From Integral Limits to a Region
A nested double integral contains a geometric description. Its outer limits give the range of one coordinate, while its inner limits say how the other coordinate varies for each fixed value of the first. Reading those limits carefully lets you identify the region, the direction of its slices, and the curves that bound it.
The order of the differentials is the first clue. In an integral with \(dy\,dx\), \(x\) is fixed while \(y\) runs between its inner limits, so the slices are vertical. With \(dx\,dy\), \(y\) is fixed while \(x\) runs between its inner limits, so the slices are horizontal.
For horizontal slices, the analogous description is \(D=\{(x,y):c\le y\le d,\ p(y)\le x\le q(y)\}\), corresponding to \(\int_c^d\int_{p(y)}^{q(y)} f(x,y)\,dx\,dy\). Here each fixed \(y\) gives a horizontal segment from \(x=p(y)\) to \(x=q(y)\).
The integrand \(f(x,y)\) does not determine the region. The bounds do. To sketch the region, first draw the outer coordinate interval and the curves in the inner bounds. Then mark which curve is the lower or left edge of each slice and check where the slices begin or end.
A Systematic Reading Method
In \(dy\,dx\), use vertical slices; in \(dx\,dy\), use horizontal slices.
This tells you which values of the fixed coordinate occur.
They give the two endpoints of each slice, as functions of the fixed coordinate.
Solve for intersections of boundary curves, evaluate the limits at the outer endpoints, and test an interior slice.
The boundary consists of the portions of the inner-limit curves that actually enclose the region, together with any edge created by an outer endpoint. Not every intersection of the full curves must belong to the region: an intersection outside the outer interval is not part of its boundary.
Worked Example: Reading Vertical Slices Between a Parabola and a Line
Interpret the bounds in \(\int_0^2\int_{x^2}^{2x} f(x,y)\,dy\,dx\). The differential order is \(dy\,dx\), so the slices are vertical. The outer limits give \(0\le x\le2\). For each such \(x\), the lower edge is \(y=x^2\) and the upper edge is \(y=2x\).
To locate where the two curves meet, set their \(y\)-values equal:
Thus \(x=0\) or \(x=2\). Substitution into either curve gives \(y=0\) when \(x=0\), and \(y=4\) when \(x=2\). The intersection points are \((0,0)\) and \((2,4)\).
Check an interior slice, for example \(x=1\). Its lower value is \(1^2=1\), and its upper value is \(2(1)=2\). Since \(1\le2\), the stated order is correct there. At the outer endpoints, the bounds also agree: at \(x=0\) both are \(0\), and at \(x=2\) both are \(4\). The region is the area between the parabola \(y=x^2\) and the line \(y=2x\), for \(0\le x\le2\), with vertical slices shrinking to points at the two intersections.
Check: The intersection points satisfy both boundary equations: \(0^2=2(0)=0\) and \(2^2=2(2)=4\). The sample slice at \(x=1\) runs from the parabola to the line, as the bounds require.
Reading Horizontal Slices
When the inner differential is \(dx\), hold \(y\) fixed and read the inner limits as the left and right endpoints. The inner limits may be written as functions of \(y\), but their boundary curves can often be described in the more familiar form \(y=\text{function of }x\). Either form represents the same curve; the slice direction tells you which coordinate is varying.
Worked Example: A Region With a Horizontal Edge
Consider \(\int_1^3\int_{y-1}^{5-y} f(x,y)\,dx\,dy\). The order \(dx\,dy\) gives horizontal slices. The outer limits say \(1\le y\le3\). For each fixed \(y\), the left endpoint is \(x=y-1\), and the right endpoint is \(x=5-y\).
Find where these endpoints coincide:
At \(y=3\), \(x=y-1=2\), so the curves meet at \((2,3)\). At the other outer endpoint \(y=1\), the left endpoint is \(x=1-1=0\), and the right endpoint is \(x=5-1=4\). These give the points \((0,1)\) and \((4,1)\), joined by the horizontal edge \(y=1\).
Check an interior slice at \(y=2\): its left endpoint is \(2-1=1\), and its right endpoint is \(5-2=3\). The slice runs from \(x=1\) to \(x=3\), so the order is valid. The boundary curves are \(x=y-1\), equivalently \(y=x+1\), and \(x=5-y\), equivalently \(y=5-x\), along with the edge \(y=1\). The two slanted edges meet at \((2,3)\).
Check: At the meeting point, both expressions give \(x=2\): \(3-1=2\) and \(5-3=2\). At \(y=1\), the full slice runs from \(x=0\) to \(x=4\), as recorded.
Outer Endpoints Can Create Boundary Edges
An outer interval does more than restrict the values of a coordinate. If the inner limits differ at an outer endpoint, that endpoint contributes an entire edge to the region. The edge is vertical when the outer coordinate is \(x\), and horizontal when it is \(y\). If the inner limits agree there, the slice has collapsed to a point instead.
Worked Example: An Outer Endpoint and an Intersection Outside the Region
Read \(\int_{-1}^1\int_{x^2}^{2-x} f(x,y)\,dy\,dx\). Since the order is \(dy\,dx\), the region has vertical slices. The outer interval is \(-1\le x\le1\), with lower curve \(y=x^2\) and upper curve \(y=2-x\).
The curves intersect where \(x^2=2-x\):
The solutions are \(x=-2\) and \(x=1\). Their corresponding \(y\)-coordinates are \(y=4\) and \(y=1\), respectively, giving intersections \((-2,4)\) and \((1,1)\). Only \((1,1)\) is in the region: \(x=-2\) is outside the allowed interval \([-1,1]\).
At the left outer endpoint \(x=-1\), the lower value is \((-1)^2=1\), while the upper value is \(2-(-1)=3\). Thus the boundary includes the vertical segment from \((-1,1)\) to \((-1,3)\). At \(x=1\), the two limits both equal \(1\), so the slice ends at \((1,1)\). As an interior check, at \(x=0\) the limits are \(0\) and \(2\), in the stated lower-to-upper order.
The region is bounded by the parabola \(y=x^2\), the line \(y=2-x\), and the vertical segment \(x=-1\), with \(x\) restricted to \([-1,1]\). Its corner points are \((-1,1)\), \((-1,3)\), and \((1,1)\).
Check: At \(x=-1\), the vertical edge endpoints match the two inner limits, \(y=1\) and \(y=3\). At \((1,1)\), both curves give \(y=1\). The other curve intersection, \((-2,4)\), cannot be on the boundary of this region because its \(x\)-coordinate violates the outer limits.
Why the Bounds Describe Exactly These Points
The region interpretation follows directly from what the nested limits specify. Assume \(g\) and \(h\) are continuous on \([a,b]\) and \(g(x)\le h(x)\) throughout that interval. For a fixed \(x_0\in[a,b]\), the inner integral ranges over precisely the \(y\)-values between \(g(x_0)\) and \(h(x_0)\). Allowing \(x_0\) to range over the whole interval collects all those vertical slices.
This equality is not an additional geometric assumption. A point belongs to the union on the left exactly when its \(x\)-coordinate is some \(x_0\in[a,b]\) and its \(y\)-coordinate lies between \(g(x_0)\) and \(h(x_0)\). Those are precisely the conditions on the right. The horizontal-slice description follows by the same reasoning, fixing \(y\) and ranging \(x\) between its inner bounds.
A Common Mistake: Assigning the Inner Limits to the Wrong Coordinate
For \(\int_0^2\int_{x^2}^{2x} f(x,y)\,dy\,dx\), a common error is to say that \(x\) runs from \(x^2\) to \(2x\), or to identify \(x^2\) and \(2x\) as left and right boundaries. That reading assigns the inner bounds to the wrong variable. The inner differential is \(dy\), so those limits give values of \(y\), not \(x\).
The correction is to use the differential order: \(x\) ranges from \(0\) to \(2\), and for each fixed \(x\), \(y\) ranges from \(x^2\) to \(2x\). The curves are therefore bottom and top boundaries for vertical slices, not left and right boundaries for horizontal slices.
A second useful check is numerical. At \(x=1\), the intended slice is \(1\le y\le2\). This confirms both the coordinate assignment and the ordering. Whenever a sketch seems inconsistent, evaluate the inner bounds at an interior value of the outer variable before proceeding.
Check Your Understanding
For each set of bounds, identify the slice direction, describe the inner boundaries, and find the key points requested.
- For \(\int_0^2\int_0^{4-x^2} f(x,y)\,dy\,dx\), state the slice direction and identify the lower and upper boundaries.
- For \(\int_0^2\int_y^{2} f(x,y)\,dx\,dy\), state the slice direction and the left and right boundaries. What are the endpoints of the slice at \(y=1\)?
- For \(\int_{-1}^1\int_0^{2-x} f(x,y)\,dy\,dx\), what boundary edge comes from the left outer endpoint \(x=-1\)?
- Why must the intersection \((-2,4)\) be excluded from the region in the third worked example?
- What does it indicate if the lower inner limit exceeds the upper inner limit for some value in the outer interval?