The surface area of revolution for the function y = |x| from the interval [-2, 2] can be calculated by evaluating the integral from [0, 2] and then doubling the result due to the symmetry of the function around the y-axis. The correct derivative dy/dx for the function is not (5/(2√x)), indicating a misunderstanding in the setup. The integral for the surface area S is given by S = 2 ∏ ∫ 5x^1/2(√(1 + ((5/(2√x))^2))) dx from x = 0 to x = 2, which should be computed before doubling.
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Jbreezy said:Homework Statement
I want the surface area of Revolution about x from [-2,2] of y = |x|
So I want to know if I can take it x from [0,2] and just multiply this result by 2?
Homework Equations
The Attempt at a Solution
Set up
dy/dx = (5/(2√x))
S = 2 ∏ ∫ 5x^1/2(√ 1 +((5/(2√x))^2)
from x = 0 , to x = 2. Then multiply this result by 2? Is this OK to do?