What is the volume of the solid produced by revolving #f(x)=cscx-cotx, x in [pi/8,pi/3] #around the x-axis?

Answer 1

A pretty ugly answer, but I got:

#V = ((48 - 16sqrt3 + 5pi)sqrt2 + 32sqrt3 - 10pi - 96sqrt(2 - sqrt2))/(24(2-sqrt2))pi#
#~~ 0.3216#

and Wolfram Alpha agrees!

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...

#= 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/8 = 22.5^@#). I just used Wolfram Alpha to save time on #csc(pi/8)# and #cot(pi/8)#.
#= 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 #sqrt2# out of. Since #sqrt6 = sqrt3sqrt2#, make sure you don't accidentally write #16# instead of #16sqrt3#!
#= color(blue)(((48 - 16sqrt3 + 5pi)sqrt2 + 32sqrt3 - 10pi - 96sqrt(2 - sqrt2))/(24(2-sqrt2))pi)#
#~~ 0.1024pi#
#~~ color(blue)(0.3216)#
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Answer 2

To find the volume of the solid produced by revolving the function ( f(x) = \csc x - \cot x ) on the interval ( \left[\frac{\pi}{8}, \frac{\pi}{3}\right] ) around the x-axis, you can use the method of cylindrical shells.

The volume ( V ) is given by the integral:

[ V = \int_{\frac{\pi}{8}}^{\frac{\pi}{3}} 2\pi x \left(\csc x - \cot x\right) , dx ]

Evaluate this integral to find the volume of the solid.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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