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Use the Theorem of Pappus to find the volume of the solid of revolution. The solid formed by revolving the region bounded by the graphs of \(y=x, y=4,\) and \(x=0\) about the \(x\) -axis

Short Answer

Expert verified
The volume of the solid of revolution is \(128/3 * \pi\) cubic units.

Step by step solution

01

Identify the bounded region

The bounded region is a right-angled triangle with vertices at the origin, (0,4) and (4,4) formed by the three equations \(y = x, y = 4, x = 0\).
02

Calculate the area and centroid

The area of the triangle (\(A\)) can be calculated using the formula \(Area = 0.5 * base * height\). Here, the base and the height are both 4 units. So, \(A = 0.5 * 4 * 4 = 8\) square units. \n\nThe centroid (\(ȳ\)) of a right triangle is 2/3 from the base to the height. Hence, \(ȳ = 2/3 * 4 = 8/3\) units.
03

Apply the Theorem of Pappus

The Pappus' Theorem states that the volume of the solid of revolution \((V)\) is \(V = 2 * \pi * ȳ * A\). Substituting the area and the centroid from Step 2, we get that \(V = 2 * \pi * (8/3) * 8\).
04

Simplify the expression

Simplify the expression to get the volume of the solid of revolution. This results in \(V = 128/3 * \pi\) cubic units.

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Key Concepts

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

Solid of Revolution
A solid of revolution is a three-dimensional shape created when a two-dimensional plane area is rotated around a line (axis) that lies in the same plane. This process is akin to spinning the area around this line as if it was on a potter's wheel, creating a symmetrical shape with smooth surfaces. For example, revolving a rectangle about one of its edges produces a cylinder, and revolving a triangle produces a cone.

When we talk about solving mathematical problems involving solids of revolution, we focus on certain parameters such as volume. The volume of such a solid can be complicated to calculate using elementary geometry. However, the Theorem of Pappus provides a way to simplify the process by relating the volume of the solid to the area of the rotating shape and the path traced by its centroid during the rotation. Specific integrals, known as volume integrals, can also be used when the shape involves curves or more complex geometries.
Centroid of a Triangle
The centroid of a triangle, often referred to in the context of 'center of mass', or 'center of gravity', is the point where the triangle's three medians intersect. A median is a line drawn from any vertex to the midpoint of the opposite side. The remarkable property of the centroid is that it divides each of the medians into two segments, with the longer one being twice the length of the shorter one, placing the centroid two-thirds of the way from the vertex to the midpoint of the opposite side.

In coordinates, if a triangle has vertices at points \(A(x_1,y_1)\), \(B(x_2,y_2)\), and \(C(x_3,y_3)\), the centroid \(G\) is given by the average of the coordinates of the vertices: \(G\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\). For a right-angled triangle, the centroid can be more simply found as the midpoint of the hypotenuse if it is revolved around one of its non-hypotenuse sides.
Volume Calculation
Volume calculation in the context of solids of revolution is a crucial aspect of understanding three-dimensional shapes and their properties. The volume signifies the amount of space enclosed within a solid. To calculate the volume of regular solids like cubes or spheres, specific formulas are used. However, for more complex geometries like the ones created by revolving a plane figure, the calculation requires different approaches, such as the disc method, the shell method, or the Theorem of Pappus.

According to the Theorem of Pappus, the volume of a solid of revolution is equal to the product of the area of the shape being revolved and the distance traveled by its centroid during the revolution (the circumference of the circle formed by the centroid's path). This theorem simplifies the volume calculation greatly, as it converts the problem into finding two primary attributes: the area of the shape and the centroid's location.
Bounded Region
In mathematics, a bounded region refers to an area that is enclosed within certain limits or boundaries. The term is often used in the context of two-dimensional areas on a plane that are confined by curves, lines, or both. Determining the boundaries of such a region is crucial, as it lays the groundwork for further calculations, including the area, the centroid, and in the context of solids of revolution – the eventual volume of the shape when it is rotated around a specified axis.

For example, in the exercise related to the Theorem of Pappus, the region of interest is shaped as a right-angled triangle bounded by the graphs of \(y=x\), \(y=4\), and \(x=0\). These lines act as the 'fences' that demarcate the edges of the triangle and establish the area that will be rotated around the x-axis to create a three-dimensional solid. Understanding and correctly identifying the bounded region are essential for applying the Theorem of Pappus effectively.

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Most popular questions from this chapter

Finding the Area of a Surface of Revolution In Exercises \(37-42,\) set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=3 x, \quad 0 \leq x \leq 3 $$

True or False? In Exercises \(83-86\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The line $$y=(1-\sqrt[3]{0.5}) x$$ divides the region under the curve $$f(x)=x(1-x)$$ on \([0,1]\) into two regions of equal area.

Pumping Gasoline In Exercises 23 and \(24,\) find the work done in pumping gasoline that weighs 42 pounds per cubic foot. (Hint: Evaluate one integral by a geometric formula and the other by observing that the integrand is an odd function.) The top of a cylindrical gasoline storage tank at a service station is 4 feet below ground level. The axis of the tank is horizontal and its diameter and length are 5 feet and 12 feet, respectively. Find the work done in pumping the entire contents of the full tank to a height of 3 feet above ground level.

Modeling Data The circumference \(C\) (in inches) of a vase is measured at three-inch intervals starting at its base. The measurements are shown in the table, where \(y\) is the vertical distance in inches from the base. $$ \begin{array}{|c|c|c|c|c|c|c|}\hline y & {0} & {3} & {6} & {9} & {12} & {15} & {18} \\ \hline C & {50} & {65.5} & {70} & {66} & {58} & {51} & {48} \\\ \hline\end{array} $$ (a) Use the data to approximate the volume of the vase by summing the volumes of approximating disks. (b) Use the data to approximate the outside surface area (excluding the base) of the vase by summing the outside surface areas of approximating frustums of right circular cones. (c) Use the regression capabilities of a graphing utility to find a cubic model for the points \((y, r),\) where \(r=C /(2 \pi) .\) Use the graphing utility to plot the points and graph the model. (d) Use the model in part (c) and the integration capabilities of a graphing utility to approximate the volume and outside surface area of the vase. Compare the results with your answers in parts ( a ) and (b).

Hydraulic Press In Exercises \(39-42,\) use the integration capabilities of a graphing utility to approximate the work done by a press in a manufacturing process. A model for the variable force \(F\) (in pounds) and the distance \(x\) (in feet) the press moves is given. $$ \text{Force} \quad \text{Interval} $$ $$ F(x)=\frac{e^{x^{2}}-1}{100} \quad 0 \leq x \leq 4 $$

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