Work To Empty Hemispherical Tank: A Calculus Problem

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Hey guys! Let's dive into a classic physics and calculus problem: calculating the work required to empty a hemispherical water tank. This is a super common question in introductory physics and calculus courses, and it's a great way to see how integrals can be used in real-world applications. We're going to break down the problem step by step, so you'll understand not just the answer, but why it's the answer.

Understanding the Problem

Our specific problem involves a hemispherical water tank with a radius of 10 meters. Imagine a giant bowl filled with water. The task is to figure out how much work is needed to pump the water out of the tank when the water level drops from 1 meter to 2 meters below the top (the "cusp") of the tank. The pump, in this case, is conveniently located right at the top of the tank. This simplifies our calculations slightly because the water needs to be lifted to that specific point.

Before we jump into the math, it's crucial to understand the concept of work in physics. Work is done when a force causes displacement. In simpler terms, it's when you push or pull something and it moves. Mathematically, work (W) is defined as the force (F) applied times the distance (d) over which it's applied: W = F * d. However, in our problem, the force isn't constant because we're dealing with a varying amount of water as the tank empties. This is where calculus comes to the rescue!

Setting Up the Calculus

To tackle this problem, we need to use integration. The basic idea is to divide the water in the tank into infinitesimally thin horizontal slices. We'll calculate the work required to lift each slice and then add up (integrate) the work done on all the slices. This is the essence of using calculus to solve problems with continuous change.

  1. Coordinate System: First, let’s set up a coordinate system. Imagine the hemisphere sitting with its flat side facing up and the curved part down. We'll place the origin (0, 0) at the center of the flat top. The y-axis will run vertically, with y = 0 at the top of the tank and y increasing downwards. This means the bottom of the hemisphere is at y = 10 meters (the radius).
  2. Slices of Water: Now, consider a thin horizontal slice of water at a depth of y meters from the top. Let's say this slice has a thickness of dy. This dy is an infinitesimally small change in y, which is perfect for our integration.
  3. Volume of a Slice: The key to this problem is figuring out the volume of this thin slice. Since it's a horizontal slice, it will be a circular disk. The volume of a disk is given by its area times its thickness. The thickness is dy, and the area is πr², where r is the radius of the circular slice.

Finding the Radius of the Slice

This is where a little bit of geometry comes in. We need to relate the radius of the slice (r) to the depth y. Think of a right triangle formed by:

  • The radius of the hemisphere (10 meters) as the hypotenuse.
  • The distance from the center of the hemisphere to the center of the slice (which is 10 - y) as one leg.
  • The radius of the slice (r) as the other leg.

Using the Pythagorean theorem (a² + b² = c²), we have:

r² + (10 - y)² = 10² r² = 10² - (10 - y)² r² = 100 - (100 - 20y + y²) r² = 20y - y²

So, the radius of the slice is r = √(20y - y²). This is a crucial step, so make sure you understand where this equation comes from.

Calculating the Volume and Weight

Now we can find the volume of the slice:

Volume (dV) = πr² dy = π(20y - y²) dy

Next, we need to find the weight of the water slice. Weight is force due to gravity (F = mg), and mass (m) is density (ρ) times volume (V). The density of water is approximately 1000 kg/m³, and the acceleration due to gravity (g) is about 9.8 m/s². Therefore:

Weight (dF) = ρg dV = 1000 kg/m³ * 9.8 m/s² * π(20y - y²) dy dF = 9800π(20y - y²) dy Newtons

Determining the Distance

The final piece of the puzzle is the distance the water slice needs to be lifted. Since the pump is at the top of the tank (y = 0), a slice at depth y needs to be lifted a distance of y meters. This is a straightforward part, but it’s important to get it right.

Calculating the Work for One Slice

Now we have all the ingredients to calculate the work (dW) required to lift a single slice:

dW = dF * distance = 9800π(20y - y²) dy * y dW = 9800π(20y² - y³) dy Joules

Integrating to Find the Total Work

To find the total work, we need to integrate dW over the limits given in the problem. The water level drops from 1 meter to 2 meters below the top of the tank, so we're integrating from y = 1 to y = 2:

W = ∫dW = ∫[1 to 2] 9800π(20y² - y³) dy W = 9800π ∫[1 to 2] (20y² - y³) dy

Now, let's evaluate the integral:

W = 9800π [ (20/3)y³ - (1/4)y⁴ ] [from 1 to 2]

W = 9800π [ (20/3)(2³) - (1/4)(2⁴) - (20/3)(1³) + (1/4)(1⁴) ] W = 9800π [ (160/3) - 4 - (20/3) + (1/4) ] W = 9800π [ (140/3) - (15/4) ] W = 9800π [ (560 - 45) / 12 ] W = 9800π [ 515 / 12 ] W ≈ 9800π * 42.9167 W ≈ 13200000 Joules or 13.2 MJ

So, the total work required to pump the water out of the tank is approximately 13.2 megaJoules. That’s a lot of energy!

Key Takeaways

  • Divide and Conquer: The key to solving these types of problems is to break them down into smaller, manageable pieces (the slices of water).
  • Calculus is Your Friend: Integration allows us to add up the work done on each slice, even though the force varies.
  • Geometry is Essential: Understanding the geometry of the situation is crucial for finding relationships between variables (like the radius of the slice and its depth).
  • Units are Important: Always keep track of your units to make sure your answer makes sense.

This problem highlights the power of calculus in solving real-world problems. Don't be intimidated by these kinds of questions. Break them down step by step, and you'll be surprised at what you can accomplish! If you have any questions, feel free to ask. Happy calculating!