Multivariable calculus

Double Integral of xy Over a Triangle: 1/8

Q: How do I know whether a region is type 1 or type 2?

Slice it. If every vertical line meets the region in one segment whose ends are graphs y = g(x), it is type 1. If every horizontal line does the same with x = h(y), it is type 2. This triangle is both, which is why either order works.

Q: What if my inner limits come out crossed, with the lower bigger than the upper?

Then the boundaries are the wrong way round and the answer comes back with the wrong sign. On 0 \le x \le 1 the line y = x sits above y = 0, so 0 is the lower limit and x is the upper one.

The double integral of xy over the triangle with 0 <= y <= x and 0 <= x <= 1 equals 1/8, or 0.125. Integrating in y first gives x^3/2, and integrating that from 0 to 1 gives 1/8. Reversing the order, with x running from y to 1, gives the same 1/8.

f(x,y) = xy

\int_0^1 \int_0^{x} xy\,dy\,dx = \frac{1}{8}

Numerically 0.125000, confirmed by quadrature on every build.

Describe the region as type 1, then integrate inside out

A type 1 region is one where $x$ runs between two numbers and, for each fixed $x$ , $y$ runs between two curves. Slice this triangle vertically: at a given $x$ in $[0,1]$ the slice starts on the $x$ -axis and stops on the line $y = x$ .

\int_0^1 \int_0^{x} xy\,dy\,dx

Do the inner integral with $x$ frozen. The variable is $y$ , so $x$ is nothing but a constant multiplier.

\int_0^{x} xy\,dy = x\left[\frac{y^2}{2}\right]_{y=0}^{y=x} = \frac{x^3}{2}

\int_0^1 \frac{x^3}{2}\,dx = \left[\frac{x^4}{8}\right]_0^1 = \frac{1}{8}

The mistake: inner limits that ignore the slanted edge

The most common wrong setup writes the inner limits as $0$ and $1$ , because the triangle sits inside the unit square. That computes the integral over the square, which is $\frac14$ , and the extra $\frac18$ is the half of the square the triangle never covered.

Notice that no algebra flags it. The inner integral still comes out clean, $\int_0^1 xy\,dy = \frac{x}{2}$ , and the outer step still runs. Only the picture catches it: the inner limits describe one vertical slice, so the upper one has to be the curve that slice ends on, which is $y = x$ , not the top of the box you drew around the region.

Inner limits may contain the outer variable. Outer limits must be numbers.
The upper inner limit is the curve bounding the slice, not the highest point of the whole region.
Substitute both inner limits into the antiderivative, the lower one included.
If a $y$ survives into the outer integral, either a limit went unsubstituted or the slices and the differentials disagree.

The other half of this mistake is drawing horizontal slices while writing $dy\,dx$ . The order of the differentials records how you sliced: $dy$ innermost means vertical slices, so the inner limits have to be the bottom and top curves.

Confirm by reversing the order

This triangle is type 2 as well. Slice horizontally: for $y$ in $[0,1]$ , $x$ runs from the line $x = y$ across to $x = 1$ .

\int_0^1 \int_{y}^{1} xy\,dx\,dy = \int_0^1 y\left[\frac{x^2}{2}\right]_{x=y}^{x=1}\,dy = \int_0^1 \frac{y - y^3}{2}\,dy = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{8}

Agreement between the two orders is not a proof, but a misread region almost never produces the same number twice. It costs a minute and catches limit errors, sign errors and dropped factors alike.

Frequently asked questions

How do I know whether a region is type 1 or type 2?

Slice it. If every vertical line meets the region in one segment whose ends are graphs $y = g(x)$ , it is type 1. If every horizontal line does the same with $x = h(y)$ , it is type 2. This triangle is both, which is why either order works.

What if my inner limits come out crossed, with the lower bigger than the upper?

Then the boundaries are the wrong way round and the answer comes back with the wrong sign. On $0 \le x \le 1$ the line $y = x$ sits above $y = 0$ , so $0$ is the lower limit and $x$ is the upper one.