use cylindrical shells method to find the volume of the solid generated when the region enclosed by the given curves is revolved about the y-axis `y=2cos(9x^2)`   , `x=0 ` , ...

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The  formula for calculating the volume of a solid rotating about `y`-axis is

`V=2piint_a^bxf(x)dx,` `0leqa<b`

In your case `f(x)=2cos(9x^2)`  `a=0,b=sqrt(pi)/6`

hence we have

`V=2piint_0^(sqrt pi/6)x cdot 2cos(9x^2)dx=|(t=9x^2),(dt=18xdx),(a_t=0),(b_t=pi/4)|=`

` ` ` ` `(2pi)/9int_0^(pi/4)costdt=(2pi)/9sint |_0^(pi/4)=(2pi)/9(sqrt2/2-0)=(sqrt2pi)/9`

` `

Hence the volume of your rotating solid is `(sqrt2 pi)/9`

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