What is the volume of the solid produced by revolving #f(x)=cscx-cotx, x in [pi/8,pi/3] #around the x-axis?
1 Answer
A pretty ugly answer, but I got:
#V = ((48 - 16sqrt3 + 5pi)sqrt2 + 32sqrt3 - 10pi - 96sqrt(2 - sqrt2))/(24(2-sqrt2))pi#
#~~ 0.3216#
Unfortunately, it's the simplest exact numerical solution, apparently, we can't make it look any nicer. :)
DISCLAIMER: The integral is not that hard, but this answer requires a lot of simplification work!
First, let's see how this graph looks.
graph{(cscx - cotx) [0.3927, 1.047, 0, 0.8]}
Here we can see it's a simple curve within this interval. Along the x-axis, the easiest way to do this is to form discs that are perpendicular to the x-axis.
The integral in general is:
#\mathbf(V = int_(pi/8)^(pi/3) pir(x)^2dx)# where
#r(x) = cscx - cotx# is the function by which the radius varies for each disc we stack of width#dx# and area#pir(x)^2# along the x-axis.
#= pi int_(pi/8)^(pi/3) (cscx - cotx)^2dx#
#= pi int_(pi/8)^(pi/3) csc^2x - 2cscxcotx + cot^2xdx#
#= pi[ int_(pi/8)^(pi/3) csc^2xdx - 2 int_(pi/8)^(pi/3) cscxcotxdx + int_(pi/8)^(pi/3) cot^2xdx]#
If you recall...
#color(green)(d/(dx)[cscx] = -cscxcotx)# , similar to how#d/(dx)[secx] = secxtanx# . So let's adjust the middle integral to make it easier to integrate.#d/(dx)[tanx] = sec^2x# , so#color(green)(d/(dx)[cotx] = -csc^2x)# . So let's also adjust the first integral.- Since
#1 + tan^2x = sec^2x# ,#color(green)(1 + cot^2x = csc^2x)# , and so, let's adjust the third integral as well.
#= pi[-int_(pi/8)^(pi/3) -csc^2xdx + 2 int_(pi/8)^(pi/3) -cscxcotxdx - int_(pi/8)^(pi/3) 1 + (-csc^2x)dx]#
Now we're ready to evaluate each one.
#= pi[-cotx + 2cscx - x - cotx]#
#= pi|[2cscx - 2cotx - x]|_(pi"/"8)^(pi"/"3)#
#= pi[(2csc(pi/3) - 2cot(pi/3) - pi/3) - (2csc(pi/8) - 2cot(pi/8) - pi/8)]#
At this point, either use your calculator, or work out the second half using half-angle formulas (since
#= pi[(2*2/sqrt3 - 2sqrt3/3 - pi/3) - (2*2/(sqrt(2 - sqrt2)) - 2(1 + sqrt2) - pi/8)]#
Use common denominators to merge some fractions, distributing negative signs over parentheses carefully.
#= pi[((4sqrt3)/3 - (2sqrt3)/3 - pi/3) - (4/(sqrt(2 - sqrt2)) - 2 - 2sqrt2 - pi/8)]#
#= pi[(2sqrt3)/3 - pi/3 - 4/(sqrt(2 - sqrt2)) + 2 + 2sqrt2 + pi/8]#
#= pi[(16sqrt3)/24 - (8pi)/24 + 48/24 + (48sqrt2)/24 + (3pi)/24 - 4/(sqrt(2 - sqrt2))]#
Here we multiply by a unit fraction to get rid of the outer radical in the denominator.
#= pi[(16sqrt3 - 5pi + 48 + 48sqrt2)/24 - 4/(sqrt(2 - sqrt2))*(sqrt(2 - sqrt2))/(sqrt(2 - sqrt2))]#
Cross-multiply to merge fractions again.
#= pi[((16sqrt3 - 5pi + 48 + 48sqrt2)(2-sqrt2))/(24(2-sqrt2)) - (96sqrt(2 - sqrt2))/(24(2 - sqrt2))]#
#= pi[((16sqrt3 - 5pi + 48 + 48sqrt2)(2-sqrt2) - 96sqrt(2 - sqrt2))/(24(2-sqrt2))]#
Expand, and then cancel out anything you can.
#= pi[(32sqrt3 - 10pi cancel(+ 96) + 96sqrt2 - 16sqrt6 + 5pisqrt2 - 48sqrt2 cancel(- 96) - 96sqrt(2 - sqrt2))/(24(2-sqrt2))]#
#= pi[(48sqrt2 - 16sqrt6 + 5pisqrt2 + 32sqrt3 - 10pi - 96sqrt(2 - sqrt2))/(24(2-sqrt2))]#
Lastly, I found terms that I could factor
#= color(blue)(((48 - 16sqrt3 + 5pi)sqrt2 + 32sqrt3 - 10pi - 96sqrt(2 - sqrt2))/(24(2-sqrt2))pi)#
#~~ 0.1024pi#
#~~ color(blue)(0.3216)#
