Question 214689
You can do this in two steps:
1). Find the area of the oval (elliptical) bottom of the dry pool.
2). Find the inside area of the oval-shaped side.
Add these together to get total inside surface area.
1) The area of an ellipse is given by:
{{{A = (pi)*a*b}}} where a = the length of the semi-major axis and b = the length of the semi-minor axis.
In this pool, the semi-major axis (a) is 107/2 = 53.5ft.
The length of the semi-minor axis (b) is 41/2 = 20.5ft.
{{{A = (pi)(53.5)(20.5)}}}
{{{highlight_green(A = 3445.5)}}}sq.ft.
2) The area of the of the iside of the elliptical-shaped pool by multplying the circumference of the pool by the depth of the pool.
The approximate circumference (C) of the elliptical pool is given by:
{{{C = 2(pi)sqrt((a^2+b^2)/2)}}}
{{{a^2 = 53.5^2}}}
{{{a^2 = 2862.25}}}
{{{b^2 = 20.5^2}}}
{{{b^2 = 420.25}}}
{{{C = 2(pi)sqrt((2862.25+420.25)/2)}}}
{{{C = 254.55}}}
Now multiply this by the depth of the pool ({{{d = 5}}}ft.)
{{{C*d = (254.55)*(5)}}}={{{highlight_green(1272.75)}}}sq.ft. This is the area of the inside wall of the pool. Add the two areas to get the total surface area (S) of the inside of the dry pool.
{{{S = 3445.5+1272.75}}}
{{{highlight(S = 4718)}}}sq.ft.