Hint
Recover y first. For the enclosed region, the horizontal line is above a positive curve, so the area and rotated volume use different integrands.
Worked solution
(a)
Integration gives \(\displaystyle y=3x^{2}-6x+C\), with \(\displaystyle C=8\) from the point. The derivative vanishes at \(\displaystyle x=1\), where \(\displaystyle y=5\). The second derivative is 6>0, so this is a minimum.
Answer: \(y=3x^2-6x+8\); minimum at (1,5).
(b)
Intersections satisfy \(\displaystyle 3x^{2}-6x+8=8\), giving \(\displaystyle x=0\) and 2. On (0,2), the line is above the curve. Thus \(\displaystyle A=\int _{0}^{2}(6x-3x^{2})dx=[3x^{2}-x^{3}]_{0}^{2}=4 \mathrm{cm}^{2}\).
Answer: \(\displaystyle 4 \mathrm{cm}^{2}\).
(c)
The outer radius is 8 and inner radius \(\displaystyle y=3x^{2}-6x+8\), which is at least 5. The washer integral is \(\displaystyle \pi \int _{0}^{2}{64-(3x^{2}-6x+8)^{2}}dx\). Its integrand expands to \(\displaystyle -9x^{4}+36x^{3}-84x^{2}+96x\). The integral is \(\displaystyle \pi {- \frac{288}{5}+ 144- 224+ 192}= \frac{272\pi }{5}\mathrm{cm}^{3}\).
Answer: \(\displaystyle \frac{272\pi }{5}\) \(\displaystyle \mathrm{cm}^{3}\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 3 marks
- 1 mark: Recover the complete curve using the point condition.
- 1 mark: Find the stationary point (1,5).
- 1 mark: Classify a minimum with a valid derivative or completed-square argument.
Alternative methods: Write \(\displaystyle y=3(x-1)^{2}+5\) to confirm the classification.
Common error: Using the given gradient itself as the curve ordinate.
(b) · 3 marks
- 1 mark: Find both intersections 0 and 2 and the correct upper curve.
- 1 mark: Set up the area as upper minus lower.
- 1 mark: Evaluate \(\displaystyle 4 \mathrm{cm}^{2}\).
Alternative methods: A complete symmetric area calculation is accepted.
Common error: Integrating the curve alone instead of the vertical separation.
(c) · 3 marks
- 1 mark: Use the correct squared outer and inner radii over [0,2].
- 1 mark: Expand and integrate the washer expression correctly.
- 1 mark: Give \(\displaystyle \frac{272\pi }{5}\mathrm{cm}^{3}\) exactly.
Alternative methods: Subtract the inner solid volume from the enclosing cylinder volume.
Common error: Squaring the area integrand (8−y) instead of using \(\displaystyle 64-y^{2}\).
Total: 9 marks.