.
#y=0# is the #x#-axis. As such, we want to revolve the area between the curve of #y=sinx#, the #x#-axis , #x=pi/2#, and #x=pi# around the #x#-axis and calculate the volume of the solid generated.
The graph below shows this area:
If we revolve this area around the #x#-axis we will get the solid shown below:
If you can imagine this solid being divided into vertical slices parallel to the #y#-axis with a extremely small thicknesses each one would look like a thin disc with the surface area of a circle and thickness of #dx#.
The circles have a radius #r# that is equal to #y=sinx# and vary in size depending on where on the #x#-axis you perform the slice.
So, because the formula for the area of a circle is #A=pir^2,#, we can calculate the area of each disc as:
#A=pisin^2x#
Now if we take the integral of this function and evaluate it between #pi/2# and #pi# we will have the volume of the solid.
This is because the integral adds the areas of infinite number of discs between the two limits together.
#V=int_(pi/2)^pipisin^2xdx=piint_(pi/2)^pi(1-cos2x)/2dx#
#V=pi/2int_(pi/2)^pidx-pi/2int_(pi/2)^picos2xdx#
#V=pi/2x-pi/2I#
#I=int_(pi/2)^picos2xdx#
Let #u=2x#
#du=2dx#
#dx=(du)/2#
Let's substitute:
#I=intcosu(du)/2=1/2intcosudu=1/2sinu=1/2sin2x#
Let's plug this in:
#V=pi/2x-pi/2*1/2sin2x=(pi/4(2x-sin2x))_(pi/2)^pi#
#V=pi/4(2pi-sin2pi-pi+sinpi)#
#V=pi/4(pi-0+0)#
#V=pi^2/4#
