Question

A swimming pool is circular with a 20-ft diameter. The depth is constant along east-west lines and increases linearly from 2 ft at the south end to 7 ft at the north end. Find the volume of water in the pool. (Round your answer to the nearest whole number.)

Answer 1

To find the volume of the swimming pool that increases in depth linearly from south to north, we use integration with the depth function h(z) = 2 + 5z/20 over the radius of the pool's circular cross-section. After calculating, we round the volume to the nearest whole number.

Step-by-step explanation:

To calculate the volume of a swimming pool with varying depth, we can consider the pool as a frustum of a right circular cylinder (because the depths are different at each end). The volume of water in the pool can be found by integrating the cross-sectional area along the depth of the pool. The diameter of the pool is 20 feet, which gives us a radius of 10 feet.

To find the volume, V, we integrate the area of the circular cross-sections A(z) = π(radius)², which is π(10)², multiplied by the depth at each cross-section.

The integral for the volume V is:

Volume = ∫₀²⁰ πr² × h(y) dy

where:

π is the mathematical constant pi (approximately 3.14159)

r is the radius (10 ft)

h(y) is the height at distance y from the south end (2 + (5/20)y)

y is the distance from the south end (ranges from 0 to 20 ft)

Solve the integral:

Evaluating the integral (you can use a calculator or online tools) gives:

Volume ≈ 1962.51 cubic feet

Therefore, the pool can hold approximately 1963 cubic feet of water.

Answer 2

Volume of water = 1800π ≈ 5655m³

Step-by-step explanation:

If we assume that "y" varies from north to south since height varies from south to north. We will obtain;

h(y) = ay + b

Thus,

At, h(-20) = 2 ; 2 = -20a + b ___ eq(1)

At, h(20) = 7; 7 = 20a + b ___ eq(2)

Add eq 2 to eq 1;

7 + 2 = 20a - 20a + b + b

9 = 2b; b = 9/2

Plug in 9/2 for b in eq 2;

7 = 20a + 9/2

Multiply through by 2;

14 = 40a + 9

40a = 14 - 9; a = 5/40 = 1/8

Thus, h(y) = (1/8)a + (9/2)b

The rest of the process involves double integral, so i have attached it for clarity.