Rotated diffuser profile
A lighting diffuser has a cross-section bounded by y = x e^(-x/2), the x-axis, x = 0 and x = 4, with x and y in centimetres. The profile is rotated about the x-axis to form a solid diffuser. The workshop needs the exact cross-sectional area and the exact volume of the solid. A technician uses a coarse numerical estimate and reports that the volume is about 5 cm³, but the design limit is 5 cm³, so the decision is borderline. The manager wants an exact check, not a rounded reading. She also asks whether extending the far boundary from x = 4 to x = 5 would keep the solid within the same 5 cm³ limit. Use integration by parts; all lengths are in cm. The curved boundary is positive on the interval.
1. Find the exact cross-sectional area bounded by the given curves.
Read the worked answer
A = ∫ from 0 to 4 x e^(-x/2) dx. Take u=x, dv=e^(-x/2)dx, so v=-2e^(-x/2). Then A = [-2x e^(-x/2)] from 0 to 4 + 2∫ from 0 to 4 e^(-x/2) dx = -8e^(-2) + 2[-2e^(-x/2)] from 0 to 4 = -8e^(-2) -4e^(-2)+4 = 4 - 12e^(-2) ≈ 4 - 1.624 = 2.376 cm².
2. Find the exact volume when rotated about the x-axis. Decide whether the x = 4 design passes the 5 cm³ limit, and whether the x = 5 extension would pass.
Read the worked answer
V = π∫ from 0 to 4 (x e^(-x/2))² dx = π∫ from 0 to 4 x²e^(-x) dx. Since d/dx[-e^(-x)(x²+2x+2)] = x²e^(-x), V = π[-e^(-x)(x²+2x+2)] from 0 to 4 = π(2 - 26e^(-4)) ≈ π(2 - 0.4762) ≈ 4.786 cm³. This is below 5 cm³, so the x=4 design passes.
For x=5, V = π(2 - 37e^(-5)) ≈ π(2 - 0.2493) ≈ 5.499 cm³, above 5 cm³. Extending to x=5 would not keep the solid within the limit.