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)#?

Question

Emily Ammerman · Accepted Answer

Please see below.

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#

Ezekiel Alexander · Answer

undefined To find the volume of the solid obtained by rotating the region bound by the curve $ y = x^2 + 1 $ and the x-axis in the interval (2, 3), we use the method of cylindrical shells.

The formula for the volume of the solid generated by revolving a curve $ y = f(x) $ around the x-axis in the interval $ [a, b] $ is given by:

$$ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx $$

In this case, the curve is $ y = x^2 + 1 $, and the interval is $ (2, 3) $. So, we have:

$$ V = 2\pi \int_{2}^{3} x \cdot (x^2 + 1) \, dx $$

We integrate this expression with respect to $ x $ over the given interval, and then we evaluate the definite integral to find the volume of the solid.

After integrating and evaluating, we will obtain the volume of the solid generated by rotating the region bound by the curve $ y = x^2 + 1 $ and the x-axis in the interval (2, 3) around the x-axis using cylindrical shells.