 Hello, this is a video about fluid pressure and force, one of the many applications of integration. First, to define pressure, it's to find us the force per unit of area over the surface of a body. So the formula for this, the pressure p of an object at a depth h and a liquid is pressure equals w times h, remember h is the depth and w is the weight density of the liquid per unit of volume. This is fluid pressure on strictly a horizontal plane. Now with regard to fluid force on a horizontal plane, we use Pascal's principle, which states that the pressure exerted by a fluid at a depth h is transmitted equally in all directions. Fluid pressure is given in terms of force per unit area or pressure p equals force over area. This equation can be arranged to obtain the fluid force on a submerged horizontal surface, so fluid force f is equal to pressure times area. We will now use this equation in an example. An example one, find the fluid force on a rectangular metal sheet measuring three feet by four feet that is submerged in six feet of water. Water has a weight density of 62.4 pounds per cubic foot. So we begin with, from Pascal's principle, we have force equals pressure times area. So please note the following before I finish up this formula. We need the following information here. We need that, okay, pressure equals, from our original given formula for pressure, it's w times h. Remember, w is weight density. Weight density is 62.4 pounds per cubic foot, so that's 62.4. And then if you look at the depth, we're submerged six feet. So my depth, my h is six, then multiply the two together and we'll end up getting 374.4 per square foot. Only other thing we need would be the area. The area a of this rectangular sheet is simply just three times four, which would be twelve feet squared or twelve square feet. So that being said, my force on this horizontal plane is going to be 374.4 times twelve. So at the end of the day, this comes out to be 4492.8 pounds. So this is a fluid force on a horizontal plane. It's pretty straightforward. Where the calculus comes into play is when we're talking about fluid force on a vertical plane, because obviously the depth, which for instance, if you had a rectangle that you submerged in water vertically, based on the depth, the bottom portion of the rectangle is going to be impacted differently by force than the top portion of the rectangle because of how it's submerged in the water. The deeper you go, the pressure increases. So we have to account for that using some calculus here. So the force F exerted by a fluid of constant weight density W against a submerged vertical plane region from Y coordinates of C to Y coordinates of D is given by the integral. You have weight density. You integrate from C to D H of Y times L of Y with respect to Y. You need H of Y, which is the depth of the fluid at Y, and you need L of Y, which is the horizontal length of the region at Y. So the best way to understand this is to actually look at an example. So we have a vertical gate in a dam that has the shape of an isosceles trapezoid eight feet across the top and six feet across the bottom with a height of five feet. What is the fluid force on the gate when the top of the gate is four feet below the surface of the water? Water has a weight density of 62.4 pounds per cubic foot. Well, good news here is I do have my formula here for fluid force on a vertical plane. I know my weight density is 62.4, but we have a little bit of work to do here. So I already have a picture of a trapezoid on the screen here, and what we're going to do now is kind of draw this orientated on a coordinate plane. So I have a gate that the top of the gate is four feet below the surface, and I need to draw that here. So I'm going to draw my XY axis and that gate is one, two, three. The top of the gate is four feet below the water surface. So there it is right there. Next, the gate is five feet tall. So think about this. So that means the bottom of the gate is going to go, okay, if the top set negative four, the bottom would have to go all the way down to negative nine. So you have four, five, six, seven, eight, nine. That's five units. Next order of business. The top is eight feet across. So I have my top, it's eight feet wide, and then I have my bottom. My bottom is six feet wide. So obviously this bottom region is going to be having more force exerted on it than the top portion because it's deeper in the water. So what I have to do here, what the ultimate goal here is, is to partition this trapezoid gate into horizontal regions or horizontal chunks or disc with thickness delta Y. So this literally this vertical trapezoid is going to be broken up into infinitely many, many pieces with thickness or I guess you could say height delta Y. All right, so one thing that we're going to do here is we need to define, okay, we have W, the weight density. We need to find H of Y. We need to find that depth. So the depth H of Y is going to be, well obviously it varies based on where you are vertically on this gate. So actually the height is given by negative Y. So look at this. Whenever the Y coordinates negative four, plug it in for the depth equation here. You got negative, negative four, giving you positive four. That means you're four feet below the water. You plug in your negative nine. Negative negative nine is nine. You're at a depth of nine feet below the water. So that's your height equation. Now as far as your length, your horizontal length equation. The horizontal length will be dictated by, okay, from this left diagonal line of the trapezoid to the right diagonal line. In this picture, I can say that the length from the Y axis to the right hand diagonal line is given by X. For instance, if you're on the X axis and you're at four, well then that tells you you're four units Y. If you're on the X axis, you're at five. That means this right portion of the trapezoid at that particular disc or that particular increment, your length is five. So how do you find the length between the Y axis and the diagonal right hand line? Well, you know you can find the equation of a line if you have two points. So that's what we're going to do here. We have three negative nine and we also have four negative four. So we have, okay, equation of the line on the right hand side. This is the diagonal line. It has to be found by using the two points four negative four and three negative nine. So we can find the slope to write the equation of the line we need the point and slope. So the slope is going to be negative nine minus negative four over three minus four. That's going to give me negative five over negative one, which is five. So now you use point slip form to write the equation of the line. You'll have, okay, Y, I'll use the point three negative nine. Y minus negative nine equals five. Open parentheses X minus three. Now what my goal here is is the thickness of the right half of my region or the length I should say is going to be given if I solve this equation for X. So what is X in terms of Y? Because that's what my integration formula is. It's in terms of Y. Y plus nine equals five X minus 15. Then you're going to go ahead and you're going to subtract the nine from both sides. Y equals five X minus 24, which means X equals Y plus 24 over five. All right, so that's ultimately what your goal is. You might have done it slightly differently, but we know that X equals Y plus 24 over five, but X only represents the right hand half of any of the given regions. Two times X would have to represent the entire length of the whole partition here. So what I have here now is the equation L of Y equals two X, which is two times Y plus 24 over five. So that's two-fifths times Y plus 24. So that's everything we need for this formula. We have our W, our weight density. We have our H of Y, which is negative Y, and then we have our L of Y. So when I calculate the integral, I have forces equal to, you have 62.4. You're integrating. H of Y would have been negative Y, and then you have your L of Y, your length, which is two-fifths, two-fifths, Y plus 24. You're going to be integrating with respect to Y. So that's our start. That's our setup. It's now time to actually go and evaluate the integral. So once we've found the components, we're then able to set up the integral, and now we're going to evaluate. So here it goes. One thing that we do want to discuss is where do the bounds come from, and that goes back to our picture that this region goes down as far as Y equals negative nine. So Y equals negative four. So let me go ahead and write this in. You're integrating with respect to Y. That's why you're looking at where is this region defined along the Y axis. Negative line to negative four. All right, now let's evaluate. This is equal to, I'm going to pull the negative sign and the two out of five out front of the integral with the 62.4. We're integrating from negative nine to negative four. And then we'll go ahead and distribute the Y into the set of parentheses that follows, which is going to give me Y squared plus 24 Y. Negative 62.4 times two over five does give you negative 24.96. We can go ahead and integrate Y squared into Y cubed over three. And then 24 Y squared will become 24 Y squared over two. Remember your bounds Y equals negative nine to Y equals negative four. So you have negative 24.96. We can simplify just a little bit. Y cubed over three plus 12 Y squared. Y equals negative nine to Y equals negative four. From here, it's a matter of plugging in your upper bound, plugging in your lower bound and then subtracting. So Y cubed over three, that's negative four cubed over three plus 12 times negative four squared. That's plugging in the upper bound minus let's plug in the lower bound. Negative nine cubed over three plus 12 times negative nine squared. That's plugging in the lower bound. And what happens when you do finally get the chance to evaluate all of this? You'll end up getting negative 24.96. 170.67 minus 729. So at the end of the day, you're going to end up getting, after evaluating this, you will get a fluid force of 13,936 pounds. And that's how you calculate fluid force on a vertical plane. So the key is to break up that vertical plane into horizontal components, each with thickness delta Y. Remember delta Y goes to zero and that's how integration is born. So there you have it. Thanks for watching.