1. Let
be the three points of intersection, i.e. the solutions to
. They are approximately



Then the area
is

since over the interval
we have
, and over the interval
we have
.

2. Using the washer method, we generate washers with inner radius
and outer radius
. Each washer has volume
, so that the volume is given by the integral

3. Each semicircular cross section has diameter
. The area of a semicircle with diameter
is
, so the volume is

4.
is continuous and differentiable everywhere, so the the mean value theorem applies. We have

and by the MVT there is at least one
such that



for integers
, but only one solution falls in the interval
when
, giving

5. Take the derivative of the velocity function:

We have
when
. For
, we see that
, while for
, we see that
. So the particle is speeding up on the interval
and slowing down on the interval
.