Calculus Notes

**galactus** · May 23rd 2008, 11:34 AM

Now, let's start on an interesting topic....fluid pressure.

If we submerge a flat piece of, say, metal horizontally in water the pressure on it is straightforward. The force on it is $F={\rho}hA$ .

Where ${\rho}=\text{weight density of the fluid}$ and

h is the depth and A is the surface area of the object.

Pressure is defined as force per unit volume. Therefore, the pressure p exerted at each point of a flat surface of area A submerged horizontally at a depth h is given by $p=\frac{F}{A}={\rho}h$

In the British system weight density is measured in pounds per cubic foot and pressure is measured in pounds per square foot.

In metric, weight density is measured in newtons per cubic meter and pressure in newtons per square meter.

One Newton per square meter is called a pascal

One Pa is approximately 0.02089 pounds per square foot.

The weight density of water is approximately 9810 newtons per cubic meter and 62.4 pounds per cubic foot in the British system.

Here is a very trivial case:

If a flat circular piece of metal of radius 2 meters is submerged horizontally in water so that the top surface is at a depth of 3 meters, then the force on top of the plate is:

Since we are using metric, we use Newtons.

$F={\rho}hA={\rho}h({\pi}r^{2})=(9810)(3)(4\pi)=117 ,720{\pi} \;\ N\approx{369,828} \;\ N$

And the pressure at each point on the plate surface is

$p={\rho}h=(9810)(3)=29,430 \;\ \frac{N}{m^{2}}=29,430 \;\ Pa$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Suppose we want to find the force on a vertical submerged surface. That is a little more involved than a simple submerged horizontal surface.

Let's introduce a vertical x-axis whose positive direction is downward and whose origin is at any convenient point.

Let's say the submerged portion of the surface extends from x=a to x=b on the x-axis. Also, suppose that a point x on the axis lies h(x)

units below the surface and that the cross section of the plate at x has width w(x). Next, divide the interval [a,b] into n subintervals with length

${\Delta}x_{1}, \;\ {\Delta}x_{2}, \;\ ......., \;\ {\Delta}x_{n}$

and that each subinterval choose an arbitrary point $x_{k}$ . We approximate the section of the plate along the kth subinterval by a rectangle

of length $w(x_{k})$ and width ${\Delta}x_{k}$ .

Because the upper and lower edges of the plate are at different depths we can't use the simple formula $F={\rho}hA$ .

But, if ${\Delta}x_{k}$ is small enough the difference in depth between the upper and lower edges is small and we can reasonably

assume the entire rectangle is concentrated at a single depth $h(x_{k})$ below the surface.

Therefore, we get: $F_{k}={\rho}h(x_{k})w(x_{k}){\Delta}x_{k}$

Where $h(x_{k})$ is the depth and $w(x_{k}){\Delta}x_{k}$ is the area of the rectangle.

Looks like a Riemann sum, huh?. Well, it is and that's where the integration comes in. We add up all those rectangles to find the total pressure.

Thus, the total force on the plate is $F=\sum_{k=1}^{n}F_{k}=\sum_{k=1}^{n}{\rho}h(x_{k}) w(x_{k}){\Delta}x_{k}$

To find the exact value of the force we take the limit as ${\Delta}x_{k}\rightarrow{0}$ .

This gives us $F=\int_{a}^{b}{\rho}h(x)w(x)dx$

So, we have the formula for fluid force:

Assume that a flat surface is submerged vertically in a liquid that has weight density ${\rho}$ and that the submerged portion

extends from x=a to x=b on a vertical x-axis. For $a\leq{x}\leq{b}$ , let w(x) be the width of the surface at x and let h(x) be the

depth at point x. Then the total fluid force on the surface is $F=\int_{a}^{b}{\rho}h(x)w(x)dx$

In the next thread we will begin some actual problems.

**colby2152** · June 20th 2008, 08:48 AM

Lesson 5.5 - Substitution

Many integrals can be tough to solve from the get go. Most of the time, the Calculus student will have to manipulate such integrands using some algebraic techniques, namely substitution. Substitution, also known as U-Substitution, is one of the most simple techniques to solve for an integral.

Initially, you start off with an integral such as $\int f(x) dx$ . When you cannot solve it outright, you will need to make a substitution where $u = u(x)$ .

You may use any variable here, but popular notation is to use the variable "u". Upon choosing your substitution, you must derive both sides of the equation with their respected variables like so: $du = u'(x) dx$

The next step is to take these actions upon the original integral.

$\int f(x) dx = \int f(x(u)) x'(u) du$

After solving for the new integral, the next step is to substitute back the original parameters and solve for the limits if is a definite integral. Substitution, like many topics in math, is best shown through a series of examples...

Example 5.5-A

$\int \frac{x}{x^2+3} dx$

What you need to recognize here are the powers. The derivative of $x^2 + 3$ is $2x$ which is of the same form ( $ax$ ) as the numerator of the integrand. We know what to do, so let's put our new found substitution technique into action!

$u = x^2 + 3$

$du = 2x dx$

$\frac{du}{2} = x dx$

We can now substitute these into the integral...

$\int \frac{du}{2u}$

$\frac{\ln|u|}{2} + C$

One last step: The original integral is in the form of x, so we need to back substitute our parameters...

$u = x^2 + 3$

$\frac{\ln|u|}{2} + C = \frac{\ln|x^2+3|}{2} + C$

Example 5.5-B

$\int xe^{x^2} dx$

Once again, identity the powers. The derivative of $x^2$ is $2x$ which is of the same form of $x$ .

$u=x^2$

$du=2x dx$

$\int \frac{e^u}{2} du$

$\frac{e^u}{2} + C = \frac{e^{x^2}}{2} + C$

Now let's try substitution that isn't involving powers upon x...

Example 5.5-C

$\int (sin(x)+4)^6 cos(x) dx$

$u = sin(x) + 4$

$du = cos(x) dx$

$\int u^6 du$

$\frac{u^7}{7}+C$

$\frac{(sin(x)+4)^7}{7}+C$

**galactus** · June 20th 2008, 09:44 AM

I almost forgot to continue my lesson on fluid pressure. I will post several problems as I get time.

Let's start with a circular window on a submarine.

A circular observation window on a submarine has a radius of 1 foot, and the center of the window is 200 feet below the water surface. What is the fluid force.

We take advantage of our symmetry. So we will locate the origin at the center of the window.

The depth at y is then $h(y) = 200-y$

The horizontal length of the window is 2x, and you can use the equation

for the circle $x^{2}+y^{2}=1$ , to solve for x as follows:

Length = $2x = 2\sqrt{1-y^{2}} = L(y)$

Now, because y varies from -1 to 1, and using 64 pounds per cubic foot as

the weight density of seawater.

$F=w\int_{a}^{b}h(y)L(y)dy$

$=64\int_{-1}^{1}(200-y)(2)\sqrt{1-y^{2}}dy$

At first glance this looks tough to solve. But if we break it in two and apply our symmetry we get:

$F=25600\int_{-1}^{1}\sqrt{1-y^{2}}dy-128\int_{-1}^{1}y\sqrt{1-y^{2}}dy$

The second integral is 0 because the integrand is odd and the limits are symmetric about the origin.

But the first integral is the area of a semicircle of radius 1, so we have:

$25600(\frac{\pi}{2})=12800{\pi}\approx{40212.4} \;\ pounds$

So, the fluid force on the window is about 40,212.4 pounds.

Not that bad, huh?.

No doubt, math like this is used when designing windows on subs or any other underwater windows.

EDIT: SORRY, THAT SHOULD BE 200-Y ON THE GRAPH, NOT JUST 200.

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