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and whose height is 
.
and a base side length of 
) to our general variables for side length 
and height of the pyramid 
. We can think of this plotted on the coordinate plane, with the width of the pyramid solid being in the 
direction.
can be squared to find a general formula for the area of the pyramid’s base 
, this can be applied to the volume using cross-sections formula 
as the next step. Note that the equation 
solved for above is in terms of 
. Our new volume function, therefore, is also in terms of 
, and we assume the pyramid’s starting height is at 
, the appropriate bounds for the integral expression are: 
.
and 
and whose cross-sections are rectangles of height 
and perpendicular to the 
axis.
axis; therefore, the volume expression will be in terms of 
. 
. By applying this formula to our general volume formula 
, we get the following: 
.
must be determined. The problem specifies the length (or height) of the rectangle cross-sections is 
. This just leaves the value of w to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between 
and 
, therefore, 
.
and 
, the base has the following domain: 
.
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