Calculus Definite Integrals: Volumes by Washer Method

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SUMMARY

The discussion focuses on calculating the volume of a solid of revolution using the Washer Method, specifically revolving the region bounded by the curves y=x^2 and y=x^0.5 around the line y=-3. The volume formula V is defined as the integral of the area A(x) from 0 to 1, where A(x) is given by π(radius^2). The correct outer radius is determined to be √x + 3, while the inner radius is x^2 + 3, clarifying the confusion regarding the addition versus subtraction of 3 in the radius calculations.

PREREQUISITES
  • Understanding of the Washer Method for calculating volumes of solids of revolution
  • Familiarity with definite integrals and their applications in volume calculations
  • Knowledge of the functions y=x^2 and y=x^0.5
  • Basic skills in sketching graphs to visualize regions and revolutions
NEXT STEPS
  • Study the derivation and application of the Washer Method in different scenarios
  • Practice calculating volumes using the Shell Method for comparison
  • Explore advanced integration techniques for more complex volume problems
  • Learn about the geometric interpretations of integrals in calculus
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations using the Washer Method, as well as educators looking for examples to illustrate these concepts.

jsun2015
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Homework Statement


Using Washer Method: Revolve region R bounded by y=x^2 and y=x^.5 about y=-3



Homework Equations


V= integral of A(x) from a to b with respect to a variable "x"
A(x)=pi*radius^2


The Attempt at a Solution


pi(integral of (x^.5-3)^2 -(x^2)^2-3) from 0 to 1 with respect to x

The answer involves x^.5+3 instead of x^.5 -3. I don't understand why you would add instead of subtract 3; y decreased from 0 to -3 when revolved around y=-3
 
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Make a drawing. See that the outer radius of your washers is ##\sqrt x + 3## and the inner is ## x^2 + 3##
 
jsun2015 said:
The answer involves x^.5+3 instead of x^.5 -3. I don't understand why you would add instead of subtract 3; y decreased from 0 to -3 when revolved around y=-3

The length of a vertical line is ##y_{upper}-y_{lower}##. Also don't forget to square your second term.
 

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