/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Volume of a Fuel Tank \(A\) tank... [FREE SOLUTION] | 91Ó°ÊÓ

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Volume of a Fuel Tank \(A\) tank on the wing of a jet aircraft is formed by revolving the region bounded by the graph of \(y=\frac{1}{8} x^{2} \sqrt{2-x}\) and the \(x\) -axis \((0 \leq x \leq 2)\) about the \(x\) -axis, where \(x\) and \(y\) are measured in meters. Use a graphing utility to graph the function and find the volume of the tank.

Short Answer

Expert verified
The volume of the tank can be obtained by computing the integral \(V= \pi \int_{0}^{2} \left(\frac{1}{8} x^{2} \sqrt{2-x}\right)^{2} dx\). The exact volume depends on the successful calculation of this integral. A mathematical software or calculator is recommended for calculating this integral.

Step by step solution

01

Understand the problem

The region bounded by the graph of the function \(y = \frac{1}{8} x^{2} \sqrt{2-x}\) and the x-axis is to be revolved about the x-axis. From this, a shape is created (which is the fuel tank). The task requires finding the volume of this tank.
02

Setup the integral

The volume of a solid of revolution around the x-axis is given by \(V=\pi \int_{a}^{b} [f(x)]^{2} d x\). In this case, \(f(x)= \frac{1}{8} x^{2} \sqrt{2-x}\) and the boundaries are a = 0, b = 2.
03

Calculate the integral

The integral to compute volume becomes \(V= \pi \int_{0}^{2} \left(\frac{1}{8} x^{2} \sqrt{2-x}\right)^{2} dx\). This integral can be computed using a mathematical calculator or software.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is a branch of mathematics that focuses on the concept of integration.
Integration is essentially the process of finding the whole given its parts, or areas and volumes in geometry.
In the context of this exercise, we use integral calculus to determine the volume of a solid of revolution, like the aircraft fuel tank.
To do this, we use a specific formula to calculate the volume of a solid formed by rotating a function around an axis. When you revolve a two-dimensional area around an axis, you get a three-dimensional shape.
Integrating the function squared and multiplying by \(\pi\) gives you the volume of this shape. It captures not just the span or two-dimensional area, but the actual space occupied within the 3D object.
It’s like not only drawing the outline of a shape but filling it in to see how much space it truly takes up.
Graphing Functions
Graphing functions is a fundamental skill that helps visualize mathematical problems.
When solving problems involving solids of revolution, it is essential to plot the function you are working with correctly.
This way, you get a clear visual on what you're revolving around an axis to get a 3D object like our fuel tank.
For the function involved in this exercise, which is \( y = \frac{1}{8} x^{2} \sqrt{2-x} \), a graph reveals how the function behaves between \( x = 0 \) and \( x = 2 \).
As \( x \) approaches 2, the term \( \sqrt{2-x} \) causes the function to decrease to zero, creating a bounded area under the curve.
Use graphing utilities, which are available in many calculators and software applications, for this.
They make it easier to plot the function accurately and explore its characteristics further.
This is particularly useful when dealing with complex functions where manual plotting could be error-prone.
Solid of Revolution
A solid of revolution is a three-dimensional object created by rotating a two-dimensional shape around an axis.
It could be around the x-axis or y-axis, but for this exercise, we're revolving around the x-axis.
This concept takes advantage of symmetry in calculus, allowing you to simplify complex volume calculations into manageable integrals.
In the case of the aircraft fuel tank example, the solid is formed by spinning the area under the function \( y = \frac{1}{8} x^2 \sqrt{2-x} \) from \( x = 0 \) to \( x = 2 \).
The result of this rotation is a "cup-like" shape with a clear boundary defined by the function itself.
By exploiting this rotational symmetry, it becomes significantly easier to determine its volume using integral calculus.
Knowing this concept provides a strong foundation that is applicable in various engineering and design fields.
Volume Calculation
Volume calculation of solids of revolution involves transforming the visual problem into a mathematical one.
Here, we use the method of disks or washers, where you consider a shape as a series of infinitesimally small disks piled together.
Each disk's volume is \( \pi [f(x)]^2 \Delta x \).
To compute the total volume, the disks' volumes are summed over the interval from \( a \) to \( b \), or in this case from 0 to 2.
This is achieved through integration, expressed as \( V = \pi \int_{0}^{2} \left(\frac{1}{8} x^{2} \sqrt{2-x}\right)^{2} dx \).
Performing this integral gives the exact volume of the 3D object.
Tools like calculators or software are often used to perform these calculations since manual computing can be cumbersome.
Understanding these principles and learning how to set up the integral effectively are key steps in tackling similar scientific and engineering problems.

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Most popular questions from this chapter

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