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 ` , `x=(sqrt(pi))/6` , `y=0`

### 1 Answer | Add Yours

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`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes