 And welcome to this quick recap of section 6.4, physics applications. We're going to begin with a recap of the big idea in this entire chapter. That is, in order to calculate a quantity that we're interested in, we can slice it up into very thin pieces, calculate the value of that piece, and then add them all up using an integral. We began by looking at the area under a curve, as in the top left here. In this case, we slice the area under the curve into very thin rectangles. Their width was called delta x, and their height was given by our function f of x. Multiplying those two quantities together, we got the area of one box, and then we used an integral to add up all of those boxes, getting an exact area. In a very similar way, we looked at the net distance traveled by an object with a given velocity, such as in the top right. Plotting the velocity curve, we sliced the area under it into very thin slices. Each of width, which had a thin width, we called delta t representing time, and a height, v of t representing velocity. Because velocity times time is distance, we were able to calculate the distance traveled by this object along one very thin slice. Again, we used an integral to add up all of these slices and get the net distance. We also did this in the case of some physics quantities, such as mass given by density. In the case of this thin rod in the bottom left, we had a variable density given by rho of x, and the volume of a very thin slice given by delta x. In order to find the mass of a slice, we multiplied the density times the volume of a thin slice and got the mass. An integral allowed us to add up all of these slices and get an exact total value for the mass. Finally, we've also looked at these in the case of three-dimensional objects. In the case of this cone in the bottom right, we sliced the cone into a very thin slice in the shape of a cylinder, and knowing both the area of the cylinder's surface and the thickness called delta x, we were able to get the volume of one thin slice. An integral let us add up all of those slices, getting the exact total volume. In each case, we multiplied two quantities that we were interested in, something given by a function and something representing the width or thickness of a very thin slice, and that gave us the new quantity that we were interested in. Now we're going to take a look at some physics quantities that we can use in a similar way. Our first one is work, which is defined as force times distance. For example, imagine lifting a bucket of water a certain distance. If the bucket has a certain constant mass of water, it has a constant weight, which is the same as a constant force. If we multiplied this by the distance that we wanted to lift it, we would get the work required to lift that water that distance. However, if the bucket is leaky, that means that the force, that is the weight of the water, varies depending on how far we've lifted it. In this case, we can't use the formula directly. However, we can apply the same tricks that we saw in the past slide. Imagine slicing the distance that we were going to lift this bucket into a very thin piece. We can call the thickness of this delta x, imagining that the x direction is going up and down in this picture. In this very thin slice, we can imagine that the bucket doesn't leak, and so the weight of water, that is the force, is constant for this very thin slice. That means that we can calculate the work to lift the water just that thin distance. So the work required to lift just this thin slice would be given by the force on this slice, which we can call a function of x. Since the water is leaking, it's a function depending on how far we've gone. And we multiply that by the distance, which we call delta x. Again, we're multiplying two quantities together, force and distance, and getting a third that we're interested in, work. That results in this result below. For a single object, such as a bucket of water, being moved along an object, an axis by a force that we'll call f of x, the total work required to lift that object is the integral from A to B of the force function with respect to x. Here A and B represent the lowest and highest points of our distance. So in our picture, we would go from A up to B. Again, we've sliced something into a very thin pieces, calculated a quantity we're interested in with that piece, and then used an integral to total that up and get an exact value. It's also possible that both force and distance can vary. For example, imagine this sump, which is a hole dug into a basement to collect water. Notice that the sump is not a nice symmetric shape, but rather its size varies from bottom to top. We want to pump all of the water in this sump out, and up to a certain level as represented by the pipes at the top here. In this case, both the distance and the force vary. If we imagine looking at a slice at the bottom of this picture, the water in that slice has a much larger distance to travel than a slice at the top. So the distance varies. In addition, the force varies. At the bottom of this, there's less water, and therefore less force. Whereas at the top of this sump, there's a greater amount of water because the sump is wider. So the force of one thin slice would be greater. In this case, in order to calculate the work done on one thin horizontal slice, which we would call delta x thick, we would have to multiply the force that is the weight of that slice times the distance that that individual slice must travel. And so we have this result. For a continuous object such as pumping water, the total work required to pump it all out is the integral from A to B of the force on one slice times the distance on one slice, dx. And again, A and B are the minimum and maximum distances that we would have to pump this water. Once again, we've added up many thin slices using an integral. Finally, we can look at force, which is here defined as pressure times area. An example where we might work with this is water pressing against the face of a dam. If this trapezoid is the face of a dam with water behind it, we can imagine slicing it into many thin horizontal pieces, such as this red rectangle. Each of the pieces is delta x thick. And on this thin slice only, we can imagine that the pressure exerted by the water is constant. That's because pressure depends on how deep you are below the top of the water, and this horizontal slice is effectively at a constant distance below the top of the water. In addition, the area of the slice is something that depends on how deep below the water we are. Using this, we could calculate the force on a slice to be the pressure on that slice times the area of that slice. And again, adding all of them up with an integral, we get the result that the total force exerted by the water on the dam is the integral from A to B of the pressure of a slice times the area of the slice. Now that we've seen these definitions, let's take a look at some of these in action.