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

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SUMMARY

The discussion focuses on solving the integral of the area defined by the curve y = 2√x when rotated about the y-axis. The user seeks assistance in completing the square for the expression x^2 + x and integrating the resulting function. The solution involves rewriting the expression as (x + 1/2)^2 - (1/2)^2 and applying the substitution w = a cosh(t), where a is determined to be 1/2. This method effectively transforms the integral into a solvable form.

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

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

\sqrt{w^2-a^2}

Then try the substitution w = a\cosh(t)
 
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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