Assume f is a nonnegative function with a continuous first derivative on [a, b]. The curve y=f(x) on [a, b] is revolved about the x-axis. Explain how to find the area of the surface that is generated.
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9. Graphical Applications of Integrals
Introduction to Volume & Disk Method
Problem 6.4.5c
Textbook Question
Let R be the region in the first quadrant bounded above by the curve y=2−x² and bounded below by the line y=x. Suppose the shell method is used to determine the volume of the solid generated by revolving R about the y-axis.

c. Write an integral for the volume of the solid using the shell method.
Verified step by step guidance1
Identify the region R bounded by the curves y = 2 - x^2 (above) and y = x (below) in the first quadrant. This region is revolved around the y-axis to form the solid.
Recall that the shell method involves integrating with respect to x when revolving around the y-axis. Each shell has a radius equal to the distance from the y-axis, which is x, and a height equal to the difference between the upper and lower functions: height = (2 - x^2) - x.
Write the volume of a typical shell as the circumference times the height times the thickness: Volume of shell = 2\pi \times (radius) \times (height) \times (thickness) = 2\pi x \big((2 - x^2) - x\big) \, dx.
Determine the limits of integration by finding the x-values where the two curves intersect in the first quadrant. Solve 2 - x^2 = x to find these points.
Set up the integral for the volume as: \(V = \int_{a}^{b} 2\pi x \big((2 - x^2) - x\big) \, dx\), where a and b are the intersection points found in the previous step.
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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 summing cylindrical shells. Each shell's volume is approximated by 2π(radius)(height)(thickness), where the radius is the distance from the axis of rotation, the height is the function value difference, and the thickness is a small change in the variable of integration.
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Setting up the Integral with Respect to x
When revolving around the y-axis, the shell radius is the x-value of the shell, and the height is the difference between the upper and lower functions, here y=2−x² and y=x. The integral sums shells from the leftmost to rightmost x-values defining the region, integrating with respect to x.
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Determining the Bounds of Integration
The bounds are found by identifying where the curves intersect in the first quadrant. Solving y=2−x² and y=x gives the limits for x, which define the interval over which the shells are integrated to find the volume.
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