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Chapter 7: Problem 29

Use the disk or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line. \(x^{1 / 2}+y^{1 / 2}=a^{1 / 2}, \quad x=0, \quad y=0\) (a) the \(x\) -axis (b) the \(y\) -axis (c) the line \(x=a\)

Short Answer

Expert verified
The volume of the solid generated revolving around the x-axis, the y-axis, and the line \(x = a\), is given by the integrals \(\int_0^a \pi * (a^{\frac{1}{2}} - x^{\frac{1}{2}})^2 \, dx\), \(2 * \pi * \int_0^a y * (a^{\frac{1}{2}} - y^{\frac{1}{2}}) \, dy\), and \(\int_0^a \pi * (a^{\frac{1}{2}} - x^{\frac{1}{2}})^2 \, dx\), respectively.

Step by step solution

01

Sketch the region

The aforementioned equations form a right triangle in the first quadrant, with sides of length \(a\), \(a\), and \(a\sqrt{2}\). So sketch out this region to visualize it.
02

Volume revolving around the x-axis

The volume of the solid generated revolving around the x-axis is calculated with the disk method. The radius of the disk is \(f(x) = a^{\frac{1}{2}} - x^{\frac{1}{2}}\) and the thickness is \(dx\), so we set up the integral for the volume as integrating from 0 to \(a\) of \(\pi * (a^{\frac{1}{2}} - x^{\frac{1}{2}})^2 \, dx\). This is due to the formula for the volume of a solid of revolution using the disc method, \(V = \int_a^b \pi[f(x)]^2 \, dx\)
03

Volume revolving around the y-axis

The volume of the solid when revolving the region about the y-axis requires the shell method because the 'height' of the shell now varies. We use \(g(y) = a^{\frac{1}{2}} - y^{\frac{1}{2}}\), where g(y) gives the height of the shell, and the thickness is \(dy\). The integral for volume is from 0 to \(a\) of \(2 * \pi * y * (a^{\frac{1}{2}} - y^{\frac{1}{2}}) \, dy\). This follows the formula for volume using the shell method, \(V = 2 \pi \int_a^b y * f(y) \, dy\)
04

Volume revolving around the line x=a

When we revolve the region about the line \(x = a\), the radius for the disk is again \(f(x) = a^{\frac{1}{2}} - x^{\frac{1}{2}}\). So the integral for this volume is \(V = \int_0^a \pi * (a^{\frac{1}{2}} - x^{\frac{1}{2}})^2 \, dx\)

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Key Concepts

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

Disk Method
The Disk Method is a technique used in calculus to find the volume of a solid of revolution. Imagine slicing a 3D object into very thin, circular disks. To find the volume, you add up (or integrate) the volumes of these disks from one end of the object to the other.

This method is best used when rotating a region around an axis to form a solid. Picture creating a stack of pancakes where each pancake is a disk!
  • Volume of each disk is calculated as the area of the circle times its thickness, given by \( \pi [f(x)]^2 \, dx\).
  • Here, \(f(x)\) is the function that the region is bounded by, forming the radius of the disk.
  • Thickness \(dx\) represents a tiny slice's thickness, integrating across limits depending on the region.
In the exercise, when the region is revolved around the x-axis, the radius is calculated using the given function and limits. This simple integration gives the entire volume.
Shell Method
Unlike the Disk Method, the Shell Method involves wrapping the region in cylindrical shells when rotating around an axis. Imagine peeling an onion where each layer represents a shell!

This approach is particularly useful when the radius and the height of the shell change along the axis of revolution.
  • Volume is calculated using the formula \(2 \pi \int_a^b y \, f(y) \, dy\).
  • The factor of \(2 \pi y\) accounts for the circumference of the shell.
  • \(f(y)\) represents the height of each shell, with \(dy\) as the vertical slice thickness.
The exercise uses this method for rotation about the y-axis, showing the adaptability of this method for different scenarios. By interchanging x and y, we adapt to the changed orientation while still achieving precise volume calculations.
Solid of Revolution
A Solid of Revolution is a 3D shape created by rotating a 2D region around an axis. Think of turning a curve into a vase-like shape; this is literally revolving a region about an axis.

These solids are common in calculus because they provide practical examples of how regions can be transformed into 3D objects.
  • Key to understanding is visualizing the transformation from 2D to 3D.
  • The axis of rotation can vary (x-axis, y-axis, other lines), affecting which method (Disk or Shell) is preferable.
  • Solids of revolution find applications in physics and engineering, modeling real-world shapes.
The exercise at hand requires visualizing how each boundary's rotation leads to different solid shapes, depending on the line of revolution.
Integral Calculus
Integral Calculus is the branch of mathematics focusing on finding the total accumulation or area under a curve. It also involves finding the antiderivative, which is reversing the process of differentiation.

When applied to solids of revolution, integral calculus helps to find volumes by summing up infinite tiny slices or shells, providing an exact volume for irregular shapes.
  • Central concept is integration, denoted by the symbols \( \int \, dx\) or \( \int \, dy\).
  • Integrals can be definite (specific limits) or indefinite (without limits).
  • The solving process involves finding an antiderivative and evaluating it within given bounds.
Integral calculus is indispensable in the exercise as it provides a way to find the exact volume of the solids formed by revolution, demonstrating its power in tackling complex geometrical problems.

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Most popular questions from this chapter

Find \(M_{x}, M_{y}\), and \((\bar{x}, \bar{y})\) for the laminas of uniform density \(\rho\) bounded by the graphs of the equations. $$ y=\sqrt{x}+1, y=\frac{1}{3} x+1 $$

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Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If the area of the region bounded by the graphs of \(f\) and \(g\) is 1, then the area of the region bounded by the graphs of \(h(x)=f(x)+C\) and \(k(x)=g(x)+C\) is also \(1 .\)

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