Volume of Solid: Answers & Clarification

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SUMMARY

The discussion focuses on calculating the volume of a solid formed by rotating the region bounded by the curves y = x and y = x² around the line y = 2. The user has successfully solved several problems but seeks clarification on problem #8, specifically using the disc/washer and shell methods. The washer method involves calculating the inner radius (2 - x) and outer radius (2 - x²), leading to the area formula of π(2 - x)² - π(2 - x²)². The shell method utilizes a thin cylinder with a radius of (2 - y) and height of (√y - y), resulting in the volume formula of 2π(2 - y)(√y - y)dy.

PREREQUISITES
  • Understanding of volume calculation using the disc/washer method
  • Familiarity with the shell method for volume integration
  • Knowledge of integration techniques in calculus
  • Ability to graph functions and identify bounded regions
NEXT STEPS
  • Study the application of the disc/washer method in volume problems
  • Learn about the shell method for calculating volumes of revolution
  • Practice integration techniques for functions involving square roots
  • Explore graphical methods for visualizing bounded regions in calculus
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Students and educators in calculus, particularly those focusing on volume calculations and methods of integration. This discussion is beneficial for anyone looking to deepen their understanding of solid geometry and volume of revolution techniques.

sonofjohn
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Thanks to the awesome help on this board, I am narrowing my problems down quickly. I have already answered problems #6, 7, 9, and 10 as b, f, c, and a respectively. I wouldn't mind if you checked them but it's not my biggest worry. Instead I have given many attempts at number 8 using the disc/washer and shell formula. The page is out of date because I have been working on it since that page has been scanned and I no longer have access to that scanner. Any clarification on the formulas and their uses in this problem would be great.

Cal1chapter6test_0004.jpg
 
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Problem 8 is "Find the volume of the solid obtained by rotating the region bounded by the curves y= x and y= x2 around the line y= 2.

First, of course, draw a picture. You should see that vertical line drawn from y= 2 will cross that region first at y= x and then at y= x2. That means that a "washer" formed by rotating the figure about y= 2 will have inner radius 2- x and outer radius 2- x2. The areas of the two circles with those radii will be [itex]\pi(2- x)^2[/itex] and [itex]\pi(2- x^2)^2[/itex]. The area of the "washer" between them will have area equal to the difference of those. Integrate that with respect to dx from x= 0 to x= 1.

If you want to do it by "shells", each shell will be a thin cylinder of thickness dy, radius 2- y, and height [itex]\sqrt{y}- y[/itex]. The surface area of that will be [itex]2\pi(2- y)(\sqrt{y}- y)[/itex] and the volume will be [itex]2\pi(2- y)(\sqrt{y}- y)dy[/itex] integrated from y= 0 to y= 1.
 

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