Problem 58
Question
An auxiliary fuel tank You are designing an auxiliary fuel tank that will fit under a helicopter's fuselage to extend its range. After some experimentation at your drawing board, you decide to shape the tank like the surface generated by revolving the curve \(y=1-\left(x^{2} / 16\right),-4 \leq x \leq 4,\) about the \(x\) -axis (dimensions in feet). a. How many cubic feet of fuel will the tank hold (to the nearest cubic foot)? b. A cubic foot holds 7.481 gal. If the helicopter gets 2 \(\mathrm{mi}\) to the gallon, how many additional miles will the helicopter be able to fly once the tank is installed (to the nearest mile)?
Step-by-Step Solution
Verified Answer
The tank holds 42 cubic feet of fuel, allowing the helicopter to fly an additional 628 miles.
1Step 1: Understanding the Volume Formula for a Solid of Revolution
To find the volume of a solid of revolution generated by rotating the curve about the x-axis, we use the formula for the volume of a solid of revolution: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]Here, the function given is \( y = 1 - \frac{x^2}{16} \) and the interval is \([-4, 4]\).
2Step 2: Setting Up the Integral for the Volume
Substitute \( f(x) = 1 - \frac{x^2}{16} \) into the volume formula:\[ V = \pi \int_{-4}^{4} \left(1 - \frac{x^2}{16}\right)^2 \, dx \]Now, we will expand the integrand \((1 - \frac{x^2}{16})^2\).
3Step 3: Simplifying the Integrand
Expand \( (1 - \frac{x^2}{16})^2 \):\( \left(1 - \frac{x^2}{16}\right)^2 = 1 - \frac{x^2}{8} + \frac{x^4}{256} \)Therefore, the integral becomes:\[ V = \pi \int_{-4}^{4} \left(1 - \frac{x^2}{8} + \frac{x^4}{256}\right) \, dx \]
4Step 4: Evaluating the Integral
Evaluate the integral:\[ V = \pi \left[ x - \frac{x^3}{24} + \frac{x^5}{1280} \right]_{-4}^{4} \]Calculate: - At \(x = 4\), it becomes \(4 - \frac{4^3}{24} + \frac{4^5}{1280}\).- At \(x = -4\), it becomes \(-4 + \frac{(-4)^3}{24} - \frac{(-4)^5}{1280}\).Compute these values to find the volume.
5Step 5: Calculating the Definite Integral
Evaluating the expression from Step 4 for \(x = 4\) and \(x = -4\):- At \(x = 4\):\(4 - \frac{64}{24} + \frac{1024}{1280} = 4 - 2.67 + 0.8 = 2.13\)- At \(x = -4\):\(-4 + \frac{-64}{24} - \frac{-1024}{1280} = -4 + 2.67 - 0.8 = -2.13\)Therefore, the definite integral is:\[ V = \pi (2.13 - (-2.13)) = \pi \times 4.26 \approx 13.38\pi \approx 42 \text{ cubic feet}\]
6Step 6: Converting Volume to Gallons
Convert the volume from cubic feet to gallons using the conversion factor:\[ 1 \text{ cubic foot} = 7.481 \text{ gallons} \]Thus, the volume in gallons is:\[ 42 \text{ cubic feet} \times 7.481 \text{ gallons/cubic foot} = 314.202 \text{ gallons} \approx 314 \text{ gallons} \]
7Step 7: Estimating Additional Miles
Since the helicopter gets 2 miles per gallon, the additional miles provided by the auxiliary tank is calculated as:\[ 314 \text{ gallons} \times 2 \text{ miles/gallon} = 628 \text{ miles} \]Therefore, the helicopter can fly approximately 628 additional miles.
Key Concepts
Volume of a SolidIntegration TechniquesDefinite IntegralFuel Efficiency Calculations
Volume of a Solid
The concept of volume is crucial in understanding how much three-dimensional space an object takes up. In calculus, when a solid is formed by revolving a region around an axis, it's called a solid of revolution. To find the volume of such a solid, one can use the disk method, which involves slicing the solid into thin disks perpendicular to the axis of revolution.
This method uses integration to sum up the volumes of these disks. The formula for the volume of a solid of revolution rotating around the x-axis is
This curve is revolved between \( x = -4 \) and \( x = 4 \), creating a symmetrical solid.
This method uses integration to sum up the volumes of these disks. The formula for the volume of a solid of revolution rotating around the x-axis is
- \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
This curve is revolved between \( x = -4 \) and \( x = 4 \), creating a symmetrical solid.
Integration Techniques
Integration, a fundamental concept in calculus, involves finding the antiderivative or the area under a curve. In practice, some functions require special techniques to integrate effectively. One common method is expanding algebraic expressions.
In our example, the function \( \left(1 - \frac{x^2}{16}\right)^2 \) needs to be expanded to simplify the integration process. After expansion, it becomes \( 1 - \frac{x^2}{8} + \frac{x^4}{256} \). This simplification helps in integrating term by term:
In our example, the function \( \left(1 - \frac{x^2}{16}\right)^2 \) needs to be expanded to simplify the integration process. After expansion, it becomes \( 1 - \frac{x^2}{8} + \frac{x^4}{256} \). This simplification helps in integrating term by term:
- The integral of \(1\) is \(x\).
- The integral of \(-\frac{x^2}{8}\) is \(-\frac{x^3}{24}\).
- The integral of \(\frac{x^4}{256}\) is \(\frac{x^5}{1280}\).
Definite Integral
The definite integral is a key concept in calculus used to calculate the total accumulation of a function over an interval. Unlike indefinite integrals, which provide a general form of the antiderivative, definite integrals have numerical values.
In this exercise, after determining the indefinite form, we evaluate the expression from \( x = -4 \) to \( x = 4 \) to get a numeric volume. This process involves calculating the antiderivative at both endpoints and finding the difference:
In this exercise, after determining the indefinite form, we evaluate the expression from \( x = -4 \) to \( x = 4 \) to get a numeric volume. This process involves calculating the antiderivative at both endpoints and finding the difference:
- At \( x = 4\), plug in the value to obtain a result.
- At \( x = -4\), plug in the value again.
- Subtract these results: \( V = \pi (f(4) - f(-4)) \).
Fuel Efficiency Calculations
Fuel efficiency calculations are essential in understanding how far a vehicle can travel on a given amount of fuel. For the helicopter in this scenario, understanding the volume of the auxiliary fuel tank allows us to estimate additional flight range.
After computing the tank's volume, we convert it to gallons using the conversion factor: 7.481 gallons per cubic foot. For instance, if the tank holds 42 cubic feet, it provides approximately 314 gallons.
Considering that the helicopter can cover 2 miles per gallon, we calculate the additional flight capacity:
After computing the tank's volume, we convert it to gallons using the conversion factor: 7.481 gallons per cubic foot. For instance, if the tank holds 42 cubic feet, it provides approximately 314 gallons.
Considering that the helicopter can cover 2 miles per gallon, we calculate the additional flight capacity:
- Multiply the total gallons by the efficiency: 314 gallons \( \times 2 \text{ miles/gallon} \) = 628 additional miles.
Other exercises in this chapter
Problem 54
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