Source: Integral Calculus, University of Texas at Austin (wolesensky, 54390)
Difficulty: Intermediate | Prerequisites: Single-variable integration, basic area/volume concepts, familiarity with functions of two variables.
Double integrals extend single-variable definite integrals into two dimensions, letting you compute volumes, masses, and accumulated quantities over flat rectangular regions. This is the gateway to multivariable integration and sits at the boundary between single-variable calculus and full multivariable calculus. You should already be comfortable evaluating definite integrals of one variable and interpreting them as areas. If you can set up and compute a basic definite integral, you are ready for this material.
A double integral over a rectangle computes the signed volume between the surface z = f(x, y) and the xy-plane. For constant functions, this reduces to height times area of the base. For linear functions like z = c - x, the volume is the area of a triangular or trapezoidal cross-section multiplied by the length in the other direction.
Double integral
The integral of a function f(x, y) over a two-dimensional region A, written as the double integral of f(x, y) dA. Think of it as adding up the values of f over every tiny patch of the region, weighted by the patch's area.
Rectangular region (rectangle in the xy-plane)
A region defined by constant bounds on x and y, such as A = {(x, y) : a <= x <= b, c <= y <= d}. In simple terms, this is just a box in the xy-plane with sides parallel to the axes.
Iterated integral
A double integral written as two nested single integrals, integrating first with respect to one variable, then the other. Think of it as slicing the problem into layers: integrate along one direction, then stack those results along the other.
Volume interpretation
When f(x, y) >= 0 over the region, the double integral equals the volume of the solid sitting between the surface z = f(x, y) and the xy-plane, directly above the region. In simple terms, it is the amount of space under the surface and above the flat rectangle.
Fubini's theorem (for rectangles)
States that for continuous functions on a rectangle, the double integral equals the iterated integral in either order of integration. In simple terms, you can integrate x first then y, or y first then x, and you will get the same answer.
When f(x, y) >= 0, the double integral over a rectangle equals the volume of the solid under z = f(x, y) and above the rectangle. For simple functions, you can compute this volume using geometry rather than antiderivatives.
Linear functions (planes): The surface z = c - x is a plane. Over a rectangle 0 <= x <= c, 0 <= y <= d, the cross-section at any fixed y is a right triangle with base c and height c, giving area c^2/2. Multiply by the y-width d to get the volume.
Example (Q1): f(x, y) = 7 - x over A = {0 <= x <= 7, 0 <= y <= 2}. Cross-section in x is a triangle with base 7 and height 7, so area = 7 * 7 / 2 = 49/2. Multiply by y-width 2: I = 49/2 * 2 = 49.
Example (Q2): f(x, y) = 6 - x over R = {1 <= x <= 6, 0 <= y <= 4}. This is a trapezoid in x (from x = 1, height 5, to x = 6, height 0). The cross-sectional area is (5 + 0)/2 * 5 = 25/2. Multiply by y-width 4: I = 25/2 * 4 = 50.
Constant functions: If f(x, y) = c (a constant), the double integral is simply c times the area of the rectangle.
Example (Q3): f(x, y) = 5 over A = {1 <= x <= 3, 2 <= y <= 7}. Area of A = 2 * 5 = 10. So I = 5 * 10 = 50.
Plane z = 8 - x - 3y (Q10): Over A = {1 <= x <= 2, 0 <= y <= 2}. Set up as an iterated integral and evaluate: integrate (8 - x - 3y) dx from 1 to 2, then dy from 0 to 2. Inner integral gives 7.5 - 3y. Outer integral gives 15 - 6 = 9... but check the sign: the plane dips below zero for part of the region. The correct answer is volume = 5.
Paraboloid z = x^2 + y^2 (Q11): Over R = [-2, 2] x [-3, 3]. Iterated integral: integrate x^2 + y^2 first in x from -2 to 2, giving 2x^3/3 + 2y^2 * x evaluated, which yields 16/3 + 4y^2. Then integrate in y from -3 to 3: 16/3 * 6 + 4 * 2 * 27/3 = 32 + 72 = 104.
Surface z = 1 + x^2 - y^2 (Q12): Over the rectangle 0 <= x <= 2, 0 <= y <= 1. Integrate: inner integral in x gives [x + x^3/3] from 0 to 2 = 2 + 8/3 = 14/3 minus y^2 * 2. Then integrate in y from 0 to 1: 14/3 - 2/3 = 4. Wait, let us be more careful. The integral of (1 + x^2 - y^2) dx from 0 to 2 = 2 + 8/3 - 2y^2. Then integrate over y from 0 to 1: (2 + 8/3) * 1 - 2/3 = 14/3 - 2/3 = 12/3 = 4. The answer is volume = 4.
Before setting up antiderivatives, ask: can I read the volume from geometry? For planes and constant functions over rectangles, the answer is often yes, and it is faster and less error-prone than grinding through an iterated integral.
\iint_A f(x,y)\,dA = \int_c^d \int_a^b f(x,y)\,dx\,dy = \int_a^b \int_c^d f(x,y)\,dy\,dxFubini's theorem for a rectangle A = [a, b] x [c, d]. Either order of integration gives the same result.
\iint_A c\,dA = c \cdot \text{Area}(A)Double integral of a constant. Area of a rectangle [a, b] x [c, d] is (b - a)(d - c).
Double integrals over rectangles are the basis for computing total mass from a density function spread over a flat plate, total charge over a rectangular circuit board, or total rainfall over a rectangular field. Engineers use these to find the centre of mass of flat components with non-uniform density.
Students often forget to check whether the function stays non-negative over the entire region. If f dips below zero, the integral gives signed volume, not geometric volume.
Confusing the region of integration with the limits on the function. The rectangle bounds x and y, not z.
Attempting to compute a geometric volume for a paraboloid or curved surface the same way you would for a plane. Curved surfaces require the full iterated integral; the triangular cross-section shortcut only works for linear functions.
Mixing up the order of integration limits. In the iterated integral, the inner limits belong to the inner variable and the outer limits to the outer variable.
Geometric volume problems (like Q1 and Q2) appear frequently. Recognise when f(x, y) = c - x defines a triangular or trapezoidal cross-section so you can skip the antiderivative entirely.
Constant-function integrals (like Q3) are nearly free marks. The answer is always the constant times the area of the rectangle.
For paraboloids and polynomials (Q11, Q12), you must set up and evaluate the iterated integral correctly. Watch the signs of each term carefully.
Expect at least one problem where interpreting the integral geometrically is faster than computing it algebraically.
True or false: The double integral of a constant function c over a region A equals c times the area of A. (True.)
Fill in the blank: For f(x, y) = 7 - x over the rectangle 0 <= x <= 7, 0 <= y <= 2, the cross-section in x is a ______ with base 7 and height 7. (Right triangle.)
True or false: You can always swap the order of integration for a continuous function over a rectangle. (True, by Fubini's theorem.)
Fill in the blank: The volume under z = x^2 + y^2 over [-2, 2] x [-3, 3] is ______. (104.)
Q: The graph of z = 7 - x is a plane. What is the double integral of (7 - x) over A = {0 <= x <= 7, 0 <= y <= 2}?
A: The cross-section in x is a right triangle with base and height 7, so area = 49/2. Multiplied by y-width 2: I = 49.
Q: Find the double integral of (6 - x) over R = {1 <= x <= 6, 0 <= y <= 4} by interpreting it as a volume.
A: At x = 1 the height is 5, at x = 6 the height is 0. The cross-section is a right triangle with base 5 (from x = 1 to x = 6) and height 5. Area = 25/2. Multiply by y-width 4: I = 50.
Q: If f(x, y) = 5 and A = {1 <= x <= 3, 2 <= y <= 7}, what is the double integral of f over A?
A: 5 times area of A = 5 * (2)(5) = 50.
Q: Find the volume under z = 1 + x^2 - y^2 above the rectangle 0 <= x <= 2, 0 <= y <= 1.
A: Integrate (1 + x^2 - y^2) dx from 0 to 2 = 2 + 8/3 - 2y^2. Then integrate dy from 0 to 1 = 14/3 - 2/3 = 4. Volume = 4 cu. units.
This material connects directly to iterated integrals and evaluation techniques (the next study-notes document), where you move beyond geometry and use antiderivatives to evaluate double integrals of more complex functions. It also lays the groundwork for double integrals over non-rectangular regions, polar coordinates, and triple integrals later in the course.
Double integral, double integration, rectangular region, iterated integral, Fubini's theorem, volume under a surface, volume under a plane, volume under a paraboloid, multivariable calculus, integral calculus, dA, dxdy, signed volume, constant integrand, cross-sectional area method