WEBVTT mathematics/calculus-ii/murray
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Today we are going to learn about an important application of integration.
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We are going to use integration to calculate the force due to hydrostatic pressure.
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Let me explain the situation there.
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The idea there is that we have some kind of think plate submerged in a liquid.
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This could be a thin plate or it could be the wall of a dam, or anything that is enclosing some fluid.
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We are trying to calculate how much force the fluid puts on that thin plate.
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The key point here is that the fluid does not put so much force on it, but when it is deeper down,
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There is more fluid piling up and pushing against that plate or against that wall and so there will be a greater force.
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We are going to calculate this using this integral formula but I have to explain what each of these terms means.
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This W outside represents the density of the fluid, which is something you would measure in kg/m³ × the acceleration due to gravity.
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That of course is always the same, and we will use the value here, we will round that to 9.8 m/s².
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We are using metric units but of course you could also use English units.
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That W when you multiply them together, what you get is something that has units of kg/m²×s²,
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That is called the weight density of the fluid.
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That is what that W represents, and that will be a constant.
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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,
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D(y) represents the depth and l(y) represents the width of the plate at that particular depth.
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So l(y) is the width, and d(y) is the depth.
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We will use this formula to calculate the total force due to hydrostatic pressure on a plate.
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Let us try an example.
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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.
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Let me start by graphing that.
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Here is the surface of the water.
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It is a semi-circular plate whose plate is level with the surface of the water and we know that the radius is 1m.
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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,
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And of course we know the acceleration due to gravity.
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What we want to do is find the width and depth of the plate at different depths.
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Let me assume that we are at depth y, so we are measuring y from the surface of the water down.
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I want to try to find the width of the plate at that particular depth.
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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.
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That width there, well, that little area just halfway across the width is going to be the sqrt(1-y²).
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The total width is l(y), which is twice that, which is 2×sqrt(1-y²).
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The depth, which is d(y), is just y itself.
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So, let us fill in our formula for hydrostatic pressure.
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Remember, W is the density of the fluid, which in this case is water × the acceleration due to gravity.
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In this case the density is 1000, the acceleration due to gravity is 9.8, and so the whole term for W is 9800.
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The total force is W, so 9800, times the integral from y = 0 to y = 1,
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Because those are the smallest and largest values of y that we are going to see in this plate.
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So integrate from y=0 to y=1, of l(y) which is 2sqrt(1-y²) × d(y), which is just y dY.
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Now we need to integrate that.
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That is not such a bad integral because we can notice that if we let u be 1 - y², then dU is -2y dY,
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Which we almost have except for the negative sign.
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We just about have 2y dY there.
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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.
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This is -9800, now the integral of the sqrt(u), you think of that as u^1/2.
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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.
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Then we want to integrate this from y=0 to y=1.
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We want to be careful about keeping the u's and y's straight.
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To make it easy I will substitute everything back into y's.
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So, we get -9800 × 2/3.
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Now u was (1-y²)¹ evaluated from y=0 to y=1.
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This is then -9800 × 2/3.
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Now if you plug in y=1, you just get 0. if you plug in y=0 then you just get -1.
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But we are subtracting that because that was the lower limit, so we get 0 - (-1), or sorry,
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If you plug in y=0 to 1-y², you just get 1 so that gets subtracted.
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So, we get 0-1, and then that negative sign cancels out with the negative sign we had on the outside.
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What we finally get is 9800 × 2/3.
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9800 × 2 = 19600/3, and remember our units here were kg/m² per s² which are Newtons,
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And so our answer here is in Newtons.
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The trick there is to make a graph of the situation that you are looking at.
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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.
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Times gravity, which is always 9.8 assuming you are using metric units.
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Then we figure out d(y) is y and l(y) is the width, that was a bit trickier,
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We had to do a bit of work with the graph to figure out that was 2 × sqrt(1-y²)
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Then we plug the whole thing into our integral formula for the force due to hydrostatic pressure.
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Then we work it through and we get our answer.
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A couple of notes here, we will be using this same example in another example later on.
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Hang on to this answer, 19600/3 Newtons, and we will come back to it later.
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In the meantime, this has been Will Murray for educator.com and we will try a couple more examples later on.