Question

The region under the curves `y=cosxsqrtsinx, 0<=x<=pi/2` is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?

Answer 1

`V=piint_0^(pi/2)(cosxsqrtsinx)^2dx=pi/3`

Explanation:

A sketch is provided:

graph{cos(x)sqrt(sin(x)) [-1.5, 1.5, -1, 4]}

As `sinx>=0` for `0<=x<=pi`, `y` is defined for the interval of interest, `0<=x<=pi/2`. As such, there will only be one volume of revolution (not two, as the question implies).

When this solid is rotated around the `x`-axis, we essentially are creating an infinite amount of cylinders with a radius given by the function value at each point `0<=x<=pi/2`.

The volume will be the sum of the volumes of these infinitesimally small cylinders, whose volumes are given individually by `piy^2dx`, where `piy^2` is the area of the face of the circle and `dx` is the infinitesimally small thickness of each cylinder.

So, the volume of the entire solid will be given by the integral of this function within the bounds, or:

`int_0^(pi/2)piy^2dx=piint_0^(pi/2)(cosxsqrtsinx)^2dx=piint_0^(pi/2)cos^2xsinxdx`

Let `u=cosx`, implying that `du=-sinxdx`. This also will cause the bounds to change from `x=0=>u=cos(0)=1` and `x=pi/2=>u=cos(pi/2)=0`:

`=-piint_0^(pi/2)cos^2x(-sinxdx)=-piint_1^0u^2du=-pi[u^3/3]_1^0`

`=-pi(0^3/3-1^3/3)=-pi(-1/3)=pi/3`