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Mathematics Advanced

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Extension 1
Level 2

Question

The diagram below shows the graph of y=ln⁡xy=\ln x.
The shaded region, bounded by y=ln⁡xy=\ln x, the line x=4x=4 and the xx-axis, is rotated 360∘360^{\circ} about the yy-axis to form a solid.
Find the volume of the solid, leaving it in simplest exact form.
Question diagram
Worked solution
Step 1: Express the curve as x=eyx=e^{y}. The shaded region corresponds to 0≤y≤ln⁡40\le y\le\ln 4, with outer radius 44 and inner radius eye^{y} (washer method about the yy-axis).
V=π∫0ln⁡4(42−(ey)2) dyV=\pi\displaystyle\int_{0}^{\ln 4}\bigl(4^{2}-(e^{y})^{2}\bigr)\,dy.
V=π∫0ln⁡4(16−e2y) dyV=\pi\displaystyle\int_{0}^{\ln 4}(16-e^{2y})\,dy.
Step 2: Integrate.
V=π[16y−e2y2]0ln⁡4V=\pi\left[16y-\dfrac{e^{2y}}{2}\right]_{0}^{\ln 4}.
V=π((16ln⁡4−e2ln⁡42)−(0−12))V=\pi\left(\left(16\ln 4-\dfrac{e^{2\ln 4}}{2}\right)-\left(0-\dfrac{1}{2}\right)\right).
V=π(16ln⁡4−162+12)V=\pi\left(16\ln 4-\dfrac{16}{2}+\dfrac{1}{2}\right).
V=π(16ln⁡4−8+12)V=\pi\left(16\ln 4-8+\dfrac{1}{2}\right).
V=16πln⁡4−15π2V=16\pi\ln 4-\dfrac{15\pi}{2}.
Answer: 16πln⁡4−15π216\pi\ln 4-\dfrac{15\pi}{2}
HSC marking criteria

Total: 3 marks

  • 1 mark — Sets up V=π∫0ln⁡4(16−e2y) dyV=\pi\displaystyle\int_{0}^{\ln 4}(16-e^{2y})\,dy (or an equivalent shell integral).
  • 1 mark — Integrates correctly (including the lower-limit contribution +12+\frac{1}{2}).
  • 1 mark — Obtains 16πln⁡4−15π216\pi\ln 4-\dfrac{15\pi}{2}.
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