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Ch. 6 - Applications of Definite Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 6 - Applications of Definite IntegralsProblem 6.PE.10b
Chapter 6, Problem 6.PE.10b

Volumes
Find the volume of the solid generated by revolving the region bounded by the parabola y² = 4x and the line y = x about
b. the y-axis

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First, identify the region bounded by the curves. The parabola is given by \(y^{2} = 4x\), which can be rewritten as \(x = \frac{y^{2}}{4}\). The line is \(y = x\), or equivalently \(x = y\). Find the points of intersection by setting \(\frac{y^{2}}{4} = y\).
Solve the equation \(\frac{y^{2}}{4} = y\) to find the intersection points. Multiply both sides by 4 to get \(y^{2} = 4y\), then rearrange to \(y^{2} - 4y = 0\), and factor as \(y(y - 4) = 0\). So, \(y = 0\) or \(y = 4\) are the limits of integration.
Since the solid is generated by revolving the region about the y-axis, use the method of cylindrical shells. The formula for the volume is \(V = \int_{a}^{b} 2\pi (\text{radius})(\text{height}) \, dy\), where the radius is the distance from the y-axis and the height is the horizontal distance between the curves.
Determine the radius and height for the shell at a given \(y\). The radius is the distance from the y-axis, which is \(x\). The height is the difference between the rightmost and leftmost \(x\) values for the region at that \(y\). Here, the right boundary is \(x = y\) (from the line) and the left boundary is \(x = \frac{y^{2}}{4}\) (from the parabola). So, height = \(y - \frac{y^{2}}{4}\).
Set up the integral for the volume: \(V = \int_{0}^{4} 2\pi y \left(y - \frac{y^{2}}{4}\right) dy\). This integral represents the volume of the solid generated by revolving the region about the y-axis.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Understanding the Region Bounded by Curves

To find the volume of a solid of revolution, first identify the region bounded by the given curves. Here, the parabola y² = 4x and the line y = x intersect, defining the limits of integration. Understanding their intersection points and the area enclosed is essential for setting up the integral correctly.
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Method of Cylindrical Shells for Volume

When revolving a region around the y-axis, the cylindrical shells method is often used. This involves integrating the volume of thin cylindrical shells with radius equal to the x-value, height given by the difference in y-values, and thickness dx. This method simplifies volume calculation when the axis of revolution is vertical.
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Setting Up and Evaluating Definite Integrals

After determining the method and limits, set up the definite integral representing the volume. This requires expressing variables appropriately, integrating with respect to x or y, and carefully evaluating the integral to find the exact volume. Mastery of integral calculus techniques is crucial for accurate computation.
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Related Practice
Textbook Question

Volumes

Find the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³ and the lines x = 1 and y = 1/2 about

c. the line x = 2

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Textbook Question

Volumes

Find the volume of the solid generated by revolving the region bounded on the left by the parabola x = y² + 1 and on the right by the line x = 5 about

c. the line x = 5

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Textbook Question

Volumes

Find the volume of the solid generated by revolving the region between the x-axis and curve y = x² ―2x about

d. the line y = 2

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Textbook Question

Areas of Surfaces of Revolution

In Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes.

_______

y = √4y ― y² , 1 ≤ y ≤ 2 ; y-axis

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