Area of y = 2 Sqrt x is rotated about the y axis with attempt

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Homework Help Overview

The discussion revolves around finding the area of the region defined by the curve y = 2√x when rotated about the y-axis. Participants are exploring integration techniques and transformations related to this problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to complete the square in the context of integration. Some participants suggest using substitutions to simplify the expression, while others question the specific parameters involved in the transformation.

Discussion Status

The discussion is ongoing, with participants providing insights into algebraic manipulation and substitution methods. There is a focus on clarifying the variables and parameters relevant to the integration process, but no consensus has been reached on a complete solution.

Contextual Notes

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Anyone? The information is presented on the uploaded jpg.
 
Write
[tex]x^2+x = (x^2+x + 1/4)-1/4=\left(x+\frac 1 2\right)^2-\left(\frac 1 2\right)^2[/tex]

Letting w = x + 1/2 gets it into the form

[tex]\sqrt{w^2-a^2}[/tex]

Then try the substitution [itex]w = a\cosh(t)[/itex]
 
What is a in your case?
 
Riazy said:
What is a in your case?

You mean in your case. a = 1/2.
 

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