Problem 43 - Calculus Single Variable

Question

A tank has the shape of a paraboloid of revolution that results from rotating about the \(y\) -axis the region that is bounded above by the horizontal line \(y=18,\) on the left by the \(y\) -axis, and on the right by the graph of \(y=2 x^{2}(x\) and \(y\) measured in feet). A pump floats on the surface of the water and pumps the water to the top of the tank. Calculate the work that is done in performing the task that is described. The tank, initially filled, is pumped until the remaining water is 3 feet deep at the center.

Step-by-Step Solution

Answer

Calculate the definite integrals, combine the results to find the total work done, then evaluate numerically to the desired precision.

Step 1: Understand the Problem

The problem involves a tank in the shape of a paraboloid. Initially, the tank is fully filled with water and is emptied until the depth of the water is 3 feet at the center. We need to calculate the amount of work done to pump the water up to the top of the tank.

Step 2: Parametrize the Shape of the Paraboloid

The equation of the parabola given is rotated about the y-axis: \[ y = 2x^2 \]. With this, the limits for integration are from 3 to 18 (initial to final water level height). The tank is bounded by y, so we write x in terms of y, as \[ x = \sqrt{\frac{y}{2}} \].

Step 3: Find the Volume of a Disk

Identify a small disk element at height y with a thickness dy. The radius of this disk is x, and thus its area is \( \pi x^2 = \pi \left( \frac{y}{2} \right) = \frac{\pi y}{2} \). The volume of the disk is the area times the thickness dy, or \( \frac{\pi y}{2} \, dy \).

Step 4: Calculate the Work Done for each Disk

The work done to move each disk to the top is the weight times the distance it needs to be lifted. The weight of the disk is \( \rho g \frac{\pi y}{2} \, dy \) where \( \rho \) is the density of water (62.4 lb/ft³), and \( g \approx 32 ft/s^2 \) (not necessarily needed here as we use gravity to relate to weight). The distance from a point y to the top is \( 18 - y \). So, the work done for each disk is: \[ dW = \rho g \frac{\pi y}{2} (18-y) \, dy \].

Step 5: Integrate Over the Height of the Water

To find the total work, integrate \( dW \) from 3 to 18: \[ W = \int_3^{18} 62.4 \cdot \frac{\pi y}{2} (18-y) \, dy \].

Step 6: Solve the Integral

Simplify and solve the integral: \[ W = 31.2\pi \int_3^{18} (18y - y^2) \, dy \]. This evaluates into two separate integrals. Calculate them individually to find: \[ W = 31.2\pi \left[ \int_3^{18} 18y \, dy - \int_3^{18} y^2 \, dy \right] \]. Once calculated, combine the results to achieve the final work value.

Step 7: Calculate the Result

Computing the definite integrals and simplifying: \[ \int_3^{18} 18y \, dy = 9y^2 \vert_3^{18}, \quad \int_3^{18} y^2 \, dy = \frac{y^3}{3} \vert_3^{18} \].Calculate these, substitute back into the main equation, and evaluate to obtain the final work done.

Key Concepts

Work in Calculus

In calculus, the concept of work is crucial for understanding how objects move and how energy is transferred. When dealing with problems like pumping water out of a tank, calculus helps us determine the amount of "work" required to perform this task. Work is essentially the force applied over a distance. In the context of our tank, this means lifting the weight of the water against gravity.

The work done is calculated by multiplying the weight of the water elements (disks) by the distance they are lifted. Here, weight is determined by the volume and density of water. The key point is that each small disk of water must be moved to the tank's top. The force needed is critical, making calculus an invaluable tool in these calculations. By breaking down the task into small elements and adding them up using integration, we get the total work done.

Integration Techniques

Integration plays a vital role in solving the work problem, particularly using the limits of integration related to physical boundaries. When we talk about integration techniques in calculus, we mean the methods used to calculate the area under the curve or the sum of infinite small parts. In this case, those small parts represent disks of water inside the tank.

• First, a small disk at height y with thickness dy is considered; its work contribution is then calculated.
• By integrating this over the water height from 3 to 18 feet, we can find the total work.

This involves setting up the integral \[W = \int_3^{18} 62.4 \cdot \frac{\pi y}{2} (18-y) \, dy\] which helps compute the total work across all disks. The use of integration is critical as it handles the continuous nature of the problem, summing up the effort for each water disk from the bottom to the top of the tank.

Paraboloid of Revolution

A paraboloid of revolution is a three-dimensional shape formed by rotating a parabola around its axis. This exercise involves a tank shaped like a paraboloid, which makes it distinct from a regular cylinder or sphere. Understanding this shape is essential, as the geometry affects how we calculate the volume and eventually the work.

• In our problem, the parabola described by the equation \[ y = 2x^2 \] rotates around the y-axis, creating the tank's walls.
• By expressing \( x \) as a function of \( y \), we find \( x = \sqrt{\frac{y}{2}} \), allowing us to define the disk's radius at any height \( y \).
• This shape influences our integration from 3 to 18 feet, representing the varying width and area over the height.

Understanding the paraboloid's properties helps solve the problem by accurately modeling the physics and geometry involved. It ensures that each small disk representing a volume slice is correctly accounted for in the total work calculation.