Problem 66 - Calculus of a Single Variable

Question

A tank on a water tower is a sphere of radius 50 feet. Determine the depths of the water when the tank is filled to one-fourth and three-fourths of its total capacity. (Note: Use the zero or root feature of a graphing utility after evaluating the definite integral

Step-by-Step Solution

Answer

The water depths for one-fourth and three-fourths capacity can be found by solving two definite integrals. Detailed solutions will depend on a student's specific method for evaluating these integrals.

Step 1: Understand the problem and set up the integral

The volume V of a sphere of radius r is \( V = \frac{4}{3}\pi r^{3} \). When the sphere is filled to one-fourth or three-fourths of its capacity, the tank's volume is \( \frac{1}{4}V \) and \( \frac{3}{4}V \) respectively. The equation for the volume of water (as a function of height h from the bottom of the tank) can be written as \( v(h) = \int_{-r}^{h-r} \pi (r^{2}-x^{2}) dx \), where x is the distance from the center of the sphere (x=0 is the center of the sphere)

Step 2: Solve the integral for 1/4 of the total volume

The tank is filled to 1/4 of its total volume, which gives \( \frac{1}{4}\frac{4}{3}\pi r^{3} = \frac{1}{3}\pi r^{3} \). Solve the integral \( \int_{-r}^{h-r} \pi (r^{2}-x^{2}) dx \) with respect to h to find h when the tank is filled to 1/4 of the total volume.

Step 3: Solve the integral for 3/4 of the total volume

Repeat the process from step 2, but this time for 3/4 of the total volume. That gives \( \frac{3}{4}\frac{4}{3}\pi r^{3} = \frac{1}{2}\pi r^{3} \). Solve the integral again to find h when the tank is filled to 3/4 of the total volume.

Key Concepts

Definite Integral

In the world of calculus, the definite integral is a powerful tool used to calculate the accumulated quantity, such as area under a curve or, as in our case, volume within a certain boundary. To understand it better, let's think of the definite integral as a method that allows us to accumulate infinitesimally small amounts of the function (like layers) between two points.
This problem involves calculating how much water is in a spherical tank, using definite integrals to find the volume filled at certain depths. The setup for the tank's volume requires integrating from one end to another, here from bottom -\((-r)\) to a certain height -(\(h-r\)) of the sphere. The integral equation -\( v(h) = \int_{-r}^{h-r} \pi (r^{2}-x^{2}) dx \) gives us the necessary framework. By solving this definite integral, we find the function representing volume with respect to the height of water in the sphere.
This function determines when the sphere reaches one-fourth or three-fourths of its full capacity. The beauty of definite integrals is in how they help us determine these relationships effectively and with mathematical accuracy.

Sphere Geometry

Geometry plays an essential role in understanding the attributes and relationships of shapes. In the context of our exercise, a sphere is a 3D object where every point on its surface is equidistant from the center. The radius, which is this distance, is crucial for understanding volume.
The volume -is calculated by the formula -\( V = \frac{4}{3}\pi r^{3} \). Spheres have properties that make their volume calculations distinct from other shapes. Thus, for a tank shaped like a sphere, determining how full it is at certain levels, such as one-fourth or three-fourths filled, involves calculating fraction parts of this total volume.
By using sphere geometry, we can derive the equation for how much water (volume) is in the tank at different depths, relying on both the spherical shape properties and calculus principles.

Graphing Utility

A graphing utility is a technological tool that enhances our ability to visualize and solve mathematical problems. In tackling the sphere volume problem, a graphing utility can simplify complex calculations and graphical representations. The root feature of these tools is particularly useful.
By inputting the derived integral functions representing water volume at different depths, you can graph them on a graphing utility. This allows you to visually analyze and confirm solutions. It simplifies finding points where the specific volumes -\( \frac{1}{4} \) and -\( \frac{3}{4} \) of total volume occur. The clear data in graphical form aids in understanding the relationships between variables like volume and depth.

• Graphing utilities reveal solutions that are hard to see visually otherwise.
• They allow easier manipulation of functions to check for errors or insights.

Thus, they are invaluable in teaching and solving mathematical concepts involving complexities such as definite integrals and geometry.