A roof over a square floor of side follows the plane , so it has height above one corner and height above the opposite corner. How much space lies between the floor and the roof? A single integral adds strips along one axis, but every point of this floor needs two coordinates, and the height changes in both directions. This lesson builds the double integral from thin columns, evaluates it as an iterated integral one variable at a time, and finds that the volume is exactly .
Key terms
Rectangular region
Definition For real numbers and , the closed rectangle is the set of points with and . Its sides are parallel to the axes and its area is . On a rectangular region every bound is a constant. The range of is the same interval for every , and the range of is for every .
In plain words A rectangle with sides along the axis directions is the simplest base for a double integral. Slicing it in either direction gives segments with the same endpoints every time, so no bound depends on the other variable.
Example The unit square has area . Cutting each side into equal pieces splits it into small squares of side and area . By contrast, the triangle of points with is not a rectangular region, because at a given the variable runs only from to . Regions of that kind are the subject of the next lesson.
Double integral
Definition Let be a bounded function on . Points and cut into the cells , with areas . Pick a sample point in each cell and form the double Riemann sum
Suppose these sums approach one number as the longest cell diagonal shrinks to , whatever sample points are chosen. Then is integrable on , and is its double integral,
Every function that is continuous on is integrable there. The symbol stands for the area of a small cell of the base.
In plain words Tile the base, stand a thin column on each tile as tall as at the chosen point, and add up the products of height and tile area. The double integral is the number these totals settle on as the tiles shrink in both directions.
Example Take on the unit square, cut into equal squares of area . The function grows in both variables, so on each cell its smallest value is at the lower left corner and its largest value is at the upper right corner. With , the four lower left corners give heights that add to , namely , , and , and the four upper right corners give , , and , which add to . So the lowest and highest sums on this grid are
For general the same count gives and , and every Riemann sum on that grid lies between them. Both bounds approach , and is continuous, hence integrable, so
Volume under a surface
Definition Let be integrable on with at every point of . The solid of points with in and has volume
If takes negative values, the integral is a signed volume. The space between the graph and the base counts as positive where the graph lies above the plane and as negative where it lies below. The volume between the graph and the base is then the integral of .
In plain words A column with base area and height has volume , so adding columns adds volume. Reading the integral as a volume needs a height that never dips below the base.
Example The roof over the unit square is continuous and nonnegative, so the double integral found above is the volume under it, in cubic units. Geometry confirms the value. The box has volume . The point reflection through its center sends to and maps the box onto itself. A point satisfies exactly when its image satisfies , so the reflection swaps the part of the box below the roof with the part above it. The two parts are congruent and share only the roof, which has no volume, so each part has volume .
Iterated integral
Definition For a function on , the iterated integral in the order is
The bracket is the inner integral, taken first in from to . Its value is then integrated by the outer integral, in from to . The differentials are read from the inside out, and each pair of limits belongs to the differential in the matching position, the innermost limits to the innermost differential. The iterated integral is defined when the inner integral exists for every in and its value is an integrable function of . The order is defined in the same way, with the roles of and exchanged. If is continuous on , both iterated integrals exist and both equal . This is Fubini’s theorem for a rectangle.
In plain words Add along one direction first, then add the results along the other. The inner integral sweeps out one slice of the solid, and the outer integral moves that slice across the base.
Example The volume under the roof is
and the integrand is continuous on the unit square, so this iterated integral equals the double integral. The inner integral, with held fixed, is
The outer integral adds these values across the square,
In the other order the inner integral is , and its integral from to is again .
Holding one variable fixed
Definition In the inner integral , the outer variable is held fixed. It is treated as a constant while runs from to , so a term without integrates like a constant, and the evaluation substitutes and only. The result
is a function of alone. When , the number is the area of the cross section of the solid by the plane perpendicular to the axis at that . The outer integral then adds these areas, as in the slicing method for volumes.
In plain words Freeze , and the problem becomes an ordinary integral in one variable. Its answer still depends on which was frozen, so it is the area of one slice and not yet the volume.
Example For the roof , freeze . Over the segment the height rises in a straight line from to , so the slice is a trapezoid of width with parallel sides and . Its area is , which is exactly the inner integral. At the slice is a triangle of area , and at its area is . A common slip is to integrate the term as if were the variable and write . Since is a constant here, its integral in is , which contributes , the area of the rectangular part of the slice.
Check yourself
- Find the volume of the solid that lies above the rectangle and below the plane .
Show the answer
Every point of the rectangle has , so the height is at least and the integral is a volume. Holding fixed,
Then . As a check, the height is linear, so its average over the rectangle is its value at the center , which is , and times the area is .
- Evaluate over in both orders. A classmate writes the order as , keeping the limits where they were. What does that expression compute?
Show the answer
In the order the inner integral is , and . In the order the inner integral is , and . The two orders agree, as they must for a continuous integrand on a rectangle.
The classmate’s expression pairs the inner limits and with , so runs over and over . That is a different rectangle, and the value is different too. The inner integral is , and
When the order changes, the limits move with their differentials.
- Evaluate . Is the result the volume of the region between the plane and the rectangle ?
Show the answer
The inner integral is , and . The result is not that volume. The plane lies below the base where and above it where , so the integral is a signed volume. The part over contributes and the part over contributes , and they cancel. The volume between the plane and the base is the integral of , which is . Reading a double integral as a volume needs a nonnegative height.