As viewed from above, a swimming pool has the shape of the ellipse given by x21600+y2400=1 the cross sections perpendicular to the ground and parallel to the y-axis are squares. find the total volume of the pool. (assume the units of length and area are feet and square feet respectively. do not put units in your answer.)

the total volume of the pool = 85,333.3ft^3

Explanation:

Since the cross sections perpendicular to the ground and parallel to the y-axis are squares.This means that the dimensions of each square-shaped slice are:

width = 2y

depth = 2y

where y is a function of x.

The thickness of each slice is equal to dx

The area of each cross-sectional slice is

(2y)(2y) = 4y^2

The volume of each cross-section is

dV = 4y^2 dx. .....1

Whe need to write y as a function of x.

This comes from the equation of the ellipse:

x^2/a^2+y^2/b^2=1

with a^2 = 1600 and b^2 = 400

Then a = 40 and b = 20

Rearranging the equation gives:

y^2 = b^2(1 - x^2/a^2)

This can be substituted into the equation for dV above: (eqn 1 above.)

dV = 4b^2(1 - x^2/a^2)dx

The volume of the pool is the sum of all of the slices, which you calculate by integrating dV. However, due to symmetry, you can calculate the volume of half of the pool, and then multiply by 2 to get the total volume. This means that you integrate from the origin of the axes (x=0) to the edge of the pool (x=a).

We will denote the “half-volume” by Vh.

Vh= integral of dV from 0 to a

Vh = [4b^2( x - x^3/3a^2) ] from 0 to a

Substituting a we have.

Vh = 4b^2(a - a/3)

Vh = (8/3)ab^2

V = 2Vh = (16/3)ab^2

Substituting the values of a and b

V = (16/3)(40 × 20^2)

V = 85,333.3ft^3

trig 15.3 cos

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach