MTH1020 · Analysis of Change
Applications of Integration: Areas, Volumes & Arc Length
Week 12 of Monash MTH1020 Analysis of Change closes the unit by turning the definite integral into a measuring tool. It computes areas between curves, the average value of a function, volumes of solids of revolution by discs/washers and cylindrical shells, arc length, and the surface area of a solid of revolution. Each application is an integral you set up from a picture and then evaluate with the Week-11 techniques, and these are examinable final-exam questions where a correct set-up and clean evaluation earn the marks.
What this chapter covers
- 01Area between curves: for f(x) ≥ g(x) on [a, b], Area = ∫ₐᵇ [f(x) − g(x)] dx; find intersection points for limits
- 02Regions integrated in y when x is a function of y; use ∫|f − g| when the top curve switches
- 03Average value of a function: (1/(b−a)) ∫ₐᵇ f(x) dx
- 04Volumes of revolution — method of discs: rotate about the x-axis, V = π ∫ₐᵇ [f(x)]² dx
- 05Volumes of revolution — cylindrical shells: rotate about the y-axis, V = ∫ₐᵇ 2π x f(x) dx
- 06Choosing discs vs shells by which axis the region is rotated about and how it touches the axis
- 07Arc length: L = ∫ₐᵇ √(1 + (f′(x))²) dx (and the parametric form)
- 08Surface area of revolution about the x-axis: A = 2π ∫ₐᵇ f(x) √(1 + (f′(x))²) dx
Volume of revolution by the disc method
- +1Choose the method and write the formula. The region between y = f(x) and the x-axis is rotated about the x-axis, so use discs: V = π ∫ₐᵇ [f(x)]² dx, with f(x) = √x and limits a = 0, b = 4.
- +1Form the integrand. [f(x)]² = (√x)² = x, so V = π ∫ from 0 to 4 of x dx.
- +1Integrate. An antiderivative of x is x²/2, so V = π [x²/2] from 0 to 4.
- +1Evaluate at the limits. V = π (4²/2 − 0²/2) = π (16/2) = π·8 = 8π ≈ 25.1 cubic units.
Key terms
- Area between curves
- For f(x) ≥ g(x) on [a, b], the area is ∫ₐᵇ [f(x) − g(x)] dx (top minus bottom). Where the upper curve switches, integrate |f − g| piecewise.
- Average value of a function
- The mean height of f over [a, b]: (1/(b−a)) ∫ₐᵇ f(x) dx. Equivalently, ∫ₐᵇ f dx equals (b−a) times this average.
- Method of discs
- For a region between y = f(x) and the x-axis rotated about the x-axis, V = π ∫ₐᵇ [f(x)]² dx — each slice is a disc of radius f(x).
- Cylindrical shells
- For a region under y = f(x) rotated about the y-axis, V = ∫ₐᵇ 2π x f(x) dx — each slice is a thin cylindrical shell of radius x and height f(x).
- Arc length
- The length of y = f(x) from a to b: L = ∫ₐᵇ √(1 + (f′(x))²) dx, valid when f′ is continuous.
- Surface area of revolution
- For y = f(x) rotated about the x-axis, A = 2π ∫ₐᵇ f(x) √(1 + (f′(x))²) dx — the radius f(x) times the arc-length element.
Applications of Integration: Areas, Volumes & Arc Length FAQ
How do I decide between discs and shells for a volume of revolution?
Look at which axis the region is rotated about and how the region meets that axis. When a region sitting against the x-axis is rotated about the x-axis, each cross-section perpendicular to the axis is a disc of radius f(x), so V = π∫[f(x)]² dx. When a region under y = f(x) is rotated about the y-axis, slicing parallel to that axis gives thin cylindrical shells of radius x and height f(x), so V = ∫2πx f(x) dx. Sketching the region and one representative slice tells you which one applies far more reliably than memorising rules.
Why do I square the function in the disc formula?
Because each slice is a disc whose area is π(radius)², and the radius is the function value f(x). So the volume element is π[f(x)]² dx, and you must square f before integrating. For y = √x that means integrating (√x)² = x, not √x — forgetting to square (or squaring after integrating) is one of the most common Week-12 exam errors.
How do I find the limits for an area between two curves?
Set the curves equal, f(x) = g(x), and solve for their intersection points — these give the limits a and b. Between them, integrate the top curve minus the bottom curve, ∫ₐᵇ [f(x) − g(x)] dx. If the curves cross inside the interval so the 'top' one switches, split the integral at the crossing and integrate |f − g| on each piece, keeping every contribution positive.
Can Sia help me set up applications-of-integration problems?
Yes. Sia can help you sketch the region, choose discs or shells, set the correct limits, and then evaluate the integral with the right Week-11 technique — step by step on your own practice questions. It explains the method and checks your reasoning; it does not do graded assessment for you, and Monash academic-integrity rules apply.
Exam move
Every Week-12 application is 'draw the picture, set up the integral, evaluate it', so practise the set-up as its own skill. For volumes, sketch the region and a representative slice to choose discs versus shells, and remember to square the radius in the disc formula. For areas, find the intersections first and integrate top minus bottom, splitting where the curves cross. Keep the arc-length and surface-area formulas on a card, since they are easy marks once set up. Because these problems combine with the Week-11 techniques, rehearse them together — a volume integral you cannot evaluate is only half the marks. This is comprehensive final-exam material, so work full problems end to end and ask Sia to check both your set-up and your evaluation.
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