The bathysphere, an early type of submarine invented in the 1930s, had an interior diameter of 56 inches and featured a steel shell 1.5 inches thick. It included three windows, each with a 10-inch diameter. To estimate the volume of steel used in the construction of the bathysphere, calculate the volume of the outer sphere (based on the outer diameter) and subtract the volume of the inner sphere (based on the inner diameter). Round the result to the nearest whole number to determine the steel volume in cubic inches.

To estimate the volume of the steel used to make the bathysphere, we can treat the bathysphere as a sphere and calculate the volume of steel as the difference between the volume of the outer sphere and the volume of the inner sphere (the air-filled part).

First, we need to calculate the radius of the outer sphere and the inner sphere.

The outer radius (R) of the bathysphere is half of the inside diameter plus the thickness of the steel: R = (56 inches / 2) + 1.5 inches = 28 inches + 1.5 inches = 29.5 inches

The inner radius (r) of the bathysphere is just half of the inside diameter: r = 56 inches / 2 = 28 inches

Next, we use the formula for the volume of a sphere, which is V = (4/3)πr^3, where r is the radius of the sphere.

Volume of the outer sphere (V_outer): V_outer = (4/3)π(29.5 inches)^3

Volume of the inner sphere (V_inner): V_inner = (4/3)π(28 inches)^3

Volume of the steel (V_steel) is the difference between V_outer and V_inner: V_steel = V_outer - V_inner

Calculating the volumes: V_outer ≈ (4/3)π(29.5)^3 ≈ (4/3)π(25698.75) ≈ 107390 cubic inches V_inner ≈ (4/3)π(28)^3 ≈ (4/3)π(21952) ≈ 91942 cubic inches

V_steel ≈ 107390 - 91942 ≈ 15448 cubic inches

Note that we also need to subtract the volume of the steel displaced by the windows, but since we don't have the exact positions and overlaps, we can provide a rough estimate assuming they don't overlap.

Each window's volume (V_window) can also be calculated using the sphere volume formula where the radius for each window is 5 inches (half of 10 inches):

V_window ≈ (4/3)π(5 inches)^3 ≈ 523.6 cubic inches

Since there are three windows: Total volume of the windows (V_windows_total) = 3 * V_window ≈ 3 * 523.6 ≈ 1570.8 cubic inches

We adjust the volume of the steel by subtracting the volume of steel displaced by the windows:

Adjusted V_steel ≈ V_steel - V_windows_total ≈ 15448 - 1570.8 ≈ 13877.2 cubic inches

Rounding to the nearest whole number: Adjusted V_steel ≈ 13877 cubic inches

Therefore, it took approximately 13877 cubic inches of steel to make the bathysphere.