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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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.
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calendarEducator since 2012
write609 answers
starTop subjects are Math, Science, and History
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)`
This makes sense but there is missing the part of obtaining the volume
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