Triangular Regions With Slanted Edges

Why Triangles Make Useful Practice Regions

A triangular region has straight edges, but setting up its integral still requires care. The vertices determine the boundary lines; the choice of vertical or horizontal slices determines how those lines become bounds. A correct setup records both the region’s full extent in the outer variable and the boundary values along each inner slice.

The main task is geometric before it is computational: identify the edges, find their equations, and locate where the relevant bounds begin and end. Once the region is described correctly, the iterated integral follows from the slicing principle for double integrals.

Definition: A triangular region is the filled-in planar region bounded by the three line segments joining three noncollinear vertices. In a vertical-slice description, $x$ ranges over an interval and each fixed-$x$ slice runs from a lower $y$-boundary to an upper $y$-boundary. In a horizontal-slice description, $y$ ranges over an interval and each fixed-$y$ slice runs from a left $x$-boundary to a right $x$-boundary.

From Vertices to Slanted Edges

To find the equation of a slanted edge through $(x_1,y_1)$ and $(x_2,y_2)$, first calculate its slope, provided $x_1\ne x_2$:

m=\frac{y_2-y_1}{x_2-x_1}.

Then use the point-slope equation $y-y_1=m(x-x_1)$ and simplify. If instead the edge is being used as a horizontal-slice boundary, solve its equation for $x$ as a function of $y$. A vertical edge has a constant $x$-coordinate and a horizontal edge has a constant $y$-coordinate; neither can be expressed as a nonconstant function of the other coordinate in the same way as a slanted edge.

The outer limits come from the region’s extreme coordinate values in the chosen slicing direction. The inner limits come from the edges met by each slice. The vertices are especially useful: they identify where edges meet and where a slice begins or ends.

Theorem: If a bounded triangular region can be described by vertical slices as $a\le x\le b$ and $g(x)\le y\le h(x)$, then for a continuous function $f$ on the region, $$ \iint_R f(x,y)\,dA = \int_a^b\int_{g(x)}^{h(x)} f(x,y)\,dy\,dx. $$ For horizontal slices described by $c\le y\le d$ and $p(y)\le x\le q(y)$, the corresponding formula is $$ \iint_R f(x,y)\,dA = \int_c^d\int_{p(y)}^{q(y)} f(x,y)\,dx\,dy. $$ The inner integral accumulates values along a slice; the outer integral adds the contributions from all slices. Each straight edge gives a linear boundary equation. Over the full outer interval, however, the upper or lower slice bound can switch from one edge to another at a vertex, so it may be piecewise linear; when that happens, split the outer integral at that vertex. For the triangle with vertices $(0,0)$, $(4,0)$, $(2,3)$, the upper bound is $y=3x/2$ for $0\le x\le2$ and $y=3(4-x)/2$ for $2\le x\le4$.

Worked Example: A Triangle with One Slanted Edge

Worked Example: Integrating Over a Right Triangle

Let $R$ have vertices $(0,0)$, $(0,2)$, and $(2,0)$. Set up and evaluate $\iint_R (x+y)\,dA$ using vertical slices.

The slanted edge joins $(0,2)$ to $(2,0)$, so its slope is

\frac{0-2}{2-0}=-1.

Using the point $(0,2)$, its equation is $y-2=-1(x-0)$, or $y=2-x$. The region extends from $x=0$ to $x=2$. For each such $x$, the slice runs from the horizontal edge $y=0$ up to $y=2-x$. Thus

\iint_R (x+y)\,dA = \int_0^2\int_0^{2-x}(x+y)\,dy\,dx.

Evaluate the inner integral:

\int_0^{2-x}(x+y)\,dy = \left[xy+\frac{y^2}{2}\right]_0^{2-x} = x(2-x)+\frac{(2-x)^2}{2}.

Expanding gives

2x-x^2+\frac{4-4x+x^2}{2} = 2-\frac{x^2}{2}.

Therefore,

\int_0^2\left(2-\frac{x^2}{2}\right)\,dx = \left[2x-\frac{x^3}{6}\right]_0^2 = 4-\frac{8}{6} = \frac{8}{3}.

Check the antiderivative: the derivative of $2x-\frac{x^3}{6}$ is $2-\frac{x^2}{2}$, the outer integrand. At $x=0$, the inner slice has height $2$; at $x=2$, it has height $0$, consistent with the vertices. The computed integral is positive, as the integrand and region are nonnegative.

Why the Bounds Describe the Triangle

The bound setup is justified by what each slice contains. For example, suppose a triangle has one vertex on the left and two edges running from that vertex toward the right. At a fixed $x$ between the leftmost and rightmost vertices, the vertical slice begins on one edge and ends on the other. If those edges have equations $y=g(x)$ and $y=h(x)$, with $g(x)\le h(x)$ throughout the interval, the points in that slice are exactly those satisfying $g(x)\le y\le h(x)$.

This describes the whole triangle because every point in it lies on one of those vertical slices, and each stated slice stays between the triangle’s two edges. The outer $x$-interval covers the triangle from its leftmost to rightmost point. The slicing formula then gives the integral: add the values across each slice by integrating in $y$, and add the slice contributions across the region by integrating in $x$. The same reasoning applies to horizontal slices, with the roles of $x$ and $y$ exchanged.

Worked Example: Two Slanted Bounds

Worked Example: Finding the Meeting Point of the Edges

Let $R$ have vertices $(0,0)$, $(0,2)$, and $(4,3)$. Set up and evaluate $\iint_R y\,dA$ using vertical slices.

The lower slanted edge joins $(0,0)$ to $(4,3)$. Its slope is $\frac{3-0}{4-0}=\frac34$, so its equation is $y=\frac{3x}{4}$. The upper slanted edge joins $(0,2)$ to $(4,3)$. Its slope is $\frac{3-2}{4-0}=\frac14$, so its equation is $y=2+\frac{x}{4}$. These edges meet where

\frac{3x}{4}=2+\frac{x}{4} \quad\Longrightarrow\quad \frac{x}{2}=2 \quad\Longrightarrow\quad x=4,\qquad y=3.

The leftmost coordinate is $0$, and the edges meet at $x=4$. Thus the vertical-slice bounds are $\frac{3x}{4}\le y\le 2+\frac{x}{4}$, for $0\le x\le4$:

\iint_R y\,dA = \int_0^4\int_{3x/4}^{2+x/4} y\,dy\,dx.

Integrate with respect to $y$:

\int_{3x/4}^{2+x/4}y\,dy = \frac12\left[\left(2+\frac{x}{4}\right)^2-\left(\frac{3x}{4}\right)^2\right] = \frac12\left(4+x+\frac{x^2}{16}-\frac{9x^2}{16}\right) = 2+\frac{x}{2}-\frac{x^2}{4}.

Now integrate with respect to $x$:

\int_0^4\left(2+\frac{x}{2}-\frac{x^2}{4}\right)\,dx = \left[2x+\frac{x^2}{4}-\frac{x^3}{12}\right]_0^4 = 8+4-\frac{64}{12} = \frac{20}{3}.

The derivative of the displayed outer antiderivative is $2+\frac{x}{2}-\frac{x^2}{4}$, as required. At $x=0$, the inner limits are $0$ and $2$; at $x=4$, both limits equal $3$. Those checks confirm that the slices start at the left vertical edge and shrink to the meeting vertex.

Choosing Horizontal Slices

A triangle can also be described with horizontal slices. In that case, the outer limits are the lowest and highest $y$-coordinates. The left and right edges must be written as $x$-values depending on $y$. The following example uses this approach directly; its bounds are determined from the vertices rather than by changing the order of a previously written integral.

Worked Example: Integrating with Horizontal Slices

Let $R$ have vertices $(1,0)$, $(5,0)$, and $(3,4)$. Evaluate $\iint_R x\,dA$ using horizontal slices.

The left slanted edge joins $(1,0)$ to $(3,4)$. Its slope as $y$ changes with $x$ is $\frac{4}{2}=2$, so $y=2(x-1)$. Solving for $x$ gives $x=1+\frac{y}{2}$. The right edge joins $(5,0)$ to $(3,4)$, with slope $\frac{4}{-2}=-2$, so $y=-2(x-5)$, or $x=5-\frac{y}{2}$. The horizontal edges range from $y=0$ to $y=4$, where the slanted boundaries meet at $x=3$. Therefore,

\iint_R x\,dA = \int_0^4\int_{1+y/2}^{5-y/2}x\,dx\,dy.

The inner integral is

\int_{1+y/2}^{5-y/2}x\,dx = \frac12\left[\left(5-\frac{y}{2}\right)^2-\left(1+\frac{y}{2}\right)^2\right] = \frac12(24-6y) = 12-3y.

Thus

\int_0^4(12-3y)\,dy = \left[12y-\frac{3y^2}{2}\right]_0^4 = 48-24 = 24.

The derivative of $12y-\frac{3y^2}{2}$ is $12-3y$. At $y=0$, the slice runs from $x=1$ to $x=5$; at $y=4$, both limits equal $3$. The inner integral is positive for $0\le y<4$ and zero at $y=4$, as expected for slices that narrow to a vertex.

A Common Mistake: Using the Wrong Line Equation

A frequent error is to calculate the slope correctly but then use it incorrectly in the line equation. In the second example, the edge from $(0,2)$ to $(4,3)$ has slope $\frac{3-2}{4-0}=\frac14$, not $\frac34$. The incorrect calculation $y=2+\frac{3x}{4}$ gives $y=5$ at $x=4$, so it does not pass through the stated vertex $(4,3)$. The correct equation $y=2+\frac{x}{4}$ gives $y=3$ there.

Always substitute both endpoints into a proposed edge equation. This quick check catches a wrong slope, an incorrect point substitution, or a line written in the wrong variable. Also check that the lower bound does not exceed the upper bound on the entire outer interval.

Slice caution: A vertical slice is parallel to a vertical edge. It is disjoint from that edge unless it has the same $x$-coordinate, in which case it coincides with the edge; it does not intersect the edge at just one endpoint. Similarly, a horizontal slice is disjoint from a horizontal edge unless it has the same $y$-coordinate, in which case it coincides with that edge. At the extreme coordinate values of a triangle, a slice may reduce to an endpoint or coincide with an edge, but those isolated outer values do not change the double integral.

A Reliable Setup Routine

For a triangular region, a short sequence of checks makes the setup more dependable:

Record the vertices.
Use them to identify the extreme coordinate values and the pairs of points that determine each edge.

Find the needed edge equations.
Calculate slopes and write each slanted boundary in the variable needed for the chosen slices.

Determine the inner limits.
For a vertical slice, identify lower and upper $y$-values; for a horizontal slice, identify left and right $x$-values.

Check vertices and order.
Substitute endpoint coordinates into the line equations and verify that the lower bound is no greater than the upper bound, or that the left bound is no greater than the right bound.

Evaluate and verify.
Integrate the inner variable first, then the outer variable. Differentiate the antiderivative and check the endpoint values.

Why This Skill Matters

Many double integrals over nonrectangular regions begin with the same geometric work: express the boundary as functions of the inner variable and make sure the outer interval covers the region. Triangles are a particularly clear setting because their edges are linear and their vertices provide direct checks.

When a setup seems doubtful, test it at the endpoints of the outer interval and at a value between them. The bounds should describe slices that begin and end on the triangle’s edges, and their order should agree with the geometry. A correct setup is not merely a pair of formulas: it is a precise description of every slice used in the integral.

Check Your Understanding

For each question, use the vertices or edge equations to justify the bounds before evaluating an integral.

A triangle has vertices $(0,0)$, $(0,4)$, and $(2,0)$. What are its vertical-slice bounds?
Find the equation of the line through $(1,2)$ and $(5,4)$, written as $y$ in terms of $x$.
A triangle has vertices $(0,0)$, $(4,0)$, and $(2,3)$. What are the equations of its two slanted edges?
For the triangle with vertices $(1,0)$, $(5,0)$, and $(3,4)$, state the horizontal-slice bounds and identify the $y$-value where the slanted edges meet.
What endpoint substitutions can you use to check a proposed slanted-edge equation?