqrt[1.6 t];
Determine the time, tmax, when the top has been reached.
In[11]:=
sol = FindRoot[r [t]== Pi/2, {t, 2}]
Out[11]=
{t -> 1.54213}
In[12]:=
tmax = t /. sol[[1]];
The area covered as a function of time...
In[13]:=
coverage[t]
Out[13]=
1250 Pi Sin[t]
--------------
      9
In[14]:=
areac[t_]  = coverage[phi]/.{phi->t};
Plot[areac[t], {t, 0, tmax},
	AxesLabel -> {"hours", "ft^2"},
	AxesOrigin -> Automatic]

Out[15]=
-Graphics-
The change in the area covered as a function of time...
In[16]:=
change[t_]  = areac'[t];
Plot[change[t], {t, 0.00001, tmax}, 
	AxesLabel -> {"hours", "ft^2/hour"},
	AxesOrigin -> Automatic]

Out[17]=
-Graphics-
(5) Suppose the painter can cover 300 square feet of roof per hour.  If the painter covers a 10 degree swath, determine his angular position as a function of time.
In[18]:=
Clear[t, cov, ph]
Define the area covered as a composition of functions: cov(phi(t)).
In[19]:=
cov[t_] = coverage[phi] /. phi -> ph[t];
We want the derivative of cov(phi(t)) with respect to t to be equal to 300.  This yields a differential equation with unknown ph(t) that is easily solved.

We call the solution phit and show its plot.
In[20]:=
sol  = Flatten[Solve[cov'[t] == 300, ph'[t]]];
In[21]:=
rhs = ph'[t] /.sol[[1]];
ans = Flatten[
	DSolve[{ph'[t] == rhs, ph[0] == 0}, ph[t], t]];
Solve::ifun: 
   Warning: Inverse functions are being used by Solve,
     so some solutions may not be found.
In[23]:=
phit[t_]  = ph[t] /. ans[[1]];
Plot[phit[t], {t, 0, 25 Pi/54},
	AxesLabel->{"hours", "radians"},
	AxesOrigin -> Automatic]

Out[24]=
-Graphics-
(6) The painter actually starts at the bottom of a swath painting at a rate of 400 square feet per hour and slows at a rate proportional to his angular position to 100 square feet per hour by the time he reaches the top of a swath due to the intricate work involved at the narrow top of the swath.  Determine his angular position as a function of time.
In[25]:=
Clear[r, rate, t, cov]
The rate as a function of the vertical angle.
In[26]:=
r[phi_] = 400 - 600 phi/Pi;
Define the area covered as a composition of functions: cov(phi(t)); and
define the rate as a composition of functions: rate(phi(t))
In[27]:=
cov[t_] = coverage[phi] /. phi -> ph[t];
In[28]:=
rate[t_] = r[phi] /. phi -> ph[t];
We want the derivative of cov(phi(t)) with respect to t to be equal to rate(t).  This yields a differential equation with unknown ph(t) that must be solved numerically.

We call the solution phit and show its plot.
In[29]:=
sol = Flatten[Solve[cov'[t] == rate[t], ph'[t]]];
In[30]:=
rhs = ph'[t] /. sol[[1]];
ans = Flatten[
	NDSolve[{ph'[t] == rhs, ph[0] == 0}, ph[t], 
		{t, 0, 1.62}]];
In[32]:=
phit[t_]  = ph[t] /. ans[[1]];
Plot[phit[t], {t, 0, 1.62},
	AxesLabel->{"minutes", "radians"},
	AxesOrigin -> Automatic]

Out[33]=
-Graphics-
ISSUES IN SOLUTION
We were unable to find any text book that talked about computing
surface area in spherical coordinates.  Formulas for computing
surface area in rectangular coordinates typically require knowledge
of the projection of the surface onto the plane over which the
integration is done.

Thus, the students must have at least a basic understanding of
the derivation of the differential volume of a cube in spherical
coordinates in order to realize that they need only look, then,
at a differential rectangle on the surface of the sphere.