How do you find the volume of the solid obtained by rotating the region bound by the curve and y=x^2+1 and x-axis in the interval (2,3)?

1 Answer
Mar 22, 2017

Please see below.

Explanation:

The volume of the solid obtained by rotating the region bound by the curve y=R(x) and x-axis in the interval (c,d) is

V=piint_c^d[R(x)]^2dx

Here, we have R(x)=x^2+1, c=2 and d=3 and volume is

piint_color(red)2^color(red)3[x^2+1]^2dx
(once we have put R(x), put c amd d as well)

= piint_color(red)2^color(red)3[x^4+2x^2+1]dx

= color(red)(pi[x^5/5+(2x^3)/3+x+C]_2^3
(once integral has been done no need to write it)

= color(red)(pi[(3^5/5+(2xx3^3)/3+3+C)-(2^5/5+(2xx2^3)/3+2+C)]

= color(red)(pi[(243/5+54/3+3+C)-(32/5+16/3+2+C)]

= color(red)(pi[(243-32)/5+(54-16)/3+(3-2)]

= color(red)(pi[211/5+38/3+1]

= color(red)(pi[(211xx3+38xx5+15)/15]

= color(red)(pi[(633+190+15)/15]

= color(red)(pi[(838)/15]=55.86pi