If #f (x)= x^2+ 1# , how do you find the volume of the solid generated by revolving the region under graph of f from x=-1 to x=1 about the x- axis?

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Answer 1 · 2017-10-21T17:24:08

#(56pi)/15#

11.73 (2. d.p.)

First we need to integrate #(x^2+1)^2#. #int(x^2+1)^2 dx#

#int(x^2+1)^2 dx= x+(1/5)*x^5+(2/3)*x^3#

So we have:

#piint_(-1)^(1) (x^2+1)^2 dx #

We have:

#pi[(x+x^5/5+(2x^3)/3+c)-(x+x^5/5+(2x^3)/3+c)]#

Plugging in 1 and -1 for our upper and lower bounds:

#pi[((1)+(1)^5/5+(2(1)^3)/3+c)-((-1)+(-1)^5/5+(2(-1)^3)/3+c)]#

#pi[(28/15)-(-28/15)]=pi(56/15)->#

#(56pi)/15#

11.73 (2. d.p.)