Area Between Curves (Functions of y)

Part 1 · Hook

Why this matters

Some regions are awkward to slice top to bottom. A pond whose banks are a parabola opening sideways and a straight wall has a bottom edge that switches curves halfway across. Turn your knife sideways: every horizontal strip runs from the same left bank to the same right bank, and one integral does the job.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Solve $y = x^3$ for $x$ .

$x = y^{1/3}$
$x = y^3$
$x = \frac{y}{3}$
$x = 3y$

Show the answer

Take the cube root of both sides: $x = \sqrt[3]{y} = y^{1/3}$ . This is the inverse function.

Correct: $x = y^{1/3}$ :
$x = y^3$ :
$x = \frac{y}{3}$ :
$x = 3y$ :

2. Where does the parabola $y = 6 - x^2$ meet the line $y = x$ ?

$x = -3$ and $x = 2$
$x = 3$ and $x = -2$
$x = \pm\sqrt{6}$
$x = 0$ and $x = 6$

Show the answer

$6 - x^2 = x$ gives $x^2 + x - 6$ $= (x + 3)(x - 2)$ $= 0$ .

Correct: $x = -3$ and $x = 2$ :
$x = 3$ and $x = -2$ :
$x = \pm\sqrt{6}$ :
$x = 0$ and $x = 6$ :

Part 4 · See it

See it first

The region between the parabola x = y² (left) and the line x = y + 2 (right), shaded from y = −1 to y = 2, where they meet at (1, −1) and (4, 2). One thin horizontal rectangle of thickness Δy has length right minus left, (y + 2) − y². A side panel gives the limits from y² = y + 2 and the area, the integral from −1 to 2 of [(y + 2) − y²] dy = 9/2. — A horizontal strip between $x = y^2$ and $x = y + 2$ : its length is right minus left, and the limits are $y$ -values. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Write each edge of the region as $x$ in terms of $y$ a horizontal strip at height $y$ runs from $x_{\text{left}}(y)$ to $x_{\text{right}}(y)$
The strip has length right minus left and thickness $\Delta y$ its area is $\left(x_{\text{right}} - x_{\text{left}}\right)\Delta y$
Stack the strips from the bottom of the region to the topthe sum becomes $\int_c^d \left(x_{\text{right}} - x_{\text{left}}\right)\,dy$
The integral is in $dy$ the limits $c$ and $d$ are the $y$ -values where the region starts and ends

Part 6 · Key ideas

Key ideas

Integrating with respect to $y$ : area $= \int_c^d \left(x_{\text{right}}(y) - x_{\text{left}}(y)\right)\,dy$ , right minus left.
Rewrite each curve as $x =$ (something in $y$ ) first: $y = \sqrt{x}$ becomes $x = y^2$ , $y = \ln x$ becomes $x = e^y$ .
Limits are $y$ -values. Find them by setting the two $x$ -expressions equal and solving for $y$ .
Choose $y$ when each side edge is one formula but the top or bottom edge changes partway across; slicing in $x$ would take two integrals.
Both ways give the same area when done correctly.

Part 7 · Misconception

A common mistake

The wrong idea: For an integral in $dy$ , you can keep the $x$ -values of the crossing points as the limits.

What actually happens: The limits belong to the variable of integration. With $dy$ , use the $y$ -coordinates of the crossings: for $x = y^2$ and $x = y + 2$ that is $y = -1$ to $y = 2$ , not $x = 1$ to $x = 4$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The region $R$ between $x = \frac{y^2}{2}$ and $x = y + 4$

1. Which integral gives the area of $R$ ?

$\int_{-2}^{4} \left[(y + 4) - \frac{y^2}{2}\right]\,dy$
$\int_{-2}^{4} \left[\frac{y^2}{2} - (y + 4)\right]\,dy$
$\int_{0}^{8} \left[\sqrt{2x} - (x - 4)\right]\,dx$
$\int_{2}^{8} \left[(y + 4) - \frac{y^2}{2}\right]\,dy$

Show the answer

Slice horizontally: each strip runs from the parabola on the left to the line on the right, so its length is right minus left, and $y$ runs from $-2$ to $4$ .

Correct: $\int_{-2}^{4} \left[(y + 4) - \frac{y^2}{2}\right]\,dy$ : Right minus left, in $dy$ , between the $y$ -values where the curves meet.
$\int_{-2}^{4} \left[\frac{y^2}{2} - (y + 4)\right]\,dy$ : Left minus right gives a negative number. The line $x = y + 4$ is the right edge.
$\int_{0}^{8} \left[\sqrt{2x} - (x - 4)\right]\,dx$ : With vertical slices the bottom edge changes at $x = 2$ : from $x = 0$ to $2$ the bottom is $y = -\sqrt{2x}$ , not $y = x - 4$ . One integral in $x$ is not enough.
$\int_{2}^{8} \left[(y + 4) - \frac{y^2}{2}\right]\,dy$ : 2 and 8 are the $x$ -coordinates of the crossings. An integral in $dy$ needs the $y$ -coordinates, $-2$ and $4$ .

2. Find the area of $R$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\int_{-2}^{4} \left(y + 4 - \frac{y^2}{2}\right)\,dy$ $= \left[\frac{y^2}{2} + 4y - \frac{y^3}{6}\right]_{-2}^{4}$ $= \frac{40}{3} - \left(-\frac{14}{3}\right)$ $= 18$ .

Answer: 18

3. A thin horizontal strip of $R$ sits at height $y$ , for $-2 \le y \le 4$ . Write its length $L(y)$ .

Type an expression for $L(y) =$ .

Show the answer

The right edge is $x = y + 4$ and the left edge is $x = \frac{y^2}{2}$ , so $L(y) = (y + 4) - \frac{y^2}{2}$ .

Answer: $L(y) =$ $y + 4 - \frac{y^{2}}{2}$

4. Integrating with respect to $y$ , what are the limits of integration for the area of $R$ ?

$y = -2$ and $y = 4$ , where $\frac{y^2}{2} = y + 4$
$x = 2$ and $x = 8$ , the $x$ -values where the curves meet
$y = 0$ and $y = 4$ , from the $x$ -axis up to the top
$x = 0$ and $x = 8$ , from the left edge to the right

Show the answer

$\frac{y^2}{2} = y + 4$ gives $y^2 - 2y - 8 = 0$ , so $(y - 4)(y + 2) = 0$ : $y = -2$ and $y = 4$ .

Correct: $y = -2$ and $y = 4$ , where $\frac{y^2}{2} = y + 4$ : In $dy$ , the limits are the lowest and highest $y$ of the region.
$x = 2$ and $x = 8$ , the $x$ -values where the curves meet: These are $x$ -values. With $dy$ the limits must be $y$ -values.
$y = 0$ and $y = 4$ , from the $x$ -axis up to the top: The region goes below the $x$ -axis, down to $y = -2$ .
$x = 0$ and $x = 8$ , from the left edge to the right: These are $x$ -values, and the integral is in $dy$ .

5. Find the area of the region bounded by $y = \ln x$ , the $y$ -axis, $y = 0$ and $y = 2$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Write the curve as $x = e^y$ . Each horizontal strip runs from the $y$ -axis ( $x = 0$ ) to $x = e^y$ , so the area is $\int_0^2 \left(e^y - 0\right)\,dy$ $= e^2 - 1$ .

Answer: $e^{2} - 1$ ≈ 6.389

6. Find the area of the region bounded by $x = y^2 - 4$ and $x = 2 - y$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$y^2 - 4 = 2 - y$ gives $y^2 + y - 6 = 0$ , so $y = -3$ or $y = 2$ . Area $= \int_{-3}^{2} \left(6 - y - y^2\right)\,dy$ $= \left[6y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-3}^{2}$ $= \frac{22}{3}$ $+ \frac{27}{2}$ $= \frac{125}{6}$ .

Answer: $\frac{125}{6}$ ≈ 20.833

Part 9 · Summary

Summary

When a region is easier to cut sideways, write its edges as

x

in terms of

y

and integrate right minus left from the lowest

y

to the highest:

A

= \int_c^d \left(x_{\text{right}} - x_{\text{left}}\right)\,dy

. The limits are

y

-values, found by setting the two edges equal.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The region RR between x=y22x = \frac{y^2}{2} and x=y+4x = y + 4

Summary

What comes next

Connections

The region $R$ between $x = \frac{y^2}{2}$ and $x = y + 4$