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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.4.32
Chapter 6, Problem 6.4.32

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis. 


{Use of Tech} y = √50 -2x², in the first quadrant; about the x-axis

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First, identify the region R bounded by the curve \(y = \sqrt{50 - 2x^2}\) in the first quadrant. Since we are in the first quadrant, both \(x\) and \(y\) are non-negative, so \(x \geq 0\) and \(y \geq 0\).
Since the solid is generated by revolving the region about the x-axis, and we are asked to use the shell method, recall that the shell method involves integrating cylindrical shells formed by slicing the region parallel to the axis of revolution. Here, the axis of revolution is horizontal (x-axis), so we will integrate with respect to \(y\).
Express \(x\) in terms of \(y\) from the given curve: starting with \(y = \sqrt{50 - 2x^2}\), square both sides to get \(y^2 = 50 - 2x^2\), then solve for \(x^2\) to find \(x^2 = \frac{50 - y^2}{2}\), and finally \(x = \sqrt{\frac{50 - y^2}{2}}\).
Set up the shell radius and height for the shell method. The radius of a shell is the distance from the shell to the axis of revolution (x-axis), which is simply \(y\). The height of the shell is the horizontal length of the region at that \(y\) value, which is \(x = \sqrt{\frac{50 - y^2}{2}}\).
Write the volume integral using the shell method formula: \(V = \int_{y=0}^{y=\sqrt{50}} 2\pi \times (\text{radius}) \times (\text{height}) \, dy = \int_0^{\sqrt{50}} 2\pi y \cdot \sqrt{\frac{50 - y^2}{2}} \, dy\). This integral represents the volume of the solid.

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Key Concepts

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

Shell Method for Volume

The shell method calculates the volume of a solid of revolution by integrating cylindrical shells. Each shell's volume is approximated by its circumference times height times thickness. This method is especially useful when the axis of rotation is parallel to the axis of the variable of integration.
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Finding Volume Using Disks

Region Bounded by Curves

Understanding the region bounded by the given curves is essential to set correct integration limits. Here, the curve y = √(50 - 2x²) in the first quadrant defines the boundary, meaning x and y are non-negative, which restricts the domain for integration.
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Finding Area When Bounds Are Not Given

Revolution About the x-axis

Revolving a region about the x-axis means the solid is generated by rotating the area around the horizontal axis. This affects the radius and height expressions in the shell method, as shells are formed perpendicular to the axis of rotation.
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Disk Method Using y-Axis
Related Practice
Textbook Question

Explain the steps required to find the length of a curve x = g(y) between y=c and y=d.

Textbook Question

60–63. Equivalent constant velocity Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.


v(t)=2 sin t, for 0≤t≤π

Textbook Question

Find the area of the region described in the following exercises.


The region bounded by y=x^2−2x+1 and y=5x−9

Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

x = 2e^√2y + 1/16e^−√2y, for 0 ≤ y ≤ ln²/√2

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Textbook Question

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.


y = 2

Textbook Question

39–44. Shell method about other lines Let R be the region bounded by y = x²,x=1, and y=0. Use the shell method to find the volume of the solid generated when R is revolved about the following lines.


y = -2