Calculus Always Slays 🔥

Today, we are going to solve a problem that involves classic application of calculus. A conical tank has a height of 10 meters and a top radius of 4 meters. The tank is filled with water up to a depth of 8 meters, leaving the top 2 meters empty. Our goal is to calculate the total work required to pump all the water to the top edge of the tank, so that it just flows out. So, can you solve it? Okay, whenever we see a work problem like this, remember the basic idea. Work is equal to force multiplied by distance. But here's the thing though. Since different layers of water have to travel different distances, we cannot simply use one value of the distance for the entire tank. Instead, we will have to divide the water into many very thin horizontal layers, then calculate the work required for one layer, and then add the work of all the layers using integration. Let us first determine the work required to move one thin layer of water. The force acting on this layer is simply its weight. Now, weight is obtained by multiplying the mass of the liquid with gravitational acceleration g. And generally in the case of liquid, we calculate its mass by multiplying the density of the liquid by its volume. Therefore, the work done for one thin layer can be written as distance multiplied by density multiplied by volume times g. Before moving ahead, we need to be careful about the units. Since all the dimensions are given in meters, we are working in SI units. The mass density of water is 1,000 kilograms per cubic meter, and g is 9.8 meters per second squared. Therefore, these terms in our expression become 1,000 multiplied by 9.8. Now let us calculate the volume of one thin layer of water. Imagine making a very thin horizontal cut through the water. This produces a thin circular disk, which can be treated as a tiny cylinder. You can think of it as a CD, which is a circular disk, and the thickness is extremely small. Now, let us place the y-axis vertically through the center of the cone with the origin at the tip, and the x-axis horizontally through the origin. This way, every horizontal slice is at a height, y. While its radius is simply the horizontal distance, x, from the center to the edge of the cone. Now the volume of a cylinder is given as pi times radius squared times its thickness. If the thickness of this water cylinder is assumed to be dy, then its volume is simply pi times x squared times dy. Great. Next, let us determine the distance this thin layer has to be lifted. As mentioned before, the layer is at a height, y, meters above the bottom of the tank. Since the total height of the tank is 10 meters, this layer must be lifted from height, y, all the way to height, 10. Therefore, the distance traveled by this layer is simply this, or 10 minus y. Putting everything together, the work required for one thin layer becomes 10 minus y, multiplied by 1,000, multiplied by 9.8, multiplied by pi x squared, multiplied by dy. Now, of course, the tank contains infinitely many such thin layers stacked one above another. The total work will therefore be the sum of the work done for every layer. Instead of adding them individually, we use integration. Before writing the integral, we only have two things left to determine, the limits of integration and the relationship between x and y. Let us first determine the limits of integration. We place the origin at the tip of the cone, so the bottom of the tank corresponds to y equals 0. The total height of the tank is 10 meters, but the top 2 meters are empty. Therefore, the water occupies only the region from y equals 0 to y equals 8. Hence, our integral will be evaluated from 0 to 8. The only unknown remaining in our expression is the radius x, because this integral is evaluated in terms of y, and thus we cannot have x in here. Since the cone is perfectly symmetric about its vertical axis, consider the entire cone. Its height is 10 meters, and its radius at the top is 4 meters. Now consider any smaller cone formed from the tip up to a height y. The height of this smaller cone is y, and its radius is x. Since both cones have the same shape, they are similar triangles. Therefore, the ratio of the radius to the height must remain the same. So we get x divided by y equals 4 divided by 10, which simplifies to 2 divided by 5. Finally, we get x equals 2 by 5 times y. We now substitute this into our work expression. Hence, the required work is the integral from 0 to 8 of this. Now, all that remains is to evaluate this integral. I will leave the calculation to you. Solve it, comment your final answer below, and if it's correct, you'll get a heart from me. Like, share, and subscribe. So good!