Plot the region R enclosed by y = cos(3x); y = cos(x); x = 0; x = pi Find all relevant intersection points. Find the volume of the solid S obtained by rotating the region R about the axis y = -1.
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To calculate the volume of rotating body around x-axis we use formula
`V_0=pi int_a^b y^2 dx`
but since we are looking fora volume of body rotating around `y=-1`, our formula becomes
`V_(-1)=pi int (y+1)^2 dx`
Also for `x in (0,pi/2)` we have `y=cosx-cos3x` and for `x in (pi/2,pi)` we have `y=cos3x-cosx` because for `x in (0,pi/2)` `cosx>cos3x` and for `x in (pi/2,pi)` it's `cos3x>cosx`.
`V_(-1)=pi(int_0^(pi/2)(cosx-cos3x+1)^2dx+int_(pi/2)^pi (cos3x-cosx+1)^2dx)=`
`=pi int_0^(pi/2)(cos^2x-cos^2 3x+1+2cosx-2cos3x-2cosxcos3x)dx+`
`pi int_(pi/2)^pi (cos^2 3x+cos^2x+1+2cos3x-2cosx-2cosxcos3x)dx=`
Now this is quite complicated, but if you carefully observe the graph of the given...
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This makes sense but there is missing the part of obtaining the volume
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