Area between curves with respect to y

Warm-up

Three quick questions from earlier lessons.

Discover

The region between $x = y^2$ and $x = 4$ is easiest to slice horizontally. A horizontal strip at height $y$ runs from the parabola to the line. How long is it?

$4 - y^2$
$y^2 - 4$

Key idea

When curves are given as $x = g(y)$ , integrate in $y$ .

If

f(y) \ge g(y)

for

c \le y \le d

, the area is

\displaystyle\int_c^d \big(f(y) - g(y)\big)\,dy

: right minus left, with $y$ -limits.

Worked example

Area between $x = y^2$ and $x = 4$ .

y

from

-2

to

2

y² = 4: y = ±2.

\displaystyle\int_{-2}^{2} (4 - y^2)\,dy

Right minus left.

= \tfrac{32}{3}

4y − y³/3.

Your turn, guided

Your turn: between $x = y^2$ and $x = 1$ .

\tfrac{4}{3}

∫ from −1 to 1 of (1 − y²) dy.

Practice ★

Find the area of the region between $x = y^2$ and $x = 9$ , integrating with respect to $y$ .
Find the area of the region between $x = y^2$ and $x = 16$ , integrating with respect to $y$ .

Which integral gives the area between $x = y^2$ and $x = 16$ ?
1. $\displaystyle\int_{-4}^{ 4 } (16 - y^2)\,dy$
2. $\displaystyle\int_0^{ 16 } (16 - y^2)\,dy$
3. $\displaystyle\int_{-4}^{ 4 } (y^2 - 16)\,dy$

Watch out!

Practice ★★

Find the area of the region between $x = y^2$ and $x = 2y + 8$ .
Find the area of the region between $x = y^2$ and $x = 3y + 4$ .

Exit ticket

Three questions to check you have got it.

Find the area of the region between $x = y^2$ and $x = 9$ , integrating with respect to $y$ .
Find the area of the region between $x = y^2$ and $x = 16$ , integrating with respect to $y$ .
Find the area of the region between $x = y^2$ and $x = 4$ , integrating with respect to $y$ .

Summary card

Area in $y$

Right minus left, $dy$ .
Limits: the $y$ -values where the curves meet.

Watch out: $x$ -values are not $y$ -limits.

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