Find the volume of revolution about the x-axis of the curve y=1/sqrt(x^2+2x+2) for 0<x<1
The volume of revolution is given by integrating Piy2 dx from 0 to 1.Squaring, y2=1/(x2+2x+2)Completing the square, we see that y=1/((x+1)2+1)Make the substitution u=x+1, so du=dx. When x is 0, respectively 1, u is 1, respectively 2. So the volume is the integral of Pi/(u2+1) du from 1 to 2. This is Piarctan(u) evaluated from 1 to 2, which is Pi*(arctan(2)-arctan(1)). In a calculator, we see this is roughly 1.011 and this is the desired volume.
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