Source: Stewart's Calculus, 8th ed., Sections 15.1, 15.2, 15.3, 10.3
Difficulty: Intermediate to Advanced | Prerequisites: Single-variable integration techniques (substitution, trig integrals), polar coordinates basics, familiarity with setting up iterated integrals.
Double integrals extend the definite integral to functions of two variables. Where a single integral computes the area under a curve, a double integral computes the volume under a surface (or, more generally, accumulates a quantity over a two-dimensional region). This set of notes covers evaluating double integrals over rectangles and more general regions, reversing the order of integration, converting to polar coordinates, computing volumes of solids, polar curves, and area in polar coordinates. These topics make up the integration half of a typical Math 126 midterm.
Double integrals are evaluated as iterated integrals (integrate one variable, then the other). When the region or integrand involves circles or x^2 + y^2, switch to polar coordinates with dA = r dr d(theta). Sometimes you need to reverse the order of integration to make the integral tractable. Volume under a surface z = f(x, y) over a region D equals the double integral of f over D.
Tags: double integral, iterated integral, Fubini's theorem, order of integration, type I region, type II region, polar coordinates, r dr dtheta, cardioid, rose curve, limacon, volume under surface
Double integral
The integral of f(x, y) over a region D in the xy-plane, written as the double integral of f dA over D. It gives the signed volume between the surface z = f(x, y) and the xy-plane.
Think of it as: the 2D version of the area-under-a-curve integral from Calc I.
Iterated integral
A double integral evaluated by integrating one variable at a time. Fubini's theorem guarantees that for continuous functions on nice regions, you can integrate in either order and get the same result.
In simple terms, you do the inner integral first (treating the other variable as a constant), then do the outer integral.
Type I region
A region described as a <= x <= b, g_1(x) <= y <= g_2(x). The y-limits depend on x.
Type II region
A region described as c <= y <= d, h_1(y) <= x <= h_2(y). The x-limits depend on y.
Think of it as: Type I means you slice vertically (integrate y first); Type II means you slice horizontally (integrate x first).
Polar coordinates
x = r cos(theta), y = r sin(theta), x^2 + y^2 = r^2. In polar, the area element is dA = r dr d(theta). The extra factor of r is essential and frequently forgotten.
Cardioid
A heart-shaped polar curve of the form r = a + a cos(theta) or r = a + a sin(theta). When the coefficients are equal, the curve passes through the origin.
In simple terms, it is the curve you get when you trace a point on the rim of a circle rolling around another circle of the same size.
Rectangles (Fubini's theorem)
For f continuous on R = [a, b] x [c, d], the double integral equals either of the iterated integrals:
Integrate from c to d with respect to y first, then from a to b with respect to x, or
Integrate from a to b with respect to x first, then from c to d with respect to y.
Worked pattern (from Problem 7a)
Double integral of (ln y) / (xy) over [1, 3] x [1, 5]:
Inner integral (with respect to y): integral of (ln y) / y dy = (ln y)^2 / 2, evaluated from 1 to 5 gives (ln 5)^2 / 2.
Outer integral (with respect to x): integral of (ln 5)^2 / (2x) dx from 1 to 3 = (ln 3)(ln 5)^2 / 2.
General regions
For non-rectangular regions, the limits of the inner integral become functions of the outer variable.
Type I: outer integral over x with constant limits, inner integral over y with limits that are functions of x.
Type II: outer integral over y with constant limits, inner integral over x with limits that are functions of y.
Choosing the order
Sketch the region D. If D is more naturally described by vertical slices (y-limits depend on x), use Type I. If horizontal slices are simpler (x-limits depend on y), use Type II. Sometimes both work; sometimes one order yields a simpler integral.
Why reverse?
Some iterated integrals cannot be evaluated in the given order because the inner integral has no closed-form antiderivative. Reversing the order often produces an integral you can compute.
Method
Sketch the region of integration from the original limits.
Re-describe the region using the other order (switch which variable has constant limits).
Rewrite the integral with the new limits.
Worked pattern (from Problem 9a)
Original: integral from 0 to 1 of (integral from x^2 to 1 of sqrt(y) sin y dy) dx.
The region is 0 <= x <= 1, x^2 <= y <= 1. Sketching: this is the area under y = 1 and above y = x^2, i.e. 0 <= y <= 1, 0 <= x <= sqrt(y).
Reversed: integral from 0 to 1 of (integral from 0 to sqrt(y) of sqrt(y) sin y dx) dy.
Inner integral: sqrt(y) sin y times sqrt(y) = y sin y.
Outer integral: integral of y sin y dy from 0 to 1 = sin 1 - cos 1 (by integration by parts).
Worked pattern (from Problem 9b)
Original: integral from 0 to 1 of (integral from sqrt(x) to 1 of sqrt(y^3 + 1) dy) dx.
Region: 0 <= x <= 1, sqrt(x) <= y <= 1, which is equivalently 0 <= y <= 1, 0 <= x <= y^2.
Reversed integral evaluates to (2/9)(2 sqrt(2) - 1).
Converting to polar
Use polar coordinates when the region D or the integrand involves x^2 + y^2, circles, or sectors.
Substitutions: x = r cos(theta), y = r sin(theta), x^2 + y^2 = r^2.
Area element: dA = r dr d(theta). Do not forget the extra r.
Setting up limits
theta ranges over the angular sweep of the region.
r ranges from the inner boundary to the outer boundary, which may be functions of theta.
Worked pattern (from Problem 7f)
Double integral of y^2 / (x^2 + y^2) dA over the annular region a^2 <= x^2 + y^2 <= b^2:
In polar: y^2 / (x^2 + y^2) = r^2 sin^2(theta) / r^2 = sin^2(theta).
Integral becomes: integral from 0 to 2 pi of (integral from a to b of sin^2(theta) r dr) d(theta).
Inner: sin^2(theta) (b^2 - a^2)/2.
Outer: (b^2 - a^2)/2 times pi = pi(b^2 - a^2)/2.
Polar curves (from Problem 10)
Key curve types to recognise:
r = -2 sin(theta): circle of radius 1 centred at (0, -1).
r = 2 + sin(3 theta): rose-like curve with three small petals added to a circle.
r = 1 + 5 sin(theta): limacon with an inner loop (because 5 > 1).
r^2 = cos(4 theta): eight-petalled rose (since 4 theta gives 8 petals for r^2 form).
r = sin(theta/2): a single closed loop.
Area in polar coordinates (from Problem 11)
Area enclosed by a polar curve r = f(theta) from alpha to beta: A = (1/2) integral from alpha to beta of r^2 d(theta).
For the region inside both cardioids r = 1 + cos(theta) and r = 1 - cos(theta): find the intersection angles, integrate the appropriate r^2 over each subregion, and use symmetry. The answer is 3 pi/2 - 4.
Volume as a double integral
The volume of the solid under z = f(x, y) and above the region D in the xy-plane is V = double integral of f(x, y) dA over D (assuming f >= 0 on D).
Worked pattern (from Problem 8a)
Volume under z = x^2 + xy^2, with 0 <= x <= 5 and -2 <= y <= 2:
V = integral from 0 to 5 of (integral from -2 to 2 of (x^2 + xy^2) dy) dx.
Inner integral: x^2(4) + x(16/3) = 4x^2 + 16x/3.
Outer integral: [4x^3/3 + 8x^2/3] from 0 to 5 = 700/3.
Volume between surfaces (from Problem 8d)
Volume bounded by the paraboloid z = 1 + 2x^2 + 2y^2 and the plane z = 7 in the first octant. The intersection is x^2 + y^2 = 3. Convert to polar in the first quadrant:
V = integral from 0 to pi/2 of (integral from 0 to sqrt(3) of (7 - 1 - 2r^2) r dr) d(theta) = 9 pi/4.
Volume under a cone (from Problem 8e)
Volume below z = sqrt(x^2 + y^2) above the ring 1 <= x^2 + y^2 <= 4. In polar, z = r:
V = integral from 0 to 2 pi of (integral from 1 to 2 of r * r dr) d(theta) = 2 pi (8/3 - 1/3) = 14 pi/3.
Key formulas
Double integral over rectangle: integral from a to b of (integral from c to d of f(x,y) dy) dx
Polar conversion: x = r cos(theta), y = r sin(theta), dA = r dr d(theta)
Area in polar: A = (1/2) integral of r^2 d(theta)
Volume: V = double integral of f(x,y) dA over D
Applications
Double integrals compute mass (when f is a density function), centre of mass, moments of inertia, and probability over two-dimensional regions. In physics, the moment of inertia of a flat plate about an axis is a double integral of (distance from axis)^2 times the density.
Polar coordinates are natural for any problem with circular symmetry: antenna radiation patterns, orbital mechanics, heat distribution in a disc-shaped object. The area formula for polar curves is how you compute the area swept by a radar beam or the cross-section of a cam.
The single most common error in polar integrals is forgetting the extra factor of r in dA = r dr d(theta). Without it, every answer is wrong.
Students often mix up the limits when reversing the order of integration. Always sketch the region first. The sketch makes the new limits visible.
When computing area in polar, students sometimes integrate r instead of r^2/2. The area formula is (1/2) integral of r^2, not integral of r.
For volume between two surfaces, the integrand is (top surface minus bottom surface), not just one of the surfaces. Check which surface is on top in the region.
Double integrals appear in at least two or three problems on a typical Math 126 midterm. Speed in setting up limits is essential.
Reversing the order of integration is a classic exam problem. If you see an inner integral you cannot evaluate, that is your cue to reverse.
Polar coordinate conversions are tested almost every exam. Practise recognising when to switch.
Volume problems often combine setting up the double integral with converting to polar.
Know the shapes of basic polar curves (cardioids, roses, limacons, circles). The exam may ask you to sketch and then compute area.
True or false: When converting a double integral to polar, dA becomes dr d(theta). (Answer: False. dA = r dr d(theta). The r factor is required.)
Fill in the blank: The area enclosed by a polar curve r = f(theta) from alpha to beta is ______. (Answer: (1/2) integral from alpha to beta of [f(theta)]^2 d(theta).)
True or false: You can always evaluate a double integral in either order. (Answer: True for continuous functions on well-behaved regions, by Fubini's theorem.)
Fill in the blank: A region where the y-limits depend on x is called a Type ______ region. (Answer: I.)
True or false: The volume under a surface z = f(x, y) over D equals the double integral of f over D, even when f is negative. (Answer: True, but then it gives signed volume, which may be negative.)
Q: Evaluate the double integral of 1/(1 + x + y) dA over R = [1, 3] x [1, 2].
A: Inner integral (with respect to y): ln(1 + x + y) evaluated from 1 to 2 = ln(3 + x) - ln(2 + x). Outer integral: evaluate from 1 to 3. Result: 6 ln 6 - 5 ln 5 - 4 ln 4 + 3 ln 3.
Q: Evaluate the integral from 0 to 1 of (integral from x^2 to 1 of sqrt(y) sin y dy) dx by reversing the order.
A: Reversed region: 0 <= y <= 1, 0 <= x <= sqrt(y). Reversed integral: integral from 0 to 1 of y sin y dy = sin 1 - cos 1.
Q: Find the volume of the solid under z = sqrt(x^2 + y^2) above the ring 1 <= x^2 + y^2 <= 4.
A: In polar, z = r. V = integral from 0 to 2 pi of (integral from 1 to 2 of r^2 dr) d(theta) = 2 pi (7/3) = 14 pi/3.
Q: Compute the double integral of arctan(x/y) dA over D = {1 <= x^2 + y^2 <= 4, 0 <= y <= x}.
A: In polar, x/y = cos(theta)/sin(theta) = cot(theta) and arctan(cot(theta)) = pi/2 - theta. Region: 1 <= r <= 2, 0 <= theta <= pi/4. Integral = (3/62) pi^2. (Note: the exact computation uses the identity arctan(cot theta) = pi/2 - theta.)
Q: Find the area enclosed by both cardioids r = 1 + cos(theta) and r = 1 - cos(theta).
A: By symmetry, compute the area in one quadrant and multiply. The total area is 3 pi/2 - 4.
Double integrals extend to triple integrals in Calc III (cylindrical and spherical coordinates), which use the same logic of converting to a coordinate system that matches the geometry.
Polar coordinates connect to parametric curves (Section 10.3) and to complex numbers in later courses. The area formula for polar curves is a special case of the general double-integral area formula.
The regions used in absolute-extrema problems (from the second study notes) are the same kinds of regions you integrate over here. Being able to describe a region both as a Type I and Type II set is useful for both topics.
double integral, iterated integral, Fubini's theorem, change order of integration, reversing order of integration, Type I region, Type II region, polar coordinates, polar integral, r dr dtheta, cardioid, limacon, rose curve, polar area, volume double integral, volume under surface, volume between surfaces, Math 126 UW, Stewart Calculus 15.1, 15.2, 15.3, 10.3