Introduction: Integrating with Respect to y

In standard integration, we sum thin vertical strips of area between a curve $y = f (x)$ and the $x$ -axis, integrating with respect to $x$ . But what if the region is bounded by the y-axis instead?

The key idea is simple: swap the roles of $x$ and $y$ . Instead of thin vertical strips, we use thin horizontal strips of width $d y$ . Each strip has length $x (y)$ , the horizontal distance from the y-axis to the curve , so the area of each strip is $x (y) d y$ .

Summing infinitely many such strips gives:

$A = \int_{a}^{b} x (y) d y$

where $a$ and $b$ are y-values (not x-values), and $x (y)$ is the curve expressed as a function of $y$ .

Note

To use this formula, you must express $x$ explicitly in terms of $y$ . If your curve is given as $y = f (x)$ , rearrange it to get $x = g (y)$ before setting up the integral.

Introduction: Integrating with Respect to y

Setting Up Area Integrals onto the y-axis

Area between a curve and the y-axis: For a curve expressed as $x = g (y)$ , the area of the region bounded by the curve, the y-axis, and the horizontal lines $y = a$ and $y = b$ is:
$A = \int_{a}^{b} g (y) d y$
provided $g (y) \geq 0$ on $[a, b]$ .

The process for setting up such an integral:

Rearrange the equation of the curve to express $x$ in terms of $y$ .
Identify the y-limits of integration from the problem (the horizontal boundaries).
Write and evaluate the integral $\int_{a}^{b} x (y) d y$ .

Warning

A very common error is to keep the limits in terms of $x$ after switching to integrating with respect to $y$ . The limits must correspond to the variable of integration. If you are integrating $d y$ , the limits are $y$ -values.

Example

Find the area bounded by the curve $x = y^{2}$ and the y-axis from $y = 0$ to $y = 2$ .

Step 1: The curve is already in the form $x (y) = y^{2}$ . ✓

Step 2: Limits are $y = 0$ to $y = 2$ . ✓

Step 3: Set up and evaluate:
$A = \int_{0}^{2} y^{2} d y = [\frac{y ^{3}}{3}]_{0}^{2} = \frac{8}{3} - 0 = \frac{8}{3}$

The area is $\frac{8}{3}$ square units.

Exam Tip

It can help to sketch the region first. Draw the horizontal strip at a general height $y$ , note where it starts (the y-axis, $x = 0$ ) and where it ends (the curve, $x = g (y)$ ). Its length is $g (y) - 0 = g (y)$ .

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Introduction: Integrating with Respect to y

Summing infinitely many such strips gives:

$A = \int_{a}^{b} x (y) d y$

where $a$ and $b$ are y-values (not x-values), and $x (y)$ is the curve expressed as a function of $y$ .

Note

To use this formula, you must express $x$ explicitly in terms of $y$ . If your curve is given as $y = f (x)$ , rearrange it to get $x = g (y)$ before setting up the integral.

Setting Up Area Integrals onto the y-axis

The process for setting up such an integral:

Rearrange the equation of the curve to express $x$ in terms of $y$ .
Identify the y-limits of integration from the problem (the horizontal boundaries).
Write and evaluate the integral $\int_{a}^{b} x (y) d y$ .

Warning

Example

Find the area bounded by the curve $x = y^{2}$ and the y-axis from $y = 0$ to $y = 2$ .

Step 1: The curve is already in the form $x (y) = y^{2}$ . ✓

Step 2: Limits are $y = 0$ to $y = 2$ . ✓

Step 3: Set up and evaluate:
$A = \int_{0}^{2} y^{2} d y = [\frac{y ^{3}}{3}]_{0}^{2} = \frac{8}{3} - 0 = \frac{8}{3}$

The area is $\frac{8}{3}$ square units.

Exam Tip

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AHL 5.17—Areas under curve onto y-axis, volume of revolution (about x and y axes)

Introduction: Integrating with Respect to y

Setting Up Area Integrals onto the y-axis

9 more sections in this topic

AHL 5.17—Areas under curve onto y-axis, volume of revolution (about x and y axes)

Introduction: Integrating with Respect to y

Setting Up Area Integrals onto the y-axis

9 more sections in this topic