Problem 16 - Thomas Calculus

Question

Gasoline in a tank A gasoline tank is in the shape of a right circular cylinder (lying on its side) of length 10 \(\mathrm{ft}\) and radius 4 ft. Set up an integral that represents the volume of the gas in the tank if it is filled to a depth of 6 ft. You will learn how to compute this integral in Chapter 8 (or you may use geometry to find its value).

Step-by-Step Solution

Answer

Set up the integral for volume as: \( 10 \times \int_{-\sqrt{7}}^{\sqrt{7}} (16 \cos^{-1} (x/4) - x\sqrt{16-x^2}) \, dx \).

Step 1: Understand the Problem

We are asked to find the volume of gasoline in a right circular cylinder tank lying on its side with a given length, radius, and depth of liquid. The tank has a radius of 4 feet and a length of 10 feet, and it is filled to a depth of 6 feet.

Step 2: Visualize the Cross Section

Visualize the cross section of the cylindrical tank as a circle with radius 4 feet, and the horizontal line 6 feet above the lowest part (bottom) of the circle. We need the area of the circular segment determined by this horizontal line.

Step 3: Set Up the Coordinate System

Place the origin of a coordinate system at the center of the circular cross section. The equation for the circle is given by: \( x^2 + y^2 = 4^2 \), or \( x^2 + y^2 = 16 \).

Step 4: Identify the Limits of Integration

The line at depth y = 6 ft intersects the circle. To find the limits of integration, calculate where this line intersects the circle using the equation: \( x^2 + (y - 4)^2 = 16 \). Solving, we find that the line intersects when \( x = \pm\sqrt{7} \).

Step 5: Find the Area of the Circular Segment

The area of interest is the sector of the circle minus the area of the triangle formed by the radius and the horizontal line at the depth of 6 ft. The central angle \( \theta \) can be calculated using inverse trigonometry: \( \theta = 2\cos^{-1}(\sqrt{7}/4) \).

Step 6: Set Up the Integral

The volume of the cylinder filled to a depth of 6 feet is the product of the length of the cylinder and the integral of the cross-sectional area function. Thus,\[\text{Volume} = \text{Length} \times \int_{-\sqrt{7}}^{\sqrt{7}} (4^2 \cos^{-1} (x/4) - x\sqrt{16-x^2}) \, dx\]

Step 7: Simplify the Integral

Multiply the integral by the length of the tank, 10 feet:\[10 \times \int_{-\sqrt{7}}^{\sqrt{7}} (16 \cos^{-1} (x/4) - x\sqrt{16-x^2}) \, dx\].This integral represents the volume of gasoline in the tank filled to a depth of 6 ft.

Key Concepts

Right Circular Cylinder

A right circular cylinder is a geometric shape that is often seen in daily life. Imagine a basic soda or soup can. When a cylinder is described as "right," it just means that the sides are perpendicular to the bases, forming a perfect 90-degree angle.
In this problem, the gasoline tank is a cylindrical shape lying on its side. It has a length of 10 feet and a radius of 4 feet. When calculating volume, these dimensions help in understanding the amount of space the tank encloses. Since it's circular in cross-section, we use specific formulas and methods, like integration, to find the precise volume of gasoline when partially filled.

Circular Segment

A circular segment is the portion of a circle enclosed by a chord and the arc it subtends. In simpler terms, it's a 'slice' of the circle, similar to a pie slice, but with a straight edge along the chord.
For this problem, we're interested in the circular segment created by a horizontal line at a depth of 6 feet from the bottom of the cylinder. The task is to find the area of this segment because it represents the portion of the tank that is filled with gasoline. Calculating this area involves geometry and integration techniques to properly account for the curved section of the circle.

Integration Limits

Integration limits define the start and end points for an integral, essentially dictating the boundaries over which you calculate the area or volume.
Here, we first set up a coordinate system centered on the tank's circular cross-section. The equation of the circle is given by: \( x^2 + y^2 = 16 \). To find where this circle is intersected by the line corresponding to the gasoline's depth (6 feet), we rewrite the equation considering the y-coordinate shift. Solving for \( x \) provides the limits, \( -\sqrt{7} \) to \( \sqrt{7} \), crucial for evaluating the segment area.

Cross-Sectional Area

The cross-sectional area in this context refers to the area of the circular segment above the line at 6 feet depth within the cylinder. It's this area that, when multiplied by the tank's length, gives the total volume of gasoline.
We calculate this area by subtracting the area of the triangle formed between the circle's center, the chord, and the diameter from the entire sector of the circle. We use the formula for a sector and apply angle calculations using inverse trigonometry, particularly \( \theta = 2\cos^{-1}(\sqrt{7}/4) \), to get the precise area.

Coordinate System Setup

Setting up the right coordinate system is crucial for solving geometry and calculus problems meticulously. We align the system's origin to the center of the circular cross-section of the cylinder to simplify calculations and apply symmetrical properties of the circle.
The circle's equation \( x^2 + y^2 = 16 \) becomes the foundation for determining the intersection points and boundaries for the integral. With this arrangement, mathematical operations like calculating integration limits, segment areas, and volumes become more straightforward and structured, leading to an accurate solution.