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Finding the volume of by rotating the region about the x-axis

A plane region is bounded by the Graph of the equation y=3-x^2 and y=abs(x).
Finding the volume of by rotating the region about the x-axis.
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2 Answers

I get the two intersections at x = (1 - √13)/2, (-1 + √13)/2; let's label them a, b (respectively).
If we evaluate ∫ab π (3 - x^2)2 dx (the greater function's volume),
then subtract ∫ab πx2 dx (the lesser function's volume),
we get π[9x - 2x3 + 1/5 x5]ab - π[1/3 x3]ab


Thank you for the help Christopher, the reason i posted this question was because i needed to know if taking the integral of the absolute value of x [abs(x)] is the same as taking the integral of x. In your explanation it seems to be the same. Am i correct??
Yes, but only because (+x)2 and (-x)2 are both the same value, namely x2. If they were not the same value, you would integrate piece-wise, first -x from a to 0, then +x from 0 to b, and subtract both "inside" volumes. In this case they just happen to simplify to the same integrand.
Hello Christopher,
Everything ok.
But the first point of intersection should be  ( -1      -sqrt(13))/2


(-1 - √13)/2 is a solution of -x2 + 3 = x. We want a solution for -x2 + 3 = -x when x < 0.
Thanks for my correction Christopher.