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Lecture Comments (12)

1 answer

Last reply by: Dr. William Murray
Thu Nov 20, 2014 3:26 PM

Post by David Llewellyn on November 19, 2014

Why in the example are you integrating between 0 and 1?  Should it not be between -1 and 0 given the convention of the xy axes or is the D function increasing in the -y direction?

1 answer

Last reply by: Dr. William Murray
Sun Nov 18, 2012 3:56 PM

Post by ahmed alzeory on November 18, 2012

oh never mind silly question

1 answer

Last reply by: Dr. William Murray
Sun Nov 18, 2012 3:54 PM

Post by ahmed alzeory on November 16, 2012

wait how is 1-0 -1 i don't get that

2 answers

Last reply by: Dr. William Murray
Sun Oct 21, 2012 5:19 PM

Post by Jason Erickson on May 23, 2012

Someone needs to answer the question by Jess Wood - I don't understand what he did either.

2 answers

Last reply by: Dr. William Murray
Sun Oct 21, 2012 5:17 PM

Post by Jess Wood on October 15, 2011

At about 1:15 how did he get x/2 as the bottom of the traingle? And what is the trig rule that allows the sqrt(3) to be in front of the (x/2)?

Hydrostatic Pressure

Main formula:

Force due to Hydrostatic Pressure =

Hints and tips:

  • Check the units used in the problem carefully. Most commonly, metric units are used for all quantities in the problem, and your answer will be in terms of Newtons, i.e. kg m/s² . Less often, English units will be used and your answer will be in terms of pounds, lb. Either way, the units should multiply and divide just as numbers do, which partially checks that you are doing the arithmetic correctly.

  • These are usually fairly complicated word problems. Always read the problem carefully and start by drawing a picture of the surface.

  • Usually the depth function D(y) is given by y, y + c, y − c, or c − y where c is some constant. Which one it is depends on how you orient your coordinates in the picture.

  • The limits a and b correspond to the highest and lowest levels. Again, this depends on how you orient your coordinates in the picture.

  • When it’s feasible, check that your answer makes sense. The amount of force due to hydrostatic pressure should always be positive!

Hydrostatic Pressure

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Important Equation 0:09
    • Main Idea
    • Different Forces
    • Weight Density Constant
    • Variables (Depth and Width)
  • Lecture Example 1 3:28
  • Additional Example 2
  • Additional Example 3

Transcription: Hydrostatic Pressure

We are working on some more examples of calculating the force due to hydrostatic pressure.0000

In this example, we are using the same situation that we had in an earlier example.0006

Let us recall that there we had a semicircular plate of radius 1.0013

We worked out last time that a depth of y, then the width there,0024

Would be l(y) = 2 × sqrt(1-y2).0034

Well we are changing things up in this example.0042

In this example, the plate is now submerged so that its diameter is below the surface of water.0043

Instead of being flush to the surface of water, we have the water up here.0053

The plate is d meters below the surface of the water.0058

Remember the whole point of force due to hydrostatic pressure is that,0063

When you submerge something deeper in the water, there is more force deeper down in the water.0070

Because you have got more water piled up on top of it.0075

There is more pressure at lower depths.0079

So, let us try to figure out how having the thing deeper in the water changes what our answer is going to be.0082

Again, we are going to set up our integral formula, which remember is w × integral from our 1 value of y to another,0090

of l(y) × d(y) dy.0103

So again, we are going to run our y from y=0 to y=1.0111

But our l(y) is still 2 × 1 - y2, because the width of the thing has not changed.0122

What has changed is the depth, the d(y).0130

What is new in this problem is the d(y),0134

Remember before was just y,0139

But now, it is y + d.0142

So I am going to set up the integral, just like before, except that we have y + d instead of y.0146

Our w, remember that was the weight density of water.0153

We got that by multiplying the density of water × the acceleration due to gravity.0157

The w was the density of water × acceleration due to gravity.0163

That is all the same.0170

That is 1000 × 9.8, that is 9800.0172

What we get here is the force is w, 9800, × the integral still from y=0, y=1.0176

Of 2 ×, instead of y, y + d.0188

× the sqrt(1-y2) dy.0194

Now if you look at this integral, it is quite similar to what we had before, in particular the y term is the same,0204

As in the earlier example.0212

We worked that out before.0216

We worked that out to be 19,600/3 Newtons.0219

That is what the y term gives you.0226

But we still have the rest of the integral to solve.0228

9800 × the integral from y=0 to y=1 of 2d × sqrt(1-y2) dy.0233

Now, this integral unfortunately is not so pleasant.0245

We have to integrate sqrt(1-y2).0252

How do we handle that?0255

We now learned several different techniques where how to do that integral.0258

You can use a trig substitution on that.0267

Where you would use y = sin(θ).0274

We have learned how to do that on the chapter on trig substitution.0279

You can also look up the answer in an integration table in the back of your textbook.0284

You will see in the back of your textbook how to find the integral of a2 - u2 du.0291

And, it will give you a formula to do that.0299

You can use either one of these techniques, trig substitution, or an integration table.0302

To do that integral.0308

What it converts into is,0310

I am just going to show the integration part,0317

The first part the nice 19,600/3, that is going to stay the same.0321

I am going to pull the 2d and then 9800 out of the integral and combine them.0324

So, we get 19,600d.0332

I am not going to go through the process of looking up this integral in a table,0340

Or going through the trig substitution, because we have already worked through those in previous lectures.0349

I am just going to go ahead and say what the answer is according to a table in a calculus book.0354

The answer for the integral of sqrt(1-y2),0360

Is y/2 × sqrt(1-y2) + 1/2 arcsin(y).0362

Remember we have to integrate this from y=0 to y=1.0380

So, we get 19,600d.0388

If we plug in y=1 to the first term, that sqrt(1-y2) goes away and gives you 0.0395

The second term gives us 1/2 arcsin(1),0404

So, 1/2 arcsin(1).0410

We have to remember what arcsin(1) is.0415

If we plug in y=0, that first term, the y/2 part of it, gives you 0.0419

Then we have -1/2 arcsin(0).0425

arcsin(0) is just 0, because sin(0) is 0.0436

arcsin(1), you have to think ok, what angle has sin(1).0442

Well, the answer is pi/2.0446

So this turns into pi/2.0448

So we get 19,600 d × pi and we have still got this 1/2 × pi/2.0454

So 1/2 × pi/2 gives us pi/4.0465

Let me now bring in this other term that we had at the beginning.0468

We get 19,600/3 + 19,600/4 is 4900 d × pi.0473

This whole answer is in terms of Newtons.0491

Let us review what made that example work.0505

First we drew a picture of the plate that was submerged in the liquid.0508

That was very important.0513

We figured out what the depth function was, that was y + d, because it made sense to run this from y=0 to y=1.0515

So our depth function was y + d.0524

We figured out what the width function was,0529

Actually we brought that over from the previous problem.0532

But that was something we figured out in the previous problem,0536

So we figured out the depth function, the width function, 0539

Then we figured out the weight density of water by taking the density of water × the acceleration due to gravity.0545

We plugged all of those into our formula for force due to hydrostatic pressure,0549

Then we worked through the integral,0555

Which was not too bad once we are armed with all of the integration techniques we learned earlier.0556

Then we got our answer.0564

Density is 300 kg/m3.0000

It says the plate is oriented vertically with 1 corner at the top flush with the surface of the fluid.0004

Again the first thing to do here is draw a picture and try to figure what is going on.0012

So, let me draw a picture.0017

Here is the top of the fluid, so the surface of the fluid.0020

The triangular plate is 2m on each side and it has a corner at the top there.0025

So there is our plate.0029

Again, we are going to use our formula for force due to hydrostatic pressure.0034

We can use the depth again to be y.0044

So d(y) = y.0050

For the width, we have to figure out what the width is at any particular value of y.0053

What we can do there is say well if the width is x, 0061

Then we can use our formulas for triangles.0066

What we know here is that x/2, if we look at that 30-60-90 triangle,0074

This is x/2 and that is y,0080

x/2 × sqrt(3) = y, using what we remember about relationships between sides in 30-60-90 triangles.0083

So, the width is x = 2y/sqrt(3).0092

So, the width, l(y) is equal to, if I rationalize that, I get 2×sqrt(3y/3).0103

Remember the w is the weight density, and that is the density of the fluid × acceleration due to gravity.0118

That is 300 × 9.8, which comes out to be 2940.0128

So, our integral is 2940 × the integral as y goes from, 0133

OK that is y=0,0144

Now, this bottom part is.0151

Ok, again we have to remember our formulas for triangles.0157

So if that is 2 meters, then half of that is 1 meter, and so the total depth is sqrt(3).0163

So y = sqrt(3).0176

Then the integral of l(y) × d(y), so 2 sqrt(3) y2/3 dy.0183

If we square the 2 sqrt(3)/3 and the 2940, we get 1960sqrt(3).0198

Times the integral from 0 to sqrt(3) of y2 dy.0209

That is 1960 × sqrt(3)/3y3.0214

The integral of y2 is y3/3.0222

Evaluate that from y=0 to sqrt(3).0226

We get 1960 × 3sqrt(3) × sqrt(3)/3.0230

That simplifies down to 1960 × 3, so 5880 Newtons.0243

I did the integration at the end there a little bit quickly.0257

The key part there is setting up the formula, finding expressions for d(y) and l(y),0260

Plugging them into the formula,0267

Then it just turns into an integration problem0270

Today we are going to learn about an important application of integration. 0000

We are going to use integration to calculate the force due to hydrostatic pressure.0004

Let me explain the situation there.0009

The idea there is that we have some kind of think plate submerged in a liquid.0012

This could be a thin plate or it could be the wall of a dam, or anything that is enclosing some fluid.0030

We are trying to calculate how much force the fluid puts on that thin plate.0038

The key point here is that the fluid does not put so much force on it, but when it is deeper down, 0044

There is more fluid piling up and pushing against that plate or against that wall and so there will be a greater force.0053

We are going to calculate this using this integral formula but I have to explain what each of these terms means.0062

This W outside represents the density of the fluid, which is something you would measure in kg/m3 × the acceleration due to gravity.0069

That of course is always the same, and we will use the value here, we will round that to 9.8 m/s2.0100

We are using metric units but of course you could also use English units.0113

That W when you multiply them together, what you get is something that has units of kg/m2×s2, 0115

That is called the weight density of the fluid.0126

That is what that W represents, and that will be a constant.0137

What this d(y) and l(y) represent are, if this is plate that is submerged in the fluid, what the d(y) and l(y) represents are at any given point, 0141

D(y) represents the depth and l(y) represents the width of the plate at that particular depth.0160

So l(y) is the width, and d(y) is the depth.0194

We will use this formula to calculate the total force due to hydrostatic pressure on a plate.0197

Let us try an example.0207

We are given a semi-circular plate of 1 m submerged in water so that its diameter is level with the surface to the water.0208

Let me start by graphing that.0217

Here is the surface of the water.0221

It is a semi-circular plate whose plate is level with the surface of the water and we know that the radius is 1m.0223

We want to find the force due to hydrostatic pressure on the plate and we are given that water has a density of 1000 kg/m3, 0233

And of course we know the acceleration due to gravity.0245

What we want to do is find the width and depth of the plate at different depths.0249

Let me assume that we are at depth y, so we are measuring y from the surface of the water down.0264

I want to try to find the width of the plate at that particular depth.0271

The way I can figure that out is to remember that this is a semi-circular plate that has radius 1, so that radius is 1.0275

That width there, well, that little area just halfway across the width is going to be the sqrt(1-y2).0285

The total width is l(y), which is twice that, which is 2×sqrt(1-y2).0298

The depth, which is d(y), is just y itself.0305

So, let us fill in our formula for hydrostatic pressure.0310

Remember, W is the density of the fluid, which in this case is water × the acceleration due to gravity.0317

In this case the density is 1000, the acceleration due to gravity is 9.8, and so the whole term for W is 9800.0329

The total force is W, so 9800, times the integral from y = 0 to y = 1,0347

Because those are the smallest and largest values of y that we are going to see in this plate.0363

So integrate from y=0 to y=1, of l(y) which is 2sqrt(1-y2) × d(y), which is just y dY.0370

Now we need to integrate that.0389

That is not such a bad integral because we can notice that if we let u be 1 - y2, then dU is -2y dY, 0395

Which we almost have except for the negative sign.0407

We just about have 2y dY there.0409

I will just pull that negative sign outside, and we get nevative 9800 × the integral from y=0 to y=1 of the sqrt(u)×dU.0413

This is -9800, now the integral of the sqrt(u), you think of that as u1/2.0440

The integral of that is u3/2, divided by 3/2, which is the same as multiplying by 2/3 so put a 2/3 in there.0450

Then we want to integrate this from y=0 to y=1.0468

We want to be careful about keeping the u's and y's straight.0474

To make it easy I will substitute everything back into y's.0477

So, we get -9800 × 2/3.0481

Now u was (1-y2)1 evaluated from y=0 to y=1.0486

This is then -9800 × 2/3.0500

Now if you plug in y=1, you just get 0. if you plug in y=0 then you just get -1.0509

But we are subtracting that because that was the lower limit, so we get 0 - (-1), or sorry, 0520

If you plug in y=0 to 1-y2, you just get 1 so that gets subtracted.0530

So, we get 0-1, and then that negative sign cancels out with the negative sign we had on the outside.0539

What we finally get is 9800 × 2/3.0545

9800 × 2 = 19600/3, and remember our units here were kg/m2 per s2 which are Newtons, 0552

And so our answer here is in Newtons.0567

The trick there is to make a graph of the situation that you are looking at.0579

In this case we had semi-circular plates, so you draw the semi-circular plates, We figure out what our W is, that is density always a constant.0586

Times gravity, which is always 9.8 assuming you are using metric units.0595

Then we figure out d(y) is y and l(y) is the width, that was a bit trickier, 0600

We had to do a bit of work with the graph to figure out that was 2 × sqrt(1-y2)0607

Then we plug the whole thing into our integral formula for the force due to hydrostatic pressure.0614

Then we work it through and we get our answer.0620

A couple of notes here, we will be using this same example in another example later on.0624

Hang on to this answer, 19600/3 Newtons, and we will come back to it later.0629

In the meantime, this has been Will Murray for and we will try a couple more examples later on.0636