What is the volume of the solid produced by revolving #f(x)=cosx, x in [0,pi] #around the x-axis?

1 Answer
Nov 26, 2016

#1/2pi^2 = 4.93# (3sf.)

Explanation:

General formula for volumes of revolution:

#V_x=int_b^apiy^2dx#
#V_y=int_b^apix^2dy#

(the pi can be taken out of the integral in either case to simplify the calculation)

#y=cosx#
#y^2=cos^2x#

The question requires solving:

#piint_0^picos^2xdx#

This is made easier using double angle formulae:

#cos2x=cos^2x-sin^2x = 2cos^2x -1#
#cos^2x=1/2(cos2x+1)#

By taking the #1/2# out of the integral it becomes:

#1/2piint_0^picos2x+1dx#

=#1/2pi[1/2sin2x+x]_0^pi#

=#1/2pi{(1/2sin2pi+pi) -(1/2sin0 + 0)}#

=#1/2pi(pi)#

=#1/2pi^2#