_{a}

^{b}π (3 - x^2)

^{2}dx (the greater function's volume),

_{a}

^{b}πx

^{2}dx (the lesser function's volume),

^{3}+ 1/5 x

^{5}]

_{a}

^{b}- π[1/3 x

^{3}]

_{a}

^{b}

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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Marked as Best Answer

I get the two intersections at x = (1 - √13)/2, (-1 + √13)/2; let's label them a, b (respectively).

If we evaluate ∫_{a}^{b} π (3 - x^2)^{2} dx (the greater function's volume),

then subtract ∫_{a}^{b} πx^{2} dx (the lesser function's volume),

we get π[9x - 2x^{3} + 1/5 x^{5}]_{a}^{b} - π[1/3 x^{3}]_{a}^{b}

SURENDRA K. | An experienced,patient & hardworking tutorAn experienced,patient & hardworking tut...

Hello Christopher,

Everything ok.

But the first point of intersection should be ( -1 -sqrt(13))/2

Surendra

(-1 - √13)/2 is a solution of -x^{2} + 3 = x. We want a solution for -x^{2} + 3 =
**-**x when x < 0.

Thanks for my correction Christopher.

Surendra

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## Comments

^{2}and (-x)^{2}are both the same value, namely x^{2}. 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.