Difficulty: Intermediate to Advanced | Prerequisites: Single-variable integration, basic polar coordinates, regions in the xy-plane
Double integrals extend single-variable integration to functions of two variables. The two main skills tested here are reversing the order of integration (which can turn an impossible integral into a doable one) and converting to polar coordinates (which simplifies integrals over circular or annular regions). You should be comfortable sketching regions in the xy-plane and converting between Cartesian and polar before working through these.
Reversing the order of integration means re-describing the same region with the other variable as the outer integral. Converting to polar coordinates replaces x² + y² with r² and dA with r dr dθ. Both techniques are about choosing the setup that makes the integral computable.
Double integral
The integral ∬_R f(x, y) dA, which computes the signed volume under the surface z = f(x, y) over a region R. Think of it as stacking up thin slices of area, each weighted by the function's height.
Iterated integral
A double integral written as two nested single-variable integrals, e.g. ∫∫ f(x, y) dy dx. The inner integral is evaluated first, treating the outer variable as a constant.
Reversing the order of integration
Rewriting ∫∫ f dy dx as ∫∫ f dx dy (or vice versa) by re-describing the region of integration. In simple terms, you are slicing the same region horizontally instead of vertically (or the reverse). This is necessary when the inner integral cannot be evaluated in closed form.
Polar coordinates
x = r cos θ, y = r sin θ, x² + y² = r². In polar, the area element becomes dA = r dr dθ. Polar coordinates are the natural choice when the region or the integrand involves x² + y².
Jacobian (for polar)
The factor r that appears in dA = r dr dθ. It accounts for the fact that polar "rectangles" are wider at larger radii. Forgetting the r is one of the most common errors in polar integrals.
Sketch the region described by the original limits.
Re-describe the region with the roles of x and y swapped: determine the new outer variable's range (constants) and the new inner variable's limits (functions of the outer variable).
Rewrite the integral with the new limits and evaluate.
Worked example: ∫₀² ∫_{x/2}¹ x/(y³ + 1) dy dx. The region is 0 ≤ x ≤ 2, x/2 ≤ y ≤ 1. Sketch: y goes from x/2 up to 1, with x from 0 to 2. The boundary y = x/2 means x = 2y. Reversing: y goes from 0 to 1, and for each y, x goes from 0 to 2y. The integral becomes ∫₀¹ ∫₀^{2y} x/(y³ + 1) dx dy. Inner integral: [x²/2]₀^{2y} / (y³ + 1) = 2y²/(y³ + 1). Outer integral: ∫₀¹ 2y²/(y³ + 1) dy. Let u = y³ + 1, du = 3y² dy. Result: (2/3) ln|y³ + 1| evaluated from 0 to 1 = (2/3) ln 2.
Convert the region's boundary equations to polar: x² + y² = r², x = r cos θ, y = r sin θ.
Replace dA with r dr dθ.
Determine the limits for r and θ from the region.
For circles not centred at the origin (like x² + y² = 4y), complete the square first or convert directly: x² + y² = 4y becomes r² = 4r sin θ, so r = 4 sin θ.
Worked example: ∬R x/(x² + y²) dA where R is inside x² + y² = 4y, outside x² + y² = 8, in the first quadrant. In polar: the integrand becomes r cos θ / r² = cos θ / r, and dA = r dr dθ, so the integrand times dA is cos θ dr dθ. The circle x² + y² = 4y becomes r = 4 sin θ. The circle x² + y² = 8 becomes r = 2√2. These intersect when 4 sin θ = 2√2, giving θ = π/4. The region in the first quadrant where r = 4 sin θ is outside r = 2√2 exists for θ from π/4 to π/2. The integral is ∫{π/4}^{π/2} ∫{2√2}^{4 sin θ} cos θ dr dθ = ∫{π/4}^{π/2} cos θ (4 sin θ − 2√2) dθ. Evaluating: [2 sin²θ − 2√2 sin θ]_{π/4}^{π/2} = (2 − 2√2) − (1 − 2) = (2 − 2√2) − (−1) = 3 − 2√2.
Polar coordinate conversion:
x = r cos θ, y = r sin θ, r² = x² + y², dA = r dr dθ
Circle through the origin:
x² + y² = 2ax becomes r = 2a cos θ; x² + y² = 2by becomes r = 2b sin θ
Double integrals compute physical quantities like mass (integrating density over a region), centre of mass, moments of inertia, and probability over two-dimensional distributions. Polar coordinates are essential in physics for problems with radial symmetry: antenna radiation patterns, gravitational fields, and heat flow through circular pipes.
The single most common error in polar integrals is forgetting the extra factor of r in dA = r dr dθ. Every polar double integral needs it.
When reversing the order of integration, students often forget to sketch the region. Without a sketch, it is very easy to write incorrect new limits.
The circle x² + y² = 4y is centred at (0, 2) with radius 2, not at the origin. In polar it becomes r = 4 sin θ, not r = 4.
Students sometimes confuse which variable gets constant limits (the outer integral) and which gets function limits (the inner integral).
⚠️ "Reverse the order of integration" problems appear regularly. The typical setup gives you an inner integral you cannot evaluate in closed form, so reversing is required.
⚠️ For polar integrals, always sketch the region and identify the correct θ range. Getting θ wrong invalidates the entire calculation.
⚠️ Show the substitution step clearly (e.g. u = y³ + 1, du = 3y² dy) for full credit on integration steps.
Fill in the blank: In polar coordinates, dA = ______. (r dr dθ)
True or false: To reverse the order of integration, you need to change the integrand. (False. You only change the limits of integration.)
Fill in the blank: The polar equation of the circle x² + y² = 4y is r = ______. (4 sin θ)
Q: Reverse the order of integration and evaluate ∫₀² ∫_{x/2}¹ x/(y³ + 1) dy dx.
A: Region: 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y. Reversed integral: ∫₀¹ ∫₀^{2y} x/(y³ + 1) dx dy = ∫₀¹ 2y²/(y³ + 1) dy = (2/3) ln 2.
Q: Evaluate ∬_R x/(x² + y²) dA where R is inside x² + y² = 4y, outside x² + y² = 8, in the first quadrant.
A: In polar, the integrand simplifies to cos θ. The region has r from 2√2 to 4 sin θ, θ from π/4 to π/2. The integral evaluates to 3 − 2√2.
Double integrals build on single-variable integration from MATH 124/125. The regions of integration are described using the same curves and surfaces from the lines-and-planes topic. Polar coordinates extend naturally to cylindrical and spherical coordinates in MATH 324 (triple integrals). The technique of reversing integration order is a special case of Fubini's theorem.
double integral, iterated integral, reversing order of integration, Fubini's theorem, polar coordinates, polar double integral, r dr dtheta, Jacobian, change of variables, area element, annular region, MATH 126, Calculus III, UW Math 126