Vase Volume: Axis and Fill Height
A potter designs a vase whose side profile, when measured in centimetres, is given by y=4−x^2 for 0≤x≤2, rotated about the y-axis to create a solid of revolution. The base is at y=0 and the rim at y=4. The potter's assistant computes the volume using the formula V=π∫_0^2 (4−x^2)^2 dx and gets 256π/15 cm^3, then claims this is the volume of the vase. A second assistant uses cylindrical shells and gets 8π cm^3. The workshop needs the actual volume, and also wants to know the volume of water when the vase is filled to height y=3. The profile is not a function of y over the entire interval? Actually x=√(4−y) for y from 0 to4. The assistants disagree. Determine which method is appropriate, compute the correct total volume, and find the fill volume up to y=3. Also decide whether the first assistant's integral computes a different geometric quantity.
1. Which assistant's value is the vase volume? Identify the geometric error in the other integral and compute the correct total volume.
Read the worked answer
The vase is rotated about the y-axis, so use horizontal washers. For a fixed y, the radius is x=√(4−y), with 0≤y≤4. Thus
V=π∫_0^4(4−y)dy=π[4y−y^2/2]_0^4=8π cm^3.
The shell method agrees:
V=∫_0^2 2πx(4−x^2)dx=8π.
The first assistant computed π∫_0^2(4−x^2)^2 dx=256π/15, which is the volume obtained by rotating the region about the x-axis, not the y-axis. The second assistant is correct.
2. Find the volume of water when filled to y=3, and explain why it is not simply (3/4) of the total volume even though 3 is three-quarters of the height 4.
Read the worked answer
At height y, the radius is √(4−y), so
V(3)=π∫_0^3(4−y)dy=π[4y−y^2/2]_0^3=15π/2 cm^3.
The total is 8π, so the fraction is (15π/2)/(8π)=15/16, not 3/4. The cross-sectional area π(4−y) shrinks to zero near the rim; the top quarter of the height, y=3 to4, contributes only
π∫_3^4(4−y)dy=π/2,
which is much less than a proportional quarter. Volume is the integral of a linearly decreasing area, so it depends quadratically on height, not linearly.