For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

### 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

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### 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-y ^{2}).*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 - y ^{2}, 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-y ^{2}) 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-y ^{2}) dy.*0233

*Now, this integral unfortunately is not so pleasant.*0245

*We have to integrate sqrt(1-y ^{2}).*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 a ^{2} - u^{2} 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-y ^{2}),*0360

*Is y/2 × sqrt(1-y ^{2}) + 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-y ^{2}) 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/m ^{3}.*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) y ^{2}/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 y ^{2} dy.*0209

*That is 1960 × sqrt(3)/3y ^{3}.*0214

*The integral of y ^{2} is y^{3}/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 problem*0270

*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/m ^{3} × 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/s ^{2}.*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/m ^{2}×s^{2},*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/m ^{3},*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-y ^{2}).*0285

*The total width is l(y), which is twice that, which is 2×sqrt(1-y ^{2}).*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-y ^{2}) × 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 - y ^{2}, 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 u ^{1/2}.*0440

*The integral of that is u ^{3/2}, 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-y ^{2})^{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-y ^{2}, 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/m ^{2} per s^{2} 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-y ^{2})*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 educator.com and we will try a couple more examples later on.*0636

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)?