IGCSE 0606 and Cross-Section Volumes: Scope and a Worked Extension
Cambridge 0606 is Additional Mathematics. This worked extension shows why a first-quadrant cross-section problem gives volume 8, not 32.
Reviewed by: Simon Clarke · IGCSE Physics Tutor · since 2015
In this article (10 sections)
Cambridge IGCSE Additional Mathematics (0606) is distinct from Cambridge IGCSE Mathematics (0580/0980) and from qualifications named Further Mathematics. Calculus content must be tied to the specific syllabus and examination year. A general solid-volume question with prescribed cross-sections should not automatically be presented as a required task for every IGCSE course.
The example below is an extension connecting graph interpretation, areas and integration. Its purpose is to explain the geometry correctly. Whether it belongs in an exam preparation plan depends on the qualification’s official syllabus.
Define the region before integrating
Take the region in the first quadrant bounded by x = 0, y = x² and y = 4. At a given height y between 0 and 4, x ranges from 0 to √y. A horizontal segment across the base therefore has length √y. The negative root −√y is outside this first-quadrant region.
Square cross-sections perpendicular to the y-axis
Suppose each horizontal segment is the side of a square cross-section. The side length is s(y) = √y, so the area is A(y) = (√y)² = y. The volume is the integral of area over height:
V = ∫₀⁴ y dy = [y²/2]₀⁴ = 8 cubic units.
The units also provide a check: a squared length gives area; multiplying area by a small height gives volume. Integrating the side length √y directly would not compute the volume of these squares.
For your IGCSE prep
Where the answer 32 comes from
If instead the base includes both sides of the parabola, a horizontal segment runs from −√y to +√y. Its length is 2√y, so a square on it has area 4y. The volume is then ∫₀⁴ 4y dy = 32 cubic units. That is a different region, including the second quadrant.
Both integrals are mathematically valid for their respective regions. Combining the phrase “first quadrant” with the width 2√y changes the problem halfway through the solution and produces the wrong answer for the stated region.
Changing the cross-section changes the area formula
If the same first-quadrant width √y is the base of a rectangle whose height is twice its base, then A(y) = 2y and the volume is 16. If it is the side of an equilateral triangle, A(y) = (√3/4)y and the volume is 2√3. Write the shape’s area formula before substituting the width.
A practical checking sequence
- Sketch and label the actual bounded region, including any quadrant restriction.
- Choose the slicing direction and write the segment endpoints.
- Subtract the endpoints to obtain width; do not add an excluded negative root.
- Convert width to area using the cross-section shape.
- Integrate over the stated interval and check cubic units.
For syllabus planning, use the 0606 Additional Mathematics document when that is your course. Do not rename it Further Mathematics or generalise one extension problem to all IGCSE students.
Frequently asked questions
What is Cambridge syllabus 0606 called?
Cambridge IGCSE Additional Mathematics.
Why is the first-quadrant volume 8?
The square side is √y, not 2√y. Its area is y, and the integral from 0 to 4 is 8.
When would 32 be correct?
When the base extends from −√y to +√y at each height, giving square area 4y. That is not the same first-quadrant region.