Hydrostatic Pressure

The book I am learning out of is Calculus Early Transcendentals 5/E Volume 1 written by James Stewart. The first section of my post is an explanation of the idea that I am trying to be able to perform myself. The second section is a description of the specific problems that I would like to solve. The third section is my work on the problems of the second section. For more details, please refer to the link which I have posted above. Happy reading. :)
----------------------------------------------------------------------
SECTION 1: AUTHOR'S EXPLANATION

8.3 Applications to Physics and Engineering

Among the many applications of integral calculus to physics and engineering, we consider two here: force due to water pressure and centers of mass. As with our previous applications to geometry (areas, volumes, and lengths) and to work, our strategy is to break up the physical quantity into a large number of small parts, approximate each small part, add the results, take the limit, and then evaluate the resulting integral.

Hydrostatic Pressure and Force

Deep-sea divers realize that water pressure increases as they dive deeper. This is because the weight of the water above them increases.

In general suppose that a tin horizontal plate with the area A square meters is submerged in a fluid density p kilograms per cubic meter at a depth d meters below the surface of the fluid as in Figure 1.

Figure 1http://i274.photobucket.com/albums/j...8-3figure1.jpg

The fluid directly above the plate has volume V = Ad, so its mass is m = pV = pAd. The force exerted by the fluid on the plate is therefore

F = mg = pgAd

where g is the acceleration due to gravity. The pressure P on the plate is defined to be the force per unit area

P = F/A = pgd

The SI unit for measuring pressure is newtons per square meter, which is called a pascal (abbreviation: $\frac{1N}{m^2} = 1 Pa$ ). Since this is a small unit, the kilopascal (kPa) is often used. For instance, because the density of water is $p = \frac{1000 kg}{m^3}$ , the pressure at the bottom of a swimming pool 2 m deep is

$P = pgd = (\frac{1000 kg}{m^3})( \frac{9.8 m}{s^2})(2 m)<br /> = 19,600 Pa = 19.6 kPa$

An important principal of fluid pressure is the experimentally verified fact that at any point in a liquid the pressure is the same in all directions. (A diver feels the same pressure on nose and both ears.) Thus, the pressure in any direction at a depth d in a fluid with mass density p is given by

$P = p g d = \delta d$

Note: When using U.S. Customary units, we write $P = p g d = \delta d$ , where $\delta = pg$ is the weight density (as opposed to p, which is the mass density.) For instance the weight density of water is $\delta = \frac{62.5 lb}{ft^3}$ .

This helps us determine the hydrostatic force against a vertical plate or wall or dam in a fluid. This is not a straightforward problem because the pressure is not constant but increases as the depth increases.

EXAMPLE 1 A dam has the shape of the trapezoid shown in Figure 2.

Figure 2http://i274.photobucket.com/albums/j...8-3figure2.jpg

The height is 20 m, and the width is 50 m at the top and 30 m at the bottom. Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.

Figure 3ahttp://i274.photobucket.com/albums/j...-3figure3a.jpg

SOLUTION We choose a vertical x-axis with origin at the surface of the water as in Figure 3(a). The depth of the water is 16 m, so we divide the interval [0,16] into sub-intervals of equal length with endpoints $x_{i}$ and we choose $x_{i}^\ast \in [ x_{i-1}, x_{i} ]$ . The ith horizontal strip of the dam is approximated by a rectangle with height Δx and width $w_i$ , where, from similar triangles in Figure 3(b),

Figure 3bhttp://i274.photobucket.com/albums/j...-3figure3b.jpg

$\frac{a}{16 - x_i^\ast} = \frac{10}{20}$

or
$a = \frac{16 - x_i^\ast}{2} = 8 - \frac{x_i^\ast}{2}$

and so
$w_i = 2(15 + a) = 2(15 + 8 - \frac{1}{2}x_i^\ast) = 46 - x_i^\ast$

If $A_i$ is the area of the ith strip, then

$A_i \approx w_i \Delta x = (46 - x_i^\ast)\Delta x$

If Δx is small, then the pressure $P_i$ on the ith strip is almost constant and we can use Equation 1 to write

$P_i \approx 1000gx_i^\ast$

The hydrostatic force $F_i$ acting on the ith strip is the product of the pressure and the area:

$F_i = P_i A_i \approx 1000gx_i^\ast ( 46-x_i^\ast ) \Delta x$

Adding these forces and taking the limit as $n\rightarrow\infty$ , we obtain the total hydrostatic force on the dam:

$F = \lim_{n\rightarrow\infty} \sum_{i=1}^n 1000 g x_i^\ast ( 46 -x_i^\ast ) \Delta x$
$= \int_0^{16} 1000 g x ( 46 - x ) dx$
$= 1000 ( 9.8 ) \int_0^{16} ( 46 x - x^2) dx$
$= 9800 \left[ 23 x^2 - \frac{x^3}{3} \right]_0^{16}$
$=4.43 \times 10^7 N$

EXAMPLE 2 Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft if the drum is submerged in water 10 ft deep.

Figure 4http://i274.photobucket.com/albums/j...8-3figure4.jpg

SOLUTION In this example it is convenient to choose the axes as in Figure 4 so that the origin is placed at the center of the drum. Then the circle has a simple equation. $x^2 + y^2 = 9$ . As in Example 1 we divide the circular region into horizontal strips of equal width. From the equation of the circle, we see that the length of the ith strip is $2 \sqrt{ 9 - ( y_i^\ast)^2}$ and so its area is

$A_i = 2 \sqrt{ 9 - ( y_i^\ast)^2} \Delta y$

The pressure on this strips is approximately

$\delta d_i = 62.5 ( 7 - y_i^\ast )$

and so the force on the strips is approximately

$\delta d_i A_i = 62.5 ( 7 - y_i^\ast ) 2 \sqrt{ 9 - ( y_i^\ast)^2 } \Delta y$

The total force is obtained by adding the forces on all the strips and taking the limit:

$F = \lim_{n\rightarrow\infty} \sum_{i=1}^n 62.5 ( 7 - y_i^\ast ) 2 \sqrt{ 9 - ( y_i^\ast)^2 } \Delta y$
$= 125 \int_{-3}^3 ( 7 - y ) \sqrt{ 9 - y^2} dy$
$= 125 \displaystyle{\cdot} 7 \int_{-3}^3 \sqrt{ 9 - y^2} dy - 125\int_{-3}^3 y \sqrt{ 9 - y^2} dy$

The second integral is 0 because the integrand is an odd function(see Theorem 5.5.7). The first integral can be evaluated using the trigonometric substitution y = 3 sin Θ, but it's simpler to observe that it is the area of a semicircular disk with radius3. Thus

$F = 875 \int_{-3}^3 \sqrt{ 9 - y^2} dy = 875 \displaystyle{\cdot} \frac{1}{2} \pi (3)^2$
$= \frac{7875 \pi}{2} \approx$ 12,370 lb
-------------------------------------------------------------------
SECTION 2: MY PROBLEMS TO SOLVE

#3 to #9: A vertical plate is submerged in water and has the indicated shape. Expain how to approximate the hydrostatic force against the end of the tank by a Riemann sum. Then express the force as an integral and evaluate it.

http://i274.photobucket.com/albums/j...3problem3a.jpg

http://i274.photobucket.com/albums/j...3problem3b.jpg

http://i274.photobucket.com/albums/j...3problem3c.jpg

http://i274.photobucket.com/albums/j...3problem3d.jpg

http://i274.photobucket.com/albums/j...3problem3e.jpg

http://i274.photobucket.com/albums/j...3problem3f.jpg

http://i274.photobucket.com/albums/j...3problem3g.jpg

The book answers for the odd numbered questions are:
3) 6000 lb
5) $6.5 \times 10^6 N$
7) $3.5 \times 10^4 N$
9) $1000 g \pi r^3 N$
-----------------------------------------------------------------------
SECTION 3: MY WORK

Being that I don't believe I am having difficulty at all with Riemann sums and that it might be too time consuming to create pictures, load them on the net, and reference them here, I'll just skip this part for you guys. :)

3) This rectangular plate is measured in feet, so we measure our P as $\delta d$ or as 62.5d. When we enter this into our integral, since we will be integrating with respect to depth, we would write this element as 62.5x. The next step is to find a way to express the width in terms of depth. In this particular problem, the width is constant no matter the depth, so we can just enter a 6 for this element. The last element of our integral will of course be our dx that represents the height of an infinitely small strip of our rectangular plate. Also, since the top of our plate is submerged 2 feet under water and the bottom of our rectangle is 6 feet under water, we will be integrating from 2 to 6. Putting all this information together we have this integral:
$\int_a^b ( p g x ) ( w ) dx$
$\int_2^6 (62.5 x) ( 6 ) dx$
$375 \left[ \frac{x^2}{2} \right]_2^6$
$375 \left[ 18 - 2 \right]$
$( 375 ) ( 16 )$
$6000 lb$

This is the correct answer for problem 3.

4) This triangle plate is measured in feet so we use 62.5x for our pressure. We can express the width as $4 - \frac{4}{3} ( x - 1 )$ . This makes our integral:
$\int_1^4 (62.5 x)(4 - \frac{4}{3} ( x - 1 ) ) dx$

which I've evaluated to:
750 lb

5) This semicircle is measured in meters so we use 9800x for our pressure. To find an expression for the width, I've taken the formula for the circle, $x_1^2 + y_1^2 = 10^2$

and first figured that if x (not $x_1$ ) is going to be the depth within the integral then:
$y_1 = -x$

so then I could substitute x for $y_1$ and solve for $2x_1$ which I figure to be the width:
$2x_1 = 2 \sqrt{100 - x^2}$

Now that I can express both the pressure and area within the integral, I can construct it:
$\int_0^{10} (9800x)( 2 \sqrt{100 - x^2} ) dx$

which I've evaluated to be:
$1.039 \times 10^7 N$

6)
$P = 1000 g x$
$w = \frac{4}{5} x$
$F = \int_0^5 (1000 g x) ( \frac{4}{5} x ) dx$
$F = \int_0^5 800 g x^2 dx$
$F = \frac{800 g}{3} \left[ x^3 \right]_0^5$
$F = \frac{ 100000 g }{3} N$

7)
$P = 62.5 x$
$w = 12 + 8x$
$F = \int_0^8 (62.5 x) (12 + 8x ) dx$

I've evaluated this to be:
109333 lb

8)
$P = 1000 g x$
$w = b + \frac{x ( a - b ) }{h}$
$F = \int_0^h (1000 g x ) ( b + \frac{ x ( a - b ) }{h} ) dx$

I've evaluated this to be:
$F = \frac{500}{3} g h^2 (2a + b) N$

9)
$P = 1000 g x$
$w = 2 \sqrt{ 2 r x - x^2 }$
$F = \int_0^2r (1000 g x ) (2 \sqrt{ 2 r x - x^2 } ) dx$

I've evaluated this to be:
$2000 g \pi r^3 N$

-------------------------------------------------------------------

In conclusion I suspect that I am getting incorrect answers whenever the width of a plate varies with the depth, however I can't really be sure where my mistakes are being made. I've been stuck on this idea for about two to three weeks. Any help would be greatly appreciated.