Question

Plot the region R enclosed byy = 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.

Accepted Answer

Blue line is graph y=cos(3x) and red is graph y=cos x. From graph you can see that intersection points are x=0,\ pi/2,\ pi.   
If you want to calculate those points you need to solve equation cos3x=cosx. 
Let's first try to write cos3x by using only cos x. In order to do that we will use addition theorem:
cos(t+s)=cos t cos s-sin t sin s 
and formula for cosine and sine of double angle:
  cos2x=cos^2x-sin^2x, sin2x=2sinxcosx.
  cos3x=cosxcos2x-sinxsin2x= 
=cosx(cos^2x-sin^2x)-sin x cdot 2 sin x cos x= 
=cos^3x-sin^2xcos x-2sin^2 x cos x=cos^3x-3sin^2x cos x= 
=cos^3x-3(1-cos^2x)cos x=cos^3x -3cos x + 3cos^3x= 
=4cos^3x-3cos x 
Now we can solve our equation cos3x=cosx. 
4cos^3x-3cosx=cosx 
4cos^3x-4cosx=0 devide equation by 4
cos^3x-cosx=0 
cosx(cos^2x-1)=0 => cosx=0 or cos^2-1=-sinx=0 
cosx=0=>x=pi/2+k pi, k in ZZ which is equal to pi/2 for x in [0,pi]. 
-sinx=0 
sinx=0=>x=k pi, k in ZZ   which is equal to 0 for k=0  and pi for k=1. 
So points of intersection are, as we concluded from the graph,   x=0,\ pi/2,\ pi. If you want to find y coordinate of intersection points only calculate cos x for each of the points and get y=1, 0, -1 respectively.

Answer

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 functions youcan see that areas of the two parts are equal hence volumes we get by rotation of those shapes are equal as well.
  2pi int_0^(pi/2)(cos^2x-cos^2 3x+1+2cosx-2cos3x-2cosxcos3x)dx= 
2pi(8/3+pi)

Answer

This makes sense but there is missing the part of obtaining the volume

Math

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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