8.5 - Finding the Area Between Curves Expressed as Functions of y

Review: Area with Respect to x

Up to now we’ve been finding area between curves with:

\text{Area} = \int_{a}^{b} [\text{top function} - \text{bottom function}]\,dx

That sweeps horizontally, left to right along the

x

-axis.

Area with Respect to y

Sometimes it’s easier (or necessary) to sweep vertically, along the

y

-axis:

\boxed{\text{Area} = \int_{c}^{d} [\text{right function} - \text{left function}]\,dy}

Key differences: - Integrate with respect to

y

, not

x

- The limits

c

and

d

are $y$ -values, not

x

-values - Subtract left from right, not bottom from top - Functions have to be written as

x = f(y)

, not

y = f(x)

Converting to $x = f(y)$ Form

Just solve each equation for

x

From $y = \sqrt{x}$ : square both sides.

y^2 = x \quad \Rightarrow \quad \boxed{x = y^2}

From $y = x - 6$ : add 6.

\boxed{x = y + 6}

From $y = 2 - x^2$ :

x^2 = 2 - y \quad \Rightarrow \quad \boxed{x = \pm\sqrt{2 - y}}

The

\pm

gives you two branches. The positive square root is the right side of the parabola, and the negative one is the left side.

Example 1: Comparing Both Methods

Find the area of the region bounded by

y = \sqrt{x}

y = 0

(from

x = 0

x = 6

y = x - 6

(from

x = 6

x = 9

), and

x = 9

8.5 - Finding the Area Between Curves Expressed as Functions of y

Method 1: with respect to x

The bottom boundary changes at

x = 6

, so we need two integrals:

\text{Area} = \int_{0}^{6} \sqrt{x}\,dx + \int_{6}^{9} [\sqrt{x} - (x-6)]\,dx

Method 2: with respect to y

Step 1: convert to $x = f(y)$ form

y = \sqrt{x} \;\Rightarrow\; x = y^2 \qquad\qquad y = x - 6 \;\Rightarrow\; x = y + 6

Step 2: identify boundaries

Right boundary: the line $x = y + 6$ (the whole way)

Left boundary: the curve $x = y^2$ (the whole way)

$y$ -limits: from $y = 0$ to $y = 3$

Step 3: set up the integral

Neither boundary changes, so we only need one integral:

\boxed{\text{Area} = \int_{0}^{3} [(y + 6) - y^2]\,dy}

That’s the advantage. With respect to

y

, one integral instead of two.

Check: both methods give the same area.

\int_{0}^{3} [(y + 6) - y^2]\,dy = \left[\frac{y^2}{2} + 6y - \frac{y^3}{3}\right]_{0}^{3} = \frac{9}{2} + 18 - 9 = \frac{27}{2}

With respect to

x

, the two integrals come out to

4\sqrt{6}

and

\frac{27}{2} - 4\sqrt{6}

, which also add up to

\frac{27}{2}

Example 2: Single Integral with Respect to y

Find the area bounded by

x = 3 - y^2

and

x = y + 1

Step 1: which function is on the right?

Test

y = 0

: - Parabola:

x = 3 - 0 = 3

(right) - Line:

x = 0 + 1 = 1

(left)

Step 2: find the $y$ -boundaries

Set them equal:

3 - y^2 = y + 1

y^2 + y - 2 = 0

(y + 2)(y - 1) = 0

y = -2

and

y = 1

Step 3: check whether the boundaries change

Right side: $x = 3 - y^2$ the whole way ✓

Left side: $x = y + 1$ the whole way ✓

Only one integral needed.

Step 4: set up

\text{Area} = \int_{-2}^{1} [(3 - y^2) - (y + 1)]\,dy = \int_{-2}^{1} (2 - y - y^2)\,dy

Step 5: evaluate

= \left[2y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-2}^{1} = \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{9}{2}

Area:

\frac{9}{2}

square units

Example 3: Multiple Integrals Required

Set up integrals both ways for the region bounded by

y = 2 - x^2

y = (x-1)^2

, and

y = \frac{3}{2}x + 1

Intersection points: -

(0, 1)

, where the line meets the upward parabola -

(0.5, 1.75)

, where the line meets the downward parabola -

(1, 0)

, the vertex of the upward parabola -

(1.36602, 0.13397)

, where the two parabolas meet

Setup with respect to x

The top boundary changes at

x = 0.5

(from the line to the downward parabola). The bottom boundary is

(x-1)^2

the whole way. Two integrals:

\text{Area} = \int_{0}^{0.5} \left[\left(\frac{3}{2}x + 1\right) - (x-1)^2\right]dx + \int_{0.5}^{1.36602} [(2 - x^2) - (x-1)^2]\,dx

Setup with respect to y

Step 1: convert to $x = f(y)$ form

From

y = 2 - x^2

x = \pm\sqrt{2-y}

We want the positive root,

x = \sqrt{2-y}

, which is the right side of the parabola.

From

y = (x-1)^2

\sqrt{y} = |x - 1| \quad \Rightarrow \quad x = 1 \pm \sqrt{y}

x = 1 + \sqrt{y}

is the right branch and

x = 1 - \sqrt{y}

is the left branch.

From

y = \frac{3}{2}x + 1

x = \frac{2}{3}(y - 1)

Step 2: analyze where the boundaries change

Moving bottom to top, from

y = 0

up to

y = 1.75

$y = 0$ to $y = 0.13397$ : right is $x = 1 + \sqrt{y}$ , left is $x = 1 - \sqrt{y}$

$y = 0.13397$ to $y = 1$ : right is $x = \sqrt{2-y}$ , left is $x = 1 - \sqrt{y}$

$y = 1$ to $y = 1.75$ : right is $x = \sqrt{2-y}$ , left is $x = \frac{2}{3}(y-1)$

Three integrals:

\text{Area} = \int_{0}^{0.13397} [(1 + \sqrt{y}) - (1 - \sqrt{y})]\,dy + \int_{0.13397}^{1} [\sqrt{2-y} - (1 - \sqrt{y})]\,dy + \int_{1}^{1.75} \left[\sqrt{2-y} - \frac{2}{3}(y-1)\right]dy

This one is deliberately nasty to show the method. Most problems are much simpler.

Choosing Between dx and dy

Use $dx$ when: - The boundaries are vertical lines - The functions are naturally in

y = f(x)

form - The top and bottom boundaries are simple

Use $dy$ when: - The boundaries are horizontal lines - The functions are naturally in

x = f(y)

form - The left and right boundaries are simpler than the top and bottom - Using

dx

would need many integrals but

dy

needs fewer

Example 1 needed two integrals with respect to

x

, but only one with respect to

y

Calculator Tips for dy Integrals

On a TI-84, even though the integral is with respect to

y

, you still enter it using X:


fnInt(expression in terms of X, X, lower y-value, upper y-value)

To calculate

\int_{0}^{3} [(y+6) - y^2]\,dy

Press MATH → 9:fnInt(

Enter fnInt((X+6) - X^2, X, 0, 3)

Press ENTER

Why does this work? The calculator’s variable button (X,T,θ,n) just represents a variable. The calculator doesn’t care what you call it.

Decision Tree: Do I Need Multiple Integrals?

With respect to $x$ : 1. Does the top boundary change? If yes, split there. 2. Does the bottom boundary change? If yes, split there. 3. You need a new integral at each change point.

With respect to $y$ : 1. Does the right boundary change? If yes, split there. 2. Does the left boundary change? If yes, split there. 3. You need a new integral at each change point.

Step-by-Step Process (with respect to y)

Step 1: Convert all equations to

x = f(y)

form

Step 2: Identify the

y

-boundaries (smallest to largest

y

-value)

Step 3: Determine right and left boundaries, testing a point to see which gives the larger

x

-value

Step 4: Check whether the boundaries change. Trace the right edge from bottom to top, then the left edge, and see if either switches curves

Step 5: Set up the integral(s):

\int_{y_1}^{y_2} [\text{right} - \text{left}]\,dy

adding a separate integral for each region where the boundaries change.

Common Mistakes to Avoid

Forgetting to convert to $x = f(y)$ . Always solve for $x$ first

Using $x$ -values as limits. For $dy$ integrals, the limits are $y$ -values

Subtracting top minus bottom instead of right minus left

Not considering both branches of a square root. $x = \pm\sqrt{\cdots}$ gives two curves: $+$ is the right side, $-$ is the left

Forgetting to check if the boundaries change. Trace both edges completely

Key Takeaways

With respect to $y$ : $\int_{c}^{d} [\text{right} - \text{left}]\,dy$

Convert equations: solve for $x$ to get $x = f(y)$ form

Limits are $y$ -values: smallest to largest $y$ -coordinate

Right minus left, not top minus bottom

Check for multiple integrals: split wherever the right or left boundary changes

Square roots give two branches: $\pm\sqrt{\cdots}$ represents both sides of a curve

Choose wisely: sometimes $dy$ needs fewer integrals than $dx$

Calculator: still use the X variable button even for $y$ integrals

Both methods work: $dx$ and $dy$ give the same area

Trace the boundaries: follow the edges of the region to spot where they change

Back Next

8.5 - Finding the Area Between Curves Expressed as Functions of y

Review: Area with Respect to x

Area with Respect to y

Converting to x=f(y)x = f(y)x=f(y) Form

Example 1: Comparing Both Methods

Example 2: Single Integral with Respect to y

Example 3: Multiple Integrals Required

Choosing Between dx and dy

Calculator Tips for dy Integrals

Decision Tree: Do I Need Multiple Integrals?

Step-by-Step Process (with respect to y)

Common Mistakes to Avoid

Key Takeaways

Converting to $x = f(y)$ Form