Math Core

Some regions are awkward to slice vertically. Their top or bottom boundary changes partway across, or the curves are given as $x$ in terms of $y$ . Turning the slices sideways, so that you integrate with respect to $y$ , often turns two integrals into one.

Right minus left

Picture thin horizontal rectangles of height $\Delta y$ . A rectangle at height $y$ stretches from the left boundary to the right boundary, so its length is $x_{\text{right}} - x_{\text{left}}$ . Adding these and letting $\Delta y \to 0$ gives the area.

Area with respect to y

If $f(y) \ge g(y)$ for $c \le y \le d$ , the area of the region between the curves $x = f(y)$ and $x = g(y)$ is

A = \int_c^d \big(f(y) - g(y)\big)\,dy = \int_c^d (\text{right} - \text{left})\,dy.

Everything is in terms of $y$ : the curves must be written as $x = \ldots$ , and the limits $c$ and $d$ are $y$ -values (bottom and top of the region), not $x$ -values.

When to switch to y

Integrate with respect to $y$ when:

the curves are given as $x = f(y)$ , such as $x = y^2$ , which is a sideways parabola and not a function of $x$ ;
a vertical slice would have a different top or bottom in different parts of the region, but a horizontal slice always has the same left and right boundaries.

Either method gives the same area whenever both are possible, because both compute the same region. The choice is about convenience: pick the direction in which every slice has the same two boundaries, so that one integral covers the whole region. On the AP exam, a question may even ask for an area "as an integral with respect to $y$ ," so be ready to set up both.

To rewrite $y = f(x)$ in the form $x = \ldots$ , solve for $x$ . For example, $y = \sqrt{x}$ with $x \ge 0$ becomes $x = y^2$ with $y \ge 0$ , and $y = x - 2$ becomes $x = y + 2$ .

The region between x = y² (left) and x = y + 2 (right). Each horizontal slice has length (y + 2) − y².Open in grapher →

Common mistake

Using $x$ -values as the limits of a $dy$ integral is the most common error here. After you switch to $y$ , find the intersection points' $y$ -coordinates, and integrate from the lowest one to the highest one.

Worked examples

Worked example: A sideways parabola and a line

Find the area of the region bounded by $x = y^2$ and $x = y + 2$ .

Solution. Intersections: $y^2 = y + 2$ gives $(y - 2)(y + 1) = 0$ , so $y = -1$ and $y = 2$ . At $y = 0$ the line gives $x = 2$ and the parabola gives $x = 0$ , so the line is on the right.

\begin{aligned} A &= \int_{-1}^{2} \big(y + 2 - y^2\big)\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2} \\ &= \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{10}{3} + \frac{7}{6} = \frac{9}{2}. \end{aligned}

In terms of $x$ , the bottom boundary of this region changes at $x = 1$ , so a $dx$ setup would need two integrals.

Worked example: One integral instead of two

Find the area of the region in the first quadrant bounded by $y = \sqrt{x}$ , $y = x - 2$ and the $x$ -axis.

Solution. With vertical slices, the bottom is the $x$ -axis for $0 \le x \le 2$ and the line for $2 \le x \le 4$ . You'd need

\int_0^4 \sqrt{x}\,dx - \int_2^4 (x - 2)\,dx = \frac{16}{3} - 2 = \frac{10}{3}.

With horizontal slices, rewrite the curves as $x = y^2$ (left) and $x = y + 2$ (right). The region runs from $y = 0$ (the $x$ -axis) up to $y = 2$ , where they meet.

A = \int_0^2 \big(y + 2 - y^2\big)\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_0^2 = 2 + 4 - \frac{8}{3} = \frac{10}{3}.

Both methods agree; the $dy$ version is one clean integral.

Worked example: A region against the y-axis

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

Solution. The $y$ -axis is the line $x = 0$ . The parabola meets it where $4 - y^2 = 0$ , so $y = \pm 2$ . The parabola is on the right.

A = \int_{-2}^{2} (4 - y^2)\,dy = \left[4y - \frac{y^3}{3}\right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{32}{3}.

Tip

Tilt your head: with $dy$ , "right" plays the role of "top" and "left" plays the role of "bottom." A quick test value of $y$ tells you which curve is on the right, just as a test $x$ tells you which is on top.