Area with Respect to the $y$-axis
Most area problems integrate left-to-right — but what if the region is easier to slice horizontally? When you flip your perspective and integrate with respect to $y$, awkward multi-piece problems often collapse to a single clean integral. Mastering this switch is essential for the Areas and Volumes module.
The curve $x = y^2$ and the line $x = 4$ enclose a region. Without computing — do you think the area is closer to 5, 11, or 22 square units? Sketch the region and write your reasoning.
Normally we slice a region into vertical strips of width $dx$ and integrate from $x = a$ to $x = b$. But for some regions — especially those described by equations of the form $x = f(y)$ — it is easier to slice into horizontal strips of height $dy$ and integrate from $y = c$ to $y = d$.
Each horizontal strip has width $x_R(y) - x_L(y)$ (right boundary minus left boundary) and height $dy$. Summing infinitely many strips gives:
$A = \displaystyle\int_c^d \bigl[x_R(y) - x_L(y)\bigr]\,dy$
If the left boundary is the $y$-axis ($x_L = 0$), this simplifies to $A = \displaystyle\int_c^d x(y)\,dy$.
Key facts
- $A = \displaystyle\int_c^d [x_R(y) - x_L(y)]\,dy$ for a region between two $x = f(y)$ curves
- Limits $c$ and $d$ are $y$-values at intersection points
- The integrand must be expressed entirely in terms of $y$
Concepts
- Why horizontal strips are the natural choice when boundaries are $x = f(y)$
- How to rewrite $y = f(x)$ as $x = g(y)$ when needed
- When $y$-integration is more efficient than $x$-integration
Skills
- Set up and evaluate $\displaystyle\int_c^d x(y)\,dy$ for standard regions
- Find intersection points and identify $x_R(y)$ and $x_L(y)$
- Decide whether to integrate with respect to $x$ or $y$ for efficiency
To integrate with respect to $y$, follow these four steps:
- Sketch the region. Label all boundaries and mark a sample horizontal strip showing $x_L$ and $x_R$.
- Find limits $c$ and $d$: solve the boundary equations simultaneously for $y$ to find the $y$-coordinates of the intersection points.
- Express boundaries in terms of $y$: rewrite each boundary as $x = (\text{function of }y)$, identifying which is $x_R$ and which is $x_L$.
- Integrate: $A = \displaystyle\int_c^d \bigl[x_R(y) - x_L(y)\bigr]\,dy$.
This mirrors the familiar $A = \displaystyle\int_a^b [y_{\text{top}}(x) - y_{\text{bottom}}(x)]\,dx$ — just with the roles of $x$ and $y$ swapped.
$y$-axis integral: (1) rearrange curve to $x=f(y)$; (2) convert limits to $y$-values; (3) $A=\int_c^d x\,dy=\int_c^d f(y)\,dy$; (4) evaluate.
Pause — copy the four-step $y$-axis integral method: (1) rearrange to $x=f(y)$; (2) find $y$-limits; (3) write $A=\int_c^d x\,dy$; (4) evaluate into your book.
Quick check: To find the area enclosed by $x = y^2$ and $x = 9$ using a $y$-integral, which limits should you use?
We just saw the four-step method for $y$-axis integrals: express $x$ as a function of $y$, swap the limits to $y$-values, integrate $x=f(y)$ with respect to $y$. That raises a question: when the region is bounded on the left by the $y$-axis (so $x_L=0$), how does the formula simplify, and what is the most common rearrangement mistake? This card answers it → $A=\int_c^d f(y)\,dy$ when the left boundary is the $y$-axis; rearrangement errors occur when students forget to invert (e.g.\ $y=x^2$ becomes $x=\sqrt{y}$, not $x=y^2$).
When the left boundary is the $y$-axis ($x_L = 0$) and the right boundary is $x = f(y)$, the formula becomes:
Example — hook answer: Find the area enclosed by $x = y^2$ and $x = 4$.
Step 1: Intersection: $y^2 = 4 \Rightarrow y = \pm 2$. Limits: $c = -2$, $d = 2$.
Step 2: $x_R = 4$, $x_L = y^2$ (since $4 \geq y^2$ on $[-2,2]$).
$= \left(8 - \tfrac{8}{3}\right) - \left(-8 + \tfrac{8}{3}\right) = \dfrac{16}{3} + \dfrac{16}{3} = \dfrac{32}{3} \approx 10.67$ sq units.
So the area is $\mathbf{\dfrac{32}{3}}$ square units — close to 11. How did your estimate compare?
When the left boundary is the $y$-axis ($x_L = 0$) and the right boundary is $x = f(y)$, the formula becomes:
Pause — copy the simplified form $A=\int_c^d f(y)\,dy$ for the area between a curve and the $y$-axis, and the rearrangement warning ($y=x^2\Rightarrow x=\sqrt{y}$ on the first quadrant) into your book.
Did you get this? True or false: to find the area between $x = y^2$ and the $y$-axis from $y = 0$ to $y = 3$, you evaluate $\displaystyle\int_0^3 y^2\,dy$.
Worked examples · 3 in a row, reveal as you go
Find the area enclosed between the curve $y = \sqrt{x}$ (i.e. $x = y^2$) and the $y$-axis from $y = 0$ to $y = 4$.
Find the area enclosed between $x = y^2$ and $x = 2 - y^2$.
Find the area between the parabola $y = x^2$ and the line $y = x$ using integration with respect to $y$.
Fill the gap: The area enclosed by $x = y^2$ and the $y$-axis from $y = 0$ to $y = 3$ is $\displaystyle\int_0^3 y^2\,dy = \Bigl[\dfrac{y^3}{3}\Bigr]_0^3 = $ square units.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when finding the area between $x = \sqrt{y}$ and $x = y$ from $y = 0$ to $y = 1$, the integrand is $\sqrt{y} - y$ (not $y - \sqrt{y}$).
Activities · practice with the ideas
Find the area enclosed between $x = y^2$ and the $y$-axis from $y = 0$ to $y = 2$.
Find the area enclosed by $x = 4 - y^2$ and the $y$-axis.
Find the area between $x = y^2$ and $x = y + 2$.
The curve $y = \ln x$ passes through $(1, 0)$ and $(e, 1)$. Rewrite it as $x = e^y$ and find the area between this curve and the $y$-axis from $y = 0$ to $y = 1$.
State one advantage of using $y$-integration over $x$-integration for the region bounded by $x = y^2 - 1$ and $x = 3$.
Odd one out: Three of these are set up correctly for a $y$-axis integral. Which one is NOT?
Earlier you estimated the area enclosed by $x = y^2$ and $x = 4$.
The exact answer is $\dfrac{32}{3} \approx 10.67$ square units. The key insight: by treating the region as horizontal strips of length $(4 - y^2)$ and height $dy$, the integral $\displaystyle\int_{-2}^{2}(4 - y^2)\,dy$ captures the full area in one calculation. No splitting required.
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 between $x = y^2$ and the $y$-axis from $y = 0$ to $y = 3$. (2 marks)
Q2. Find the area enclosed between $x = y^2$ and $x = y + 2$. (3 marks)
Q3. The region $R$ is bounded by $y = \ln x$, the $y$-axis, $y = 0$ and $y = 2$. Write $y = \ln x$ as $x = e^y$ and find the area of $R$ by integrating with respect to $y$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $A = \displaystyle\int_0^2 y^2\,dy = \Bigl[\dfrac{y^3}{3}\Bigr]_0^2 = \dfrac{8}{3}$ sq units.
2. $4 - y^2 = 0 \Rightarrow y = \pm 2$. $A = \displaystyle\int_{-2}^{2}(4-y^2)\,dy = \Bigl[4y - \dfrac{y^3}{3}\Bigr]_{-2}^{2} = \dfrac{32}{3}$ sq units.
3. $y^2 = y + 2 \Rightarrow y^2 - y - 2 = 0 \Rightarrow y = 2$ or $y = -1$. $A = \displaystyle\int_{-1}^{2}(y+2-y^2)\,dy = \Bigl[\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}\Bigr]_{-1}^{2} = \dfrac{9}{2}$ sq units.
4. $A = \displaystyle\int_0^1 e^y\,dy = [e^y]_0^1 = e - 1 \approx 1.718$ sq units.
5. The parabola $x = y^2 - 1$ is not a function of $x$ (it has two $x$-branches), so $x$-integration would require splitting at $y = 0$. A single $y$-integral from $y = -2$ to $y = 2$ (found from $y^2 - 1 = 3$) handles the whole region at once.
Q1 (2 marks): $A = \displaystyle\int_0^3 y^2\,dy = \Bigl[\dfrac{y^3}{3}\Bigr]_0^3 = 9$ sq units [1 integrate, 1 evaluate].
Q2 (3 marks): $y^2 = y+2 \Rightarrow y = -1, 2$ [1]. $A = \displaystyle\int_{-1}^{2}(y+2-y^2)\,dy = \Bigl[\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}\Bigr]_{-1}^{2} = (2+4-\tfrac{8}{3}) - (\tfrac{1}{2}-2+\tfrac{1}{3}) = \dfrac{9}{2}$ sq units [2].
Q3 (3 marks): $x = e^y$ [1]. $A = \displaystyle\int_0^2 e^y\,dy = [e^y]_0^2 = e^2 - e^0 = e^2 - 1 \approx 6.39$ sq units [2].
Five timed questions on $y$-axis integrals. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering $y$-axis integration questions. Lighter alternative to the boss.
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