These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. The context and data in Question 6 are invented. Assumptions: no calculator except in Questions 4 and 6; angles in radians; exact answers unless a calculator is allowed, then three decimal places. Notation: π ∫ (a to b) [f(x)]² dx means π times the definite integral of [f(x)]² from x = a to x = b. Every solid is formed by one full turn about the stated axis. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
The region bounded by y = eˣ, the x-axis, x = 0 and x = 1 is spun about the x-axis. Which expression gives the volume of the solid?
- (A) π ∫ (0 to 1) e^(x²) dx
- (B) 2π ∫ (0 to 1) e^(2x) dx
- (C) π [∫ (0 to 1) eˣ dx]²
- (D) π ∫ (0 to 1) e^(2x) dx
Answer and explanation
Answer: (D). The radius at x is eˣ, so each disc has area π(eˣ)² = πe^(2x). Integrate along the x-axis from 0 to 1. (Its value is π(e² − 1)/2 ≈ 10.036, but you were only asked for the expression.)
- (A) squares the exponent instead of the function: (eˣ)² = e^(2x), not e^(x²).
- (B) uses 2π, which belongs to the circumference 2πr, not the disc area πr².
- (C) squares the integral instead of the radius. Each slice must be squared before you add the slices.
Question 2 (multiple choice · core)
The region bounded by y = 1/x, the x-axis, x = 1 and x = 4 is spun about the x-axis. What is the volume of the solid?
- (A) 3/4
- (B) 3π/4
- (C) 15π/16
- (D) π ln 4
Answer and explanation
Answer: (B). V = π ∫ (1 to 4) (1/x)² dx = π ∫ (1 to 4) x⁻² dx = π[−1/x] from 1 to 4 = π(−1/4 + 1) = 3π/4.
- (A) leaves out π, so it is the integral of the squared radius, not a volume.
- (C) puts the limits into the integrand 1/x² instead of into the antiderivative: π(1 − 1/16) = 15π/16.
- (D) forgets to square the radius: π ∫ (1 to 4) (1/x) dx = π ln 4.
Question 3 (multiple choice · core)
The region bounded by y = √x, the y-axis and the line y = 2 is spun about the y-axis. What is the volume of the solid?
- (A) 8π/3
- (B) 8π
- (C) 32π/5
- (D) 64π/5
Answer and explanation
Answer: (C). The region lies between the y-axis and the curve, so slice horizontally. At height y the curve is at x = y² (from y = √x), so the radius is y². The region runs from y = 0 to y = 2. V = π ∫ (0 to 2) (y²)² dy = π ∫ (0 to 2) y⁴ dy = π(32/5) = 32π/5.
- (A) forgets to square the radius: π ∫ (0 to 2) y² dy = 8π/3.
- (B) is π ∫ (0 to 4) (√x)² dx. That spins the region under the curve (a different region) about the x-axis (a different axis).
- (D) uses 2π instead of π.
Question 4 (multiple choice · calculator · core)
The region bounded by y = x e^(−x), the x-axis and the line x = 3 is spun about the x-axis. To three decimal places, what is the volume of the solid?
- (A) 0.235
- (B) 0.737
- (C) 1.473
- (D) 2.516
Answer and explanation
Answer: (B). The curve meets the x-axis at x = 0 and is positive on (0, 3], so the region touches the axis from x = 0 to x = 3. V = π ∫ (0 to 3) (x e^(−x))² dx ≈ 0.737.
- (A) leaves out π: ∫ (0 to 3) x² e^(−2x) dx ≈ 0.235.
- (C) uses 2π instead of π.
- (D) forgets to square the radius: π ∫ (0 to 3) x e^(−x) dx ≈ 2.516.
Question 5 (constructed response · core)
Let R be the region in the first quadrant bounded by y = 4 − x², the x-axis and the y-axis.
(a) Find the volume of the solid formed when R is spun about the x-axis. (b) Find the volume of the solid formed when R is spun about the y-axis. (c) The answers to (a) and (b) are different. Explain why, by describing the radius of a typical disc in each case.
Worked solution and suggested Marlbridge rubric
(a) R runs from x = 0 to x = 2 (where 4 − x² = 0) and touches the x-axis all the way. Radius 4 − x².
V = π ∫ (0 to 2) (4 − x²)² dx = π ∫ (0 to 2) (16 − 8x² + x⁴) dx = π[16x − 8x³/3 + x⁵/5] from 0 to 2 = π(32 − 64/3 + 32/5) = 256π/15 (about 53.617).
(b) R also touches the y-axis all the way, from y = 0 to y = 4. Solve y = 4 − x² for x (x ≥ 0): x = √(4 − y). Radius √(4 − y).
V = π ∫ (0 to 4) (√(4 − y))² dy = π ∫ (0 to 4) (4 − y) dy = π[4y − y²/2] from 0 to 4 = π(16 − 8) = 8π (about 25.133).
(c) In (a) each disc is vertical and its radius is the height of R above the x-axis, up to 4. In (b) each disc is horizontal and its radius is the width of R from the y-axis, at most 2. The two solids have different shapes, so their volumes differ. Spinning the same region about different axes does not give the same solid.
| Point | What earns it |
|---|---|
| 1 | Correct integral for (a): π ∫ (0 to 2) (4 − x²)² dx, with π, the square and limits |
| 1 | Value 256π/15 |
| 1 | Correct integral for (b) in y: π ∫ (0 to 4) (4 − y) dy (or with √(4 − y) squared), with y-limits 0 and 4 |
| 1 | Value 8π, and an explanation in (c) that the radius is a vertical distance in one case and a horizontal distance in the other |
A response that uses dx with limits 0 to 4 in (b) earns neither of the (b) marks.
Question 6 (constructed response · calculator · core)
A candle maker pours wax into a mould. The solid candle is the shape formed by spinning a region about the vertical y-axis. At height y centimetres above the base, the candle’s radius is r(y) centimetres. Selected values are in the table. (The data are invented.)
| y (cm) | 0 | 4 | 8 | 12 | 16 |
|---|---|---|---|---|---|
| r(y) (cm) | 3 | 4.5 | 5 | 4.5 | 3 |
(a) Write an integral expression, in terms of r, for the volume of the candle. Include units. (b) Use a trapezoidal sum with the four subintervals in the table to estimate the volume of the candle. (c) The candle maker models the radius by r(y) = 3 + y/2 − y²/32 for 0 ≤ y ≤ 16. Check that this model agrees with the table, then use it to find the volume of the candle. (d) Interpret the answer to (c) in context.
Worked solution and suggested Marlbridge rubric
(a) Each horizontal slice is a disc of radius r(y) and thickness dy. V = π ∫ (0 to 16) [r(y)]² dy cubic centimetres.
(b) The slice areas πr² at the five heights are 9π, 20.25π, 25π, 20.25π and 9π cm². Each subinterval is 4 cm wide.
Trapezoidal sum = (4/2)[9π + 2(20.25π) + 2(25π) + 2(20.25π) + 9π] = 2π(149) = 298π ≈ 936.195 cm³.
(c) Check: r(0) = 3; r(4) = 3 + 2 − 0.5 = 4.5; r(8) = 3 + 4 − 2 = 5; r(12) = 3 + 6 − 4.5 = 4.5; r(16) = 3 + 8 − 8 = 3. All five match.
V = π ∫ (0 to 16) (3 + y/2 − y²/32)² dy ≈ 961.746 cm³ (exactly 4592π/15).
(d) The candle contains about 961.746 cubic centimetres of wax (just under one litre), if the model is accurate. The table estimate in (b) is about 2.7% lower; with only five readings the trapezoids cannot follow the curve exactly.
| Point | What earns it |
|---|---|
| 1 | Integral in (a) with π, r(y) squared, dy and limits 0 to 16 |
| 1 | Trapezoidal sum in (b) using areas πr² (not radii) with the correct weights; 298π or 936.195 |
| 1 | Checks the model against the table and sets up π ∫ (0 to 16) (3 + y/2 − y²/32)² dy |
| 1 | 961.746 with units cm³, interpreted as the volume of wax in the candle |
A trapezoidal sum of the radii alone, multiplied by π at the end, earns no mark for (b): squaring must happen at each height.
Question 7 (constructed response · stretch)
For b > 0, let R_b be the region bounded by y = 2√x, the x-axis and the line x = b.
(a) Show that the solid formed by spinning R_b about the x-axis has volume V(b) = 2πb². (b) Find the value of b for which this volume is 18π. (c) Take b = 3. A student says the volume when R₃ is spun about the y-axis is π ∫ (0 to 2√3) (y²/4)² dy. Explain why this is not correct, and name the method that is needed.
Worked solution and suggested Marlbridge rubric
(a) R_b touches the x-axis from x = 0 to x = b. The radius is 2√x, so V(b) = π ∫ (0 to b) (2√x)² dx = π ∫ (0 to b) 4x dx = π[2x²] from 0 to b = 2πb².
(b) 2πb² = 18π gives b² = 9, so b = 3 (b > 0).
(c) R₃ lies to the right of the curve x = y²/4 (from y = 2√x), between the curve and the line x = 3. It does not touch the y-axis except at the origin. The student’s integral uses radius y²/4: that is the volume of the solid formed by spinning the other region, between the y-axis and the curve, from y = 0 to y = 2√3. For R₃, each horizontal slice is a disc of radius 3 with a hole of radius y²/4 cut out: a washer. The washer method (Topic 8.11) is needed.
| Point | What earns it |
|---|---|
| 1 | Correct integral π ∫ (0 to b) (2√x)² dx and simplification to 2πb² |
| 1 | b = 3, with the negative root rejected |
| 1 | Explains that R₃ does not touch the y-axis, so there is a gap between the region and the axis |
| 1 | Identifies that the student’s integral spins the region between the y-axis and the curve, and that washers (outer radius 3, inner radius y²/4) are needed |
How did you do?
- Q1, Q2 or Q4 wrong: reread “The disc method” and Worked example 1 in the study guide. Square the radius, keep π, never 2π.
- Q3 or Q5(b) wrong: redo Worked example 2 and “Revolving about the y-axis”: slice horizontally, write x in terms of y, use y-limits.
- Q6 wrong: remember that a table gives radii, and each slice’s area is π times the radius squared. Worked example 3 shows the calculator notation.
- Q7(c) wrong: see the third point under “The disc method”: discs need the region to touch the axis.
Then tick off the topic checklist.