Find the volume generated by rotating y=xsin(pi*x), y=0, x=2, x=3, around the line x=1.

Question

Find the volume generated by rotating y=xsin(pi*x), y=0, x=2, x=3, around the line x=1.

Answer 1

In this problem, it is easier to use the shell method (as opposed to the disc method in which the maximum value of y would need to be found). The formula for the shell method is

V = π∫_a^bxf(x)dx

when revolving around the x-axis. Here, we are revolving the curve around the x=1 axis so the radius of revolution will be shorter by one. Thus, the formula becomes

V = π∫₂³xf(x-1)dx

   = π∫₂³x((x-1)sin(x-1))dx

   = π∫₂³ (x² – x)sin(x-1)dx

   = π∫₂³ x²sin(x-1)dx - π∫₂³ xsin(x-1)dx

   = π[2cos(x-1) – x²cos(x-1) – 2sin(x-1)]₂³ – π[sin(x-1) – xcos(x-1)]₂³

   ≈ 6.084π – 2.397π

   ≈ 3.687π

   ≈ 11.583

Integration by parts and substitution was used in both integrals.

Note: The best first step in problems like these is to draw a graph, but that was not possible on this venue due to the character limitation.

Answer 2

corrected--a lot. Something was not right, when I began analyzing more

it was apparent the previous response was in error

using shell method for computing volume

integrand is 2π(x)(x+1)sin(π(x+1))dx

the integration is from x=1 to x=2

why??,,,,,,,(x) is the r part of the shell and needs to go from 1 to 2 from axis of rotation.

the function xsin(x) now is moved negative so it's zero is -1 from rotation axis, (x+1) is zero at x=-1

calculating integral

V=-2x²cos(π(x+1))+(4/π)*xsin(π(x+1))+(4/π²)cos(π(x+1))-2xcos(π(x+1))+(2/π)sin(π(x+1)

from x=1 to x=2

V≈ 15.1 cubic units

Note: the same answer is obtained with integrand 2π(x-1)[(x)sin(πx)]dx and integrate from x=2 to x=3.

Here (x-1) is the r part of the shell.