I'm studing for a test next week, and out of 40 or so practice
problems, I'm having trouble with these.  I keep getting stuck, so
steps would help. Thanks! :)

-A swimming pool with vertical sides has circular glass plates
covering the lights.  The water is 10ft deep, the radius of the
circular glass plates is 1 foot.  If the top of the circular plate is
one foot beneath the surface of the water, what is the total force on
each glass plate? (use 62.5 lb/ft^2 for water's density)

-Let R be the plane region bounded by the curve y=e^x, the line y=3,
and the Y-axis. Find the total surface area of the solid generated by
rotating R about the X-axis. (That includes the area of all surfaces
generated.  If this can be done partly by geometry instead of
calculus, that is OK).

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Clarification of Question by tobster85-ga on 14 Apr 2005 17:54 PDT

Thanks for all the help. I'll go through the answers tonight and
finalize this question.

As for the second question...

http://tutorial.math.lamar.edu/AllBrowsers/2414/SurfaceArea.asp

I worked it out on paper, and it's sort of tricky.  As for it being on
a test, I assume you won't have a computer/calculator that can do
integrals for you.  If you can, it turns into quite a trivial problem.
 If not, I believe there are some simplifications involving changes of
variables that will help yield a solution.

As far as the first question, Pressure = Force / Area.

At sea level, you will feel 1 Atmosphere (1 ATM) of pressure--14.7 psi
(lb/in^2).  If you go to 33ft below the surface, your body will feel
twice that or 2 ATM.

So, an object one foot under the water will feel (14.7 / 33) = .445 psi extra.

This means that the top of the light at one foot will experience 15.1
psi of pressure.  The area of your light is pi x radius squared, or
put in inches...

pi (12 in)^2 = 144pi in^2.  

If Pressure is Force / Area, then Force = PxA which is:

F = 15.1 lb/in^2 (144pi in^2) = Approx 6827 lbs of total force on the plate

***NOTE***: This would assume that the ENTIRE glass plate is at a
depth of one foot--i.e. horizontal.  But it isn't.  It's vertical. 
Hence you will need to do some sort of calculation over the interval
1ft to 2ft.  Unfortunately, I'm not mathematically savvy enough to
provide that formula...perhaps mathtalk or someone else can provide
the answer.

Good luck,

Herkdrvr

1) About the force on a circular glass plate...

Here is one way---the origin (0,0) is the center of a glass plate, and
this is 2 ft below the water surface. This is to maintain the circle
(glass plate) as x^2 +y^2 = 1.

Hydrostatic pressure is proportional to the depth. The deeper a point,
the greater is the pressure.
The force on any area of the plate is (pressure at that area) times (the area).
We take a horizontal strip of the plate that is dy in height, making
this strip to be (2x)*dy in area. This strip is our dA. The force on
this dA is (pressure)*(2x dy).

The dA is at (2 -y) depth. So the pressure there is (density *depth) =
62.5*(2-y) psf.

water density is 62.5 pcf, or 62.5 pounds per cubic foot

So the force, dF, on dA is (62.5)(2-y)(2x dy)
Or,
dF = (62.5)(2-y)(2x)dy = 125x(2-y)dy.

we express the x into y-terms, because of the dy,
x^2 +y^2 = 1
x = sqrt(1 -y^2)
So,
dF = 125sqrt(1 -y^2)(2-y)dy  ....(i)

The boundaries of dy are from y=1 to y=(-1), so,

F = (125)INT.(1 -> -1)[sqrt(1-y^2) *(2-y)]dy  ....(ii)

You can do the integration by online integrators.

----------------------
If you want do do it longhand,

F = (125)INT.(1 -> -1)[sqrt(1-y^2) *(2-y)]dy  ....(ii)
F = (125)INT.(1 -> -1)[{2*sqrt(1-y^2) dy} -{y*(sqrt(1-y^2) dy}]
F = (250)INT.(1 -> -1)[sqrt(1-y^2)]dy -(125)INT.(1 -> -1)[y*(sqrt(1-y^2)]dy
Using the Table of Integrals,
F = [(250){(1/2)(y)sqrt(1-y^2) +(1/2)arcsin(y)}
-(125){(1/3)*(1-y^2)^(3/2)}](1 -> -1)

F = [(250){(1/2)(-1)sqrt(1-1) +(1/2)arcsin(-1)} -(125){(1/3)*(1-1)^(3/2)}]
    -[(250){(1/2)(1)sqrt(1-1) +(1/2)arcsin(1)} -(125){(1/3)*(1-1)^(3/2)}]

F = [250{0 +(1/2)(-pi/2)} -0] -[250{0 +(1/2)(pi/2)} -0]
F = -250pi/4 -250pi/4
F = -125pi lbs, or about 392.7 pounds  

F came out as negative because we maintained y is positive going up
and negative going down, all relative to the origin (0,0). So since
the depth is going down, then we got negative F.

Therefore, the total force acting on each glass plate is 125pi or
392.7 pounds.  ....answer.

============
2) About a generated surface areas....

Draw and analyze the figure on paper.
By rotating R about the x-axis, there are 3 surfaces formed:

>>>a circular cylinder due to the y=3.
Surface area, A1 = 2pi*r*h
radius is 3
heigth is x when y=3.  
(...y = e^x; ln(y) = x*ln(e);  ln(y) = x;  so when y=3, x = ln(3)...)
So,
A1 = 2pi(3)(ln(3)) = 6ln(3)*pi....about 20.708 sq.units.

>>>an area between two concentric circles due to the y-axis.
Surface area, A2 = pi(R^2 -r^2)
Outer circle has Radius = 3
Inner circle has radius = 1  ...because when x=0, or at the y-axis, y
= e^x = e^0 = 1.
So,
A2 = pi(3^2 -1^2) = pi(9 -1) = 8pi ....about 25.133 sq.units

>>> a curved area due to y=e^x from y=1 to y=3. Or, from x=0 to x=ln(3).
Surface area, A3 = summmation of all rings from radius 1 to radius 3.
We will use integration here.
Imagine a ring whose radius is y and whose width at the circumference is dx.
Or, a cylinder of radius y and height of dx.
The surface area of this ring/cylinder is (2pi*y)*dx.
That is our dA.
dA = 2pi*y dx
The boundaries of dx are from x=0 to x=ln(3).
So,
A3 = (2pi)INT.(0-->ln(3))[y]dx
We convert the y into x-terms,
A3 = (2pi)INT.(0-->ln(3))[e^x]dx
A3 = (2pi)[e^x](0-->ln(3))
A3 = (2pi){e^(ln(3)) -e^0}
A3 = (2pi){3 -1}
A3 = 4pi ....about 12.566 sq.units

Therefore, the tota surface area that is generated, A, is
A = A1 +A2 +A3
A = 6ln(3)*pi +8pi +4pi
A = about 18.6pi  ...about 58.4 sq.units  ....answer.

Regarding your first problem - you have to solve a integral obviously.

The integral I found was as follows...
rho = density
g = gravity
h = height (distance from surface of water)
dh = your integrating w/ respect to h
integrate: rho*g*h*2*(1-(h-2)^2)^(1/2) dh from 1 to 3.
The answer is the force.  I suppose you could move rho*g*2 in front of
the integral because they are constants.

How did I get the integral.
rho*g*h = the pressure at a height h in the pool.
the 2*(1-(h-2)^2)^(1/2)dh is just to calculate the area of the circle.
I got it from (x-2)^2+y^2 = 1.

The beauty of engineering-related math is that for a lot of practical
real world problems, you don't actually need to use calculus to solve
it.  In practical applications, a lot of objects are symmetric, so you
can use the principle of symmetry to help you out.

For the case of your swimming pool problem, you can again use this
principle of symmetry.  The whole point is that you want to break the
circle up into a million little blocks of equal area.  We'll say their
area is each dA.  Now, you want to take the force acting on each
little block and sum them up.  This process is of course called
integration.  But, for a symmetrical case, like what you have here,
you have just as many dAs above the centerline of the circle as you do
below the centerline of the circle.  Since hydrostatic pressure (water
pressure) is based only on depth, you can simply assume that the
entire force is acting as a point load at the center of area of your
object.

In fact, you can always assume that the entire force is acting at the
center of area.  But, funding the center of area requires integration
for a non-symmetrical object, so you're in the same boat either way.

So, what does this mean?  Well, the center of your circle is located 2
feet below water level.

Pressure @ some depth = density * depth* gravitational acceleration 
(this equation is true for any fluid)
Pressure = 62.5 lb-mass/ft^3 * 2 ft * 32.2 ft/sec^2
Pressure = 4025 lb-mass/ft-sec^2

Recall on earth, 1 lb-mass = 1 lb force = 1 lb ft/sec^2

Pressure = 4025 lb/ft^2

Force = pressure * area

Force = 4025 lb/ft^2 * (pi) r^2 = 4025 * (pi) * 1^2 = 4025*pi lbs.

One hint for your test; unless the professor says you have to use
integration for these types of problems, I recommend you apply the
principle of symmetry, and clearly write "APPLY PRINCIPLE OF
SYMMETRY..BLAH BLAH BLAH" on your exam.  If you're not strong in
integration, there is a lot of potential for error, and both ways, you
get the same answer, so it's to your benefit to go with the simpler
method.

Whooops - I looked at my answer and realized it was awfully high...

The tricky thing about English units is that the unit of weight, the
lb, is used often as a measure of weight and mass.  I too fell into
the trap, so here's the real answer!

You mention the density of water is 62.5 lb/ft^3 (you mentioned ft^2,
but I assume you meant ft^3)

This is lb-mass/ft^3, because density is always given as mass/volume. 
Because of this, the gravitational acceleration (32.2) is already
incorporated into this number.

Thus, the true pressure is
Pressure = Density (in units lb/ft^3) * Height (in units feet)
Pressure = 62.5 * 2 = 125 lb/ft^2

Multiply by A = (pi)*r^2, and you get 125 pi, or approximately 392.7 lb

Regarding herkdrvr's comment, you would probably assume that the area
behind the plate (where the actual light bulb is located) is not in a
vaccuum, so it has an equal an opposite air(atmoshperic) pressure in
there.  Otherwise, if it were a vaccuum, you would have to add
atmospheric pressure to that number to get the true pressure acting on
the plate.