Mixed Application Problems
A river engineer needs to know the volume of water flowing through a cross-section that is bounded by two curves, while a biologist models a population with a logistic differential equation. Both problems require you to seamlessly combine areas, volumes and differential equations — the three pillars of Module 9 — in a single solution. This lesson trains that multi-step thinking.
Consider the region enclosed by $y = x$ and $y = x^2$. Before using any formula — which curve is on top between their intersection points, and how would you set up the area integral? Sketch and write your reasoning below.
Every mixed calculus problem rewards two habits: sketch first, then classify (which technique?), then set up the integral precisely before evaluating. Rushing to algebra without a diagram is the single biggest cause of errors in multi-step questions.
The sketch-and-classify strategy: (1) draw the curves and shade the region, (2) identify the technique (area between curves, washer, disk, or DE application), (3) write the integral in full before touching a calculator.
Area: $A = \int_a^b [f(x)-g(x)]\,dx$ · Volume: $V = \pi\int_a^b [f(x)^2-g(x)^2]\,dx$
Key facts
- Area between $f$ and $g$: $A = \displaystyle\int_a^b [f(x)-g(x)]\,dx$ when $f \geq g$
- Washer volume: $V = \pi\displaystyle\int_a^b [R(x)^2 - r(x)^2]\,dx$
- DE solution method: separate variables, integrate, apply initial condition
Concepts
- Why you must sketch the region before choosing the technique
- How combining area and volume in one problem requires careful limit identification
- Why the constant of integration and initial conditions matter in DE application problems
Skills
- Solve multi-step problems combining areas, volumes, and differential equations
- Identify intersections, sketch regions, and select the correct integral setup
- Apply the general strategy: sketch, classify, integrate, verify
Mixed application problems combine two or three techniques in one question. The four-step strategy never fails:
- Sketch. Draw both curves (or the region), mark all intercepts and intersection points.
- Classify. Identify each sub-task: is it an area, a volume, or a DE? Note which formula applies.
- Set up. Write the integral or DE in full with correct limits before evaluating.
- Verify. Check units, sign, and reasonableness. Area must be positive; volume must be positive.
Worked through the hook: For $y = x$ and $y = x^2$:
- Intersections: $x = x^2 \Rightarrow x(x-1)=0$, so $x=0$ and $x=1$.
- On $[0,1]$: $x \geq x^2$ (check with $x=\tfrac{1}{2}$: $\tfrac{1}{2} > \tfrac{1}{4}$).
- Area: $A = \displaystyle\int_0^1 (x - x^2)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$.
- Volume (washer about $x$-axis): $V = \pi\displaystyle\int_0^1 (x^2 - x^4)\,dx = \pi\left[\dfrac{x^3}{3}-\dfrac{x^5}{5}\right]_0^1 = \pi\left(\dfrac{1}{3}-\dfrac{1}{5}\right) = \dfrac{2\pi}{15}$.
Mixed application problems combine two or three techniques in one question. The four-step strategy never fails:
Pause — copy the four-step multi-step strategy (identify technique, write formula, compute, verify) with a note on when to use the complement shortcut into your book.
Quick check: The curves $y = \sqrt{x}$ and $y = x^2$ intersect at $x = 0$ and $x = 1$. On $[0,1]$, which expression correctly gives the enclosed area?
We just saw the four-step multi-step strategy: identify the technique (area, volume, DE), extract the relevant formula, compute sub-parts sequentially, verify units and sign. That raises a question: when a single region is both integrated for area and rotated for volume, what is the exact pair of formulas you evaluate, and how do you reuse the same limits? This card answers it → $A=\int_a^b[f-g]\,dx$ and $V=\pi\int_a^b([f]^2-[g]^2)\,dx$; limits $a,b$ are the same intersection points for both.
Some HSC problems ask you to find both the area of a region and the volume when that region is rotated. The key is that the integration limits are shared — you find them once from the intersection sketch and reuse them for both calculations.
Example: The region between $y = \sqrt{x}$ and $y = x^2$ for $0 \leq x \leq 1$.
- Area: $A = \displaystyle\int_0^1 (\sqrt{x}-x^2)\,dx = \left[\dfrac{2}{3}x^{3/2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{2}{3} - \dfrac{1}{3} = \dfrac{1}{3}$.
- Volume about $x$-axis (washer): $V = \pi\displaystyle\int_0^1 (x - x^4)\,dx = \pi\left[\dfrac{x^2}{2}-\dfrac{x^5}{5}\right]_0^1 = \pi\!\left(\dfrac{1}{2}-\dfrac{1}{5}\right) = \dfrac{3\pi}{10}$.
Some HSC problems ask you to find both the area of a region and the volume when that region is rotated. The key is that the integration limits are shared — you find them once from the intersection sketch and...
Pause — copy the combined area-volume pair: $A=\int_a^b[f-g]\,dx$ and $V=\pi\int_a^b([f]^2-[g]^2)\,dx$, noting that limits $a,b$ are identical for both into your book.
Did you get this? True or false: when revolving the region between $y = \sqrt{x}$ and $y = x^2$ about the $x$-axis using the washer method, the outer radius is $R(x) = \sqrt{x}$ and the inner radius is $r(x) = x^2$.
Worked examples · 3 in a row, reveal as you go
The region $R$ is enclosed by $y = 4 - x^2$ and $y = x + 2$. (a) Find the area of $R$. (b) Find the volume when $R$ is rotated about the $x$-axis.
A curve passes through $(0, 1)$ and satisfies $\dfrac{dy}{dx} = 3x^2$. Find the area between the curve and the $x$-axis from $x=0$ to $x=2$.
Find the area enclosed between $x = y^2$ and $x = 2 - y^2$.
Fill the gap: The volume generated when the region between $y = x$ and $y = x^2$ ($0 \leq x \leq 1$) is rotated about the $x$-axis is $V = \pi\displaystyle\int_0^1(x^2 - x^4)\,dx = \dfrac{}{15}\pi$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the volume obtained by rotating the region between $y = x$ and $y = x^2$ about the $x$-axis can be written as $\pi\displaystyle\int_0^1 (x - x^2)^2\,dx$.
Activities · practice with the ideas
Find the area enclosed by $y = x^2$ and $y = 4$. Show the full integration setup.
The region $R$ is enclosed by $y = \sqrt{x}$ and $y = x^2$. Find the volume when $R$ is rotated about the $x$-axis using the washer method.
A curve satisfies $\dfrac{dy}{dx} = 6x^2 - 2$ and passes through $(1, 4)$. Find the curve, then find the area between the curve and $y = 0$ from $x = 0$ to $x = 2$.
Find the area enclosed by $x = y^2 - 1$ and $x = 3 - y^2$. (Integrate with respect to $y$.)
The region between $y = \cos x$ and $y = \sin x$ on $[0, \pi/4]$ is rotated about the $x$-axis. Write (do not evaluate) the integral for the volume using the washer method, clearly identifying $R(x)$ and $r(x)$.
Odd one out: Three of these statements about the region bounded by $y = x$ and $y = x^2$ on $[0,1]$ are correct. Which one is NOT?
Earlier you sketched $y = x$ and $y = x^2$ and considered which technique to use for area and volume.
The region is bounded between $x=0$ and $x=1$, with $y=x$ on top. Area $= \dfrac{1}{6}$. For volume, the washer method requires $R^2 - r^2 = x^2 - x^4$ — not $(x-x^2)^2$. The distinction between these two expressions is the most common source of error in this type of question.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the area enclosed by $y = x^2$ and $y = 4$. (2 marks)
Q2. The region $R$ is enclosed by $y = \sqrt{x}$ and $y = x^2$. Find the volume generated when $R$ is rotated about the $x$-axis. Leave your answer in exact form. (3 marks)
Q3. A curve satisfies $\dfrac{dy}{dx} = 3x^2 - 3$ and passes through $(2, 3)$. Find the equation of the curve, then find the area between the curve and the $x$-axis from $x = 0$ to $x = 1$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Intersections: $x=\pm2$. $A = \int_{-2}^2(4-x^2)\,dx = \left[4x-\frac{x^3}{3}\right]_{-2}^2 = (8-\frac{8}{3})-(-8+\frac{8}{3}) = \frac{32}{3}$ sq units.
2. $V = \pi\int_0^1(x-x^4)\,dx = \pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1 = \pi(\frac{1}{2}-\frac{1}{5}) = \frac{3\pi}{10}$.
3. $y = 2x^3-2x+C$; $y(1)=4 \Rightarrow 2-2+C=4 \Rightarrow C=4$, so $y=2x^3-2x+4$. On $[0,2]$: $y \geq 0$ (check: $y(0)=4, y(1)=4, y(2)=16$). $A = \int_0^2(2x^3-2x+4)\,dx = [\frac{x^4}{2}-x^2+4x]_0^2 = 8-4+8 = 12$.
4. Intersections: $y=\pm\sqrt{2}$. $A = \int_{-\sqrt{2}}^{\sqrt{2}}(4-2y^2)\,dy = [4y-\frac{2y^3}{3}]_{-\sqrt2}^{\sqrt2} = 2(4\sqrt2-\frac{2(2\sqrt2)}{3}) = \frac{16\sqrt2}{3}$.
5. $V = \pi\int_0^{\pi/4}(\cos^2x-\sin^2x)\,dx = \pi\int_0^{\pi/4}\cos 2x\,dx$.
Q1 (2 marks): $x^2=4 \Rightarrow x=\pm2$ [1]. $A = \int_{-2}^2(4-x^2)\,dx = \frac{32}{3}$ sq units [1].
Q2 (3 marks): $R(x)=\sqrt x$, $r(x)=x^2$; limits $0$ to $1$ [1]. $V=\pi\int_0^1(x-x^4)\,dx$ [1] $= \pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1 = \frac{3\pi}{10}$ [1].
Q3 (3 marks): $y=x^3-3x+C$; $y(2)=8-6+C=3 \Rightarrow C=1$, so $y=x^3-3x+1$ [1]. On $[0,1]$: check $y(0)=1>0$, $y(1)=-1<0$ — split at root. Root: $x^3-3x+1=0$ in $(0,1)$ ≈ $0.347$; for an exact answer use numerical value or note that $A = \int_0^1|x^3-3x+1|\,dx$ [1]; evaluating gives $A = \frac{5}{4} - (-1) = \ldots$ (full exact working expected) [1].
Five timed questions combining areas, volumes and differential equations. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering mixed application questions. Lighter alternative to the boss.
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