Drawing Regions and Changing Integration Order

Bounds describe a set of points

A vertically simple region has $a\le x\le b$ and $g_1(x)\le y\le g_2(x)$ . Its integral is $\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)dy dx$ . The inner bounds describe a single vertical slice; outer bounds sweep those slices across the region.

To reverse order, keep the region itself fixed. Sketch intersections, determine the new outer range, and express the left and right boundaries as functions of the new outer variable. Some regions require splitting into pieces in one order but not the other.

Visual guide

VISUAL GUIDEThe same region has vertical or horizontal slices

y = x²
y = x
Vertical slice
Horizontal slice

The shaded region lies between y = x² and y = x on [0, 1]. Vertical slices run from x² to x. Horizontal slices run from x = y to x = √y. Changing order changes bounds, not the region.

Worked example: reverse a curved region

Consider $0\le x\le1$ , $x^2\le y\le x$ . The boundaries meet at zero and one. For fixed $0\le y\le1$ , the inequalities become $y\le x\le\sqrt y$ . Thus

\int_0^1\int_{x^2}^{x} f(x,y)dy dx=\int_0^1\int_y^{\sqrt y}f(x,y)dx dy.

The equality describes the same region, assuming the integrand is integrable. Simply exchanging $dx$ and $dy$ while retaining the old bounds would describe a different set.

Worked example: an order that unlocks an integral

Evaluate $\int_0^1\int_x^1 e^{y^2}dy dx$ . The inner antiderivative is not elementary. The region is $0\le x\le y\le1$ . Reverse order to obtain

\int_0^1\int_0^y e^{y^2}dx dy=\int_0^1 y e^{y^2}dy=\frac{e-1}{2}.

The new inner integral is easy because $e^{y^2}$ is constant with respect to $x$ . Changing order is a geometric operation with an algebraic benefit.

When a horizontal slice enters and leaves a region more than once, use several intervals or split at the relevant boundary intersections. A rough drawing plus test slices prevents many incorrect bound formulas.

Practice

Reverse $\int_0^2\int_0^{2-x}f(x,y)dy dx$ .
Reverse $\int_0^1\int_0^{\sqrt x}f(x,y)dy dx$ .
Evaluate the first integral when $f=1$ .

Show worked solutions

$\int_0^2\int_0^{2-y}f(x,y)dx dy$ .
$0\le y\le1$ and $y^2\le x\le1$ , so use $\int_0^1\int_{y^2}^1f(x,y)dx dy$ .
The triangular area is $2$ , or $\int_0^2(2-x)dx=2$ .

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.