Unit 8 · Topic 8.5 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice, plus 1 from earlier lessons

AP Calculus AB 8.5: Finding the Area Between Curves Expressed as Functions of y

Q: Do I always have to rewrite y = f(x) as x = g(y)?

Only when you slice horizontally. A dy integral needs each edge as x in terms of y. If solving for x is hard, that is a sign to slice with dx instead.

Q: Can the limits of a dy integral be negative?

Yes. They are heights, and a region can sit below the x-axis. The region between x = 6 - y2 and x = y runs from y = -3 to y = 2.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 8: 10 to 15% of the multiple-choice section

What you'll be able to do

Calculate the area of a region in the plane with a definite integral, slicing with respect to $x$ or with respect to $y$ , whichever the region suits.

Key vocabulary

Term	What it means	The exam trap
Area of a region in the plane	The size of a flat region enclosed by curves, found with a definite integral of the length of a slice across it.	An area is never negative, so a negative result points to a setup mistake, not a negative area.
Function of $y$	A rule $x = g (y)$ that gives one $x$ -value for each height $y$ ; its graph is read sideways.	An edge given as $y$ in terms of $x$ has to be rewritten before it can go inside a $d y$ integral, and a curve can split into two halves when you do.
Definite integral with respect to $y$	$\int_{c}^{d} h (y) d y$ , the integral of a function whose input is $y$ , from $y = c$ to $y = d$ .	Write the $d y$ every time. Without it, or with $d x$ in its place, the integral says something else.

The parabola x = 6 minus y squared, opening to the left from its vertex at (6, 0), and the line x = y. They meet at (-3, -3) and (2, 2), and the region between them is shaded. A horizontal slice at y = 1 runs from the line at x = 1 to the parabola at x = 5. The shaded area is 20.83.

Part 1 Think first How big is the pond?

Commit to a guess first. Checking it against the answer is what makes the method stick.

Predict first

A surveyor walks north from the south tip of a pond. Every few steps she measures straight across the pond, west to east. At $y$ meters north of the south tip the pond is $4 y - y^{2}$ meters wide, until it closes at its north tip, 4 meters north. About how large is the pond's surface?

In short. The area between two curves written as $x = f (y)$ and $x = g (y)$ , with $f (y) \geq g (y)$ , is the definite integral of right minus left, $\int_{c}^{d} (f (y) - g (y)) d y$ .

Each horizontal slice has length right minus left, and its thickness is a small change in $y$ .
The limits $c$ and $d$ are the heights where the region starts and stops.
Slicing with respect to $x$ or $y$ gives the same area, so pick the one that needs fewer integrals.

Part 2 Learn it Slicing the other way

5a. Horizontal slices: right minus left

In Topic 8.4 each vertical slice was top minus bottom, added up with $d x$ . Some regions are easier to cut the other way. A curve written as $x = f (y)$ gives an $x$ -value at every height $y$ .

Cut the region into thin horizontal slices. A slice at height $y$ runs from the left curve to the right curve, so its length is right minus left, $f (y) - g (y)$ when $f (y) \geq g (y)$ , and its thickness is a small change in $y$ . Adding the slices from the lowest height $c$ to the highest height $d$ gives

A = \int_{c}^{d} (f (y) - g (y)) d y .

The limits are heights: where the region starts and stops, often where the curves meet.

In the graph at the top, $x = 6 - y^{2}$ and $x = y$ meet where $6 - y^{2} = y$ , which gives $y = - 3$ and $y = 2$ . At $y = 0$ the parabola is at $x = 6$ and the line at $x = 0$ , so the parabola is on the right, and

A = \int_{- 3}^{2} ((6 - y^{2}) - y) d y = \frac{125}{6} \approx 20.83.

One height inside the region is enough to tell which curve is on the right, as long as the curves do not cross between $c$ and $d$ .

Quick check · Calculate

No calculator. Find the area of the region bounded by the curves $x = y^{2}$ and $x = 6 y - y^{2}$ .

Your answer

5b. Writing an edge as a function of y

Edges are often given as $y$ in terms of $x$ . To slice horizontally you need each edge as $x$ in terms of $y$ , so solve for $x$ :

the line $y = x - 6$ becomes $x = y + 6$ ;
the curve $y = \sqrt{x}$ becomes $x = y^{2}$ , for $y \geq 0$ only, because a square root is never negative;
the curve $y = e^{x}$ becomes $x = \ln y$ .

Two kinds of edge need a second look. A vertical line such as $x = 9$ is already a function of $y$ : it is the same left or right edge at every height. A horizontal line such as the $x$ -axis, $y = 0$ , is not an edge you subtract. It is where the slices stop, so it gives a limit of the integral.

A curve such as $y = x^{2}$ solves to $x = \pm \sqrt{y}$ , which is two functions of $y$ : the right half $x = \sqrt{y}$ and the left half $x = - \sqrt{y}$ . Before you use one, check which half actually borders your region.

Quick check · Match

No calculator. Each curve is written with $y$ in terms of $x$ . Tap a curve, then the same curve written with $x$ in terms of $y$ .

5c. Same area, either variable

A region's area does not depend on how you slice it. Take the region between $x = y^{2}$ and the line $x = 4$ . With horizontal slices,

\int_{- 2}^{2} (4 - y^{2}) d y = \frac{32}{3} .

With vertical slices, the top edge is $y = \sqrt{x}$ and the bottom edge is $y = - \sqrt{x}$ , so

\int_{0}^{4} (\sqrt{x} - (- \sqrt{x})) d x = \frac{32}{3} .

Since both give the same number, choose the slices that make the work shorter. Picture a slice sweeping across the region.

If every vertical slice runs between the same two curves, one integral with respect to $x$ does it.
If the top or the bottom edge changes partway across, a $d x$ setup needs one integral for each piece. Try horizontal slices: if each one runs between the same left and right curves, one integral with respect to $y$ does it.

The same test works in reverse. A region whose left or right edge changes partway up is often simpler with $d x$ .

Quick check · Multiple choice

No calculator. Let $R$ be the region bounded by $y = x^{3}$ , the horizontal line $y = 8$ and the $y$ -axis. Which integral with respect to $y$ gives the area of $R$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Subtracting left minus right, or writing top minus bottom out of habit, in a $d y$ integral.	The length of a horizontal slice is the right edge's $x$ -value minus the left edge's. The other order gives the area with the wrong sign.	Test one height inside the region and see which curve has the larger $x$ -value. That one goes first.
Using the $x$ -coordinates of the meeting points as the limits of a $d y$ integral.	The variable of integration is $y$ , so the limits are the lowest and highest heights of the region.	Solve for the $y$ -values where the edges meet, or read the heights of the corner points.
Putting an edge's $y = f (x)$ formula into a $d y$ integral without solving it for $x$ .	The integrand has to give a slice's length at height $y$ , which needs each edge as $x$ in terms of $y$ .	Solve each edge for $x$ first, and keep only the half of a curve that borders the region.

Part 3 Watch it done A region that wants horizontal slices

Watch each step, then do the last one yourself.

Find the area of the region bounded by $y = \sqrt{x}$ , the line $y = x - 6$ and the $x$ -axis.

Step 1 · Choose the slices
A vertical slice ends on the $x$ -axis for $0 < x < 6$ but on the line $y = x - 6$ for $x > 6$ , so $d x$ needs two integrals. Every horizontal slice runs from the curve on the left to the line on the right, so use $d y$ .
Step 2 · Write each edge as x in terms of y
$y = \sqrt{x}$ becomes $x = y^{2}$ for $y \geq 0$ , and $y = x - 6$ becomes $x = y + 6$ . The $x$ -axis, $y = 0$ , is the bottom of the region.
Step 3 · Find the limits
The edges meet where $y^{2} = y + 6$ , so $(y - 3) (y + 2) = 0$ and $y = 3$ or $y = - 2$ . The region runs from $y = 0$ up to $y = 3$ , the height of the corner $(9, 3)$ .
Why this step?
Why is the lower limit 0 and not $- 2$ ?
The bottom edge of this region is the $x$ -axis, where $y = 0$ . At $y = - 2$ the line meets $x = y^{2}$ at $(4, - 2)$ , which is on the lower half of that parabola, $y = - \sqrt{x}$ , and not on $y = \sqrt{x}$ . That point is not on this region at all.
Step 4 · Set up right minus left
At $y = 1$ the line is at $x = 7$ and the curve at $x = 1$ , so the line is on the right. $A = \int_{0}^{3} ((y + 6) - y^{2}) d y = {[\frac{y^{2}}{2} + 6 y - \frac{y^{3}}{3}]}_{0}^{3}$ .
Your step · Calculate

No calculator. Finish it. The area of the region bounded by $y = \sqrt{x}$ , $y = x - 6$ and the $x$ -axis is ${[\frac{y^{2}}{2} + 6 y - \frac{y^{3}}{3}]}_{0}^{3}$ . What is the area?

Your answer

Part 4 Try it Two new regions

Try it · Calculate

Graphing calculator required. Let $R$ be the region bounded by the curves $x = 4 - y^{2}$ and $x = e^{y}$ . Find the area of $R$ . Give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. Let $R$ be the region bounded by $y = x^{2}$ and $y = 2 x$ . Its area is $\int_{0}^{2} (2 x - x^{2}) d x$ . A classmate says the same area is $\int_{0}^{4} (\sqrt{y} - \frac{y}{2}) d y$ . Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

Free response on the exam is written by hand, so work this on paper. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. On a park map, with $x$ and $y$ measured in meters, the west shore of a pond follows the curve $x = y^{2}$ and a straight dam along its east side follows the line $x = 2 y + 8$ . The region $R$ enclosed by the shore and the dam is the surface of the pond.

(a) Find the area of the pond's surface. Show the integral you use and its antiderivative. Indicate units of measure.

(b) Write, but do not evaluate, an expression involving one or more integrals with respect to $x$ that gives the same area. Explain why a single integral with respect to $x$ is not enough.

(c) A rope along the horizontal line $y = k$ divides the pond's surface into two parts of equal area. Write an equation involving an integral with respect to $y$ that $k$ satisfies, and verify that $k = 1$ satisfies it.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) The shore and the dam meet where $y^{2} = 2 y + 8$ , so $(y - 4) (y + 2) = 0$ and $y = - 2$ or $y = 4$ . At $y = 0$ the dam is at $x = 8$ and the shore at $x = 0$ , so the dam is on the right.

A = \int_{- 2}^{4} ((2 y + 8) - y^{2}) d y = {[y^{2} + 8 y - \frac{y^{3}}{3}]}_{- 2}^{4} = \frac{80}{3} - (- \frac{28}{3}) = 36

The pond's surface is 36 square meters.

(b) Solved for $y$ , the shore is $y = \sqrt{x}$ above and $y = - \sqrt{x}$ below, and the dam is $y = \frac{x - 8}{2}$ . The corners are $(4, - 2)$ and $(16, 4)$ .

\int_{0}^{4} (\sqrt{x} - (- \sqrt{x})) d x + \int_{4}^{16} (\sqrt{x} - \frac{x - 8}{2}) d x

One integral is not enough because the bottom edge changes at $x = 4$ : to the left of $x = 4$ a vertical slice ends on the lower half of the shore, $y = - \sqrt{x}$ , and to the right of it the slice ends on the dam.

(c) Half of the area is 18, so $\int_{- 2}^{k} ((2 y + 8) - y^{2}) d y = 18$ . With $k = 1$ , ${[y^{2} + 8 y - \frac{y^{3}}{3}]}_{- 2}^{1} = \frac{26}{3} - (- \frac{28}{3}) = 18$ , so $k = 1$ satisfies the equation.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds the limits $y = - 2$ and $y = 4$ and writes $\int_{- 2}^{4} ((2 y + 8) - y^{2}) d y$ , right minus left.
Shows the antiderivative $y^{2} + 8 y - \frac{y^{3}}{3}$ and gets 36 square meters.
Writes $\int_{0}^{4} 2 \sqrt{x} d x + \int_{4}^{16} (\sqrt{x} - \frac{x - 8}{2}) d x$ or an equal expression, with limits 0, 4 and 16.
Explains that the bottom edge of a vertical slice changes at $x = 4$ , from $y = - \sqrt{x}$ to the dam.
Writes $\int_{- 2}^{k} ((2 y + 8) - y^{2}) d y = 18$ or an equivalent equation, such as $\int_{k}^{4} ((2 y + 8) - y^{2}) d y = 18$ , and verifies that it holds at $k = 1$ .

Part 6 Check yourself A reservoir on a map

The two case questions use the same table. Answer them from it.

On a map of a reservoir, with $x$ and $y$ in meters, $y$ is the distance north of the dam. At height $y$ the east shore is at $x = f (y)$ and the west shore is at $x = g (y)$ , where $f$ and $g$ are continuous and $f (y) > g (y)$ for $0 \leq y \leq 120$ . The reservoir's surface is the region between the two shores from $y = 0$ to $y = 120$ .

Positions of the two shores
$y$ (meters)	$f (y)$ (meters)	$g (y)$ (meters)
0	50	10
30	80	0
60	95	5
90	90	20
120	60	40

Case question 1 · Calculate

No calculator. Use a trapezoidal sum with the four subintervals shown in the table to approximate the area of the reservoir's surface, in square meters.

Your answer

Case question 2 · Multiple choice

No calculator. Which expression gives the exact area of the reservoir's surface, in square meters?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the area of the region bounded by $x = y^{2} - 1$ and the line $x = 3$ . Tap the first line with a mistake.

Check yourself · Sort it

No calculator. How many integrals does each region need? Tap a card, then a bin.

One integral in $y$ , two in $x$

One integral in $x$ , two in $y$

One integral either way

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

To find an area with horizontal slices, solve each edge for in terms of $y$ , then integrate the curve minus the curve with respect to $y$ .

Part 7 Keep it Bring back earlier ideas

This lesson stands on vertical slices from Topic 8.4 and on reading a definite integral as signed area from Topic 6.6, where a piece below the axis counts as negative. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Which integral gives the area of the region bounded by $y = x^{2} - 4$ and $y = 2 - x$ ?

From 6.6 Applying Properties of Definite Integrals · Calculate

No calculator. Use geometry to evaluate $\int_{1}^{7} (x - 3) d x$ .

Your answer

In three lines

For curves written as $x = f (y)$ and $x = g (y)$ , a horizontal slice has length right minus left.
The area is $\int_{c}^{d} (f (y) - g (y)) d y$ , with $c$ and $d$ the lowest and highest heights of the region.
Slicing with $d x$ or $d y$ gives the same area; choose the one whose slices always end on the same two curves.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When you set up a $d y$ integral, write each edge as $x$ in terms of $y$ , say which one is on the right, and use heights as the limits. If a part asks you to write but not evaluate, the integral with its integrand, both limits and $d y$ is the whole answer.

Questions students ask

How do I know whether to integrate with respect to

x

y

Picture slices sweeping across the region. If every vertical slice ends on the same two curves, use $d x$ . If the top or bottom edge changes partway across but every horizontal slice ends on the same left and right curves, use $d y$ . Either way gives the same area.

Do I always have to rewrite

y = f (x)

x = g (y)

Only when you slice horizontally. A $d y$ integral needs each edge as $x$ in terms of $y$ . If solving for $x$ is hard, that is a sign to slice with $d x$ instead.

Why right minus left, and not top minus bottom?

A horizontal slice lies along one height, so its two ends differ only in $x$ . Its length is the larger $x$ -value minus the smaller, which is the right edge minus the left edge.

Can the limits of a

d y

integral be negative?

Yes. They are heights, and a region can sit below the $x$ -axis. The region between $x = 6 - y^{2}$ and $x = y$ runs from $y = - 3$ to $y = 2$ .

What if the curve on the right changes partway up?

Then one integral of right minus left is not enough. Split the region at the height where the curves cross and add an integral for each piece. Topic 8.6 does this.

Course alignment, for teachers

AP Calculus AB topic 8.5, Unit 8: Applications of Integration.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first How big is the pond?

Part 2 Learn it Slicing the other way

5a. Horizontal slices: right minus left

5b. Writing an edge as a function of y

5c. Same area, either variable

Part 3 Watch it done A region that wants horizontal slices

Part 4 Try it Two new regions

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A reservoir on a map

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic