Vase Volume from a Rotated Profile and the Mean-Radius Trap
A potter is designing a cylindrical-symmetry vase by rotating the curve r(x)=a+1/(x+1) centimetres about the x-axis, for 0≤x≤4, where x is height in cm and r is radius. The parameter a>0 is chosen so the interior volume is exactly 500 cm^3. The potter's assistant estimates capacity by taking the mean radius over the height, forming a cylinder of height 4 cm, and multiplying by π. The assistant argues this is accurate because the radius changes smoothly; the potter is unsure whether the estimate is too high or too low. You are asked to find a, evaluate the volume integral carefully, and assess the cylinder approximation. Use centimetres throughout and give the final volume in litres where appropriate. The profile is measured to the inner wall; wall thickness is ignored. The integral may be left in exact form until the final numerical value.
1. Set up and evaluate the volume integral to find a. Give a to three decimal places.
Read the worked answer
Using disks, V=π∫ from 0 to 4 of [a+1/(x+1)]^2 dx. Expand: [a+1/(x+1)]^2=a^2+2a/(x+1)+1/(x+1)^2. Integrate term by term: ∫0^4 a^2 dx=4a^2, ∫0^4 2a/(x+1) dx=2a ln5, ∫0^4 1/(x+1)^2 dx=[-1/(x+1)]0^4=1-1/5=4/5. Thus V=π(4a^2+2a ln5+4/5). Setting V=500 gives 4a^2+2a ln5+4/5=500/π. Numerically, 500/π≈159.155, so 4a^2+3.218876a-158.355=0. Solving the positive root: a=[-3.218876+sqrt(3.218876^2+4(4)(158.355))]/8≈5.902. So a≈5.902 cm.
2. The assistant uses the mean radius over the height to form a cylinder of height 4 cm. Decide whether this underestimates or overestimates the true volume, and quantify the error using the value of a found above.
Read the worked answer
The mean radius is rbar=(1/4)∫0^4 [a+1/(x+1)] dx=a+(ln5)/4. The assistant's cylinder volume is Vcyl=π(4)rbar^2=4πrbar^2. The true volume is π∫r^2 dx. By Cauchy-Schwarz or Jensen for the convex function u^2, (1/4)∫r^2 dx ≥ [(1/4)∫r dx]^2, with equality only if r is constant. Since r is strictly decreasing here, the cylinder estimate underestimates the true volume. Using a≈5.902, rbar≈5.902+0.402=6.304. Then Vcyl≈4π(6.304)^2≈499.5 cm^3, while the true volume is 500 cm^3. The underestimate is about 0.5 cm^3, or roughly 0.1%. So the approximation is very accurate but not exact.
3. If a is measured with an uncertainty of ±0.05 cm, use dV/da to estimate the resulting uncertainty in the volume. Comment on whether the 500 cm^3 target is sensitive to this measurement error.
Read the worked answer
Differentiate V(a)=π(4a^2+2a ln5+4/5): dV/da=π(8a+2ln5). At a≈5.902, 8a≈47.216 and 2ln5≈3.219, so dV/da≈π(50.435)≈158.45 cm^3 per cm. For Δa=±0.05, the linear estimate is ΔV≈158.45(0.05)≈7.9 cm^3. Thus the capacity could be about 492 to 508 cm^3, an uncertainty of roughly 1.6%. The target is moderately sensitive: a small radius-parameter error produces a measurable volume error, so the potter should control a to better than ±0.05 cm if 500 cm^3 must be met closely.