Calculus: Find the volume of the solid generated by revolving the region about the y-axis.?

Firstly , sketch the region .. To easily see what it looks like . using a TI-84 or its equivalent, just try y1 =√(sin(8x)) along the x axis , from x = 0 to π/16

This is a quarter of a cos-like curve, crawling up the y-axis,
starting at (1,0) and curving backwards to (0,π/16).

Since its defined as x = f(y) then its easiest to use ∫...dy and washers about the y-axis.
In this case, the washer has no center hole, so its really a circular slab.
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Solution :
Draw an integrating volume , with thickness "dy" around the y-axis
The washer volume is dV = π(R^2 -r^2) dy
The outer radius is R= √sin(8y) so R^2 = [√sin(8y)]^2 = sin(8y)

( as long as we have not gone past the first loop where cos becomes negative, and the square root is not defined... which we have not !! )
The inner radius , r , is just "0"

so dV =π(R^2 -r^2) dy = π[sin(8y)]dy from 0 to π/16
Integrating , :
V = ∫ dV = π∫sin(8y) dy from 0 to π/16
V = π[ -cos(8y) /8 ] from 0 to π/16

Evaluating top and bottom :
V = π[-cos(8π/16) / 8 +cos (0)/8]
V = π(0 + (1/8) (1)] = π/8 <----- exact answer
or
V= 0.392699... <---- numerical answer

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Confirmation :
On something like a TI-84 one can calculate
ans = π fnInt( sin(8x) ,x , 0 , π/16) = 0.39269...

or
One can use an internet TI-84 program , called VOLUME
which gives a nice sketch of the problem , with limits , and the rotating rectangle , and an answer , also V = 0.392699