Source: Math 126 Spring 2023 Final Exam, University of Washington
Tags: double integral, reversing order of integration, polar coordinates, r dr dθ, region of integration, MATH 126, Calculus III
Difficulty: Intermediate to Advanced | Prerequisites: Single-variable integration, trigonometric identities, sketching regions
Double integrals extend single-variable integration to functions of two variables, letting you compute areas, volumes, and averages over 2D regions. Two core exam skills are reversing the order of integration (swapping dx dy for dy dx or vice versa) and converting Cartesian integrals to polar form. Both require you to sketch the region of integration first. If you are not yet comfortable sketching curves like y = x/2 or x² + y² = 4y, practise that before tackling these problems.
Reversing the order of integration means re-describing the same region with the other variable as the outer integral. Polar coordinates simplify integrals over circular or radial regions by replacing dA with r dr dθ.
Double integral
The integral ∬_R f(x, y) dA, which sums f over a 2D region R. Think of it as stacking up thin slices of volume (or weighted area) across the region.
Order of integration
Whether you integrate with respect to x first (then y) or y first (then x). The choice affects the limits, not the final answer. Some orders make the integral much easier to evaluate.
Reversing the order of integration
Rewriting ∫∫ f dx dy as ∫∫ f dy dx (or vice versa) by re-describing the region's bounds. This often turns an impossible integral into a solvable one.
Polar coordinates
x = r cos θ, y = r sin θ, and dA = r dr dθ. Useful whenever the region or the integrand involves x² + y². The extra factor of r in dA is essential and often forgotten.
Jacobian (in polar conversion)
The factor r in dA = r dr dθ. It accounts for the fact that polar "rectangles" are not actual rectangles, so area elements scale with distance from the origin.
Start by sketching the region. Identify the curves that bound it.
Re-express the bounds with the other variable as the outer integral.
Example: evaluate ∫₀² ∫_{x/2}¹ x/(y³+1) dy dx.
Original region: 0 ≤ x ≤ 2, x/2 ≤ y ≤ 1. The line y = x/2 is the same as x = 2y.
Reversed: 0 ≤ y ≤ 1, 0 ≤ x ≤ 2y.
New integral: ∫₀¹ ∫₀^{2y} x/(y³+1) dx dy.
Inner integral: [x²/2]₀^{2y} / (y³+1) = 2y²/(y³+1).
Substitution u = y³+1, du = 3y² dy: ∫₁² (2/3)(1/u) du = (2/3) ln 2.
Convert when the region or integrand involves x² + y².
Replace x² + y² with r², x with r cos θ, y with r sin θ, and dA with r dr dθ.
Key conversions for circles:
x² + y² = 8 becomes r = √8 = 2√2.
x² + y² = 4y becomes r² = 4r sin θ, so r = 4 sin θ.
Example: evaluate ∬_R x/(x²+y²) dA, where R is inside x² + y² = 4y, outside x² + y² = 8, first quadrant.
In polar: integrand = (r cos θ)/r² = cos θ/r. With dA = r dr dθ, the integrand becomes cos θ.
Intersection: 4 sin θ = √8, so sin θ = √2/2, meaning θ = π/4.
Region: θ from π/4 to π/2, r from √8 to 4 sin θ.
Integral: ∫{π/4}^{π/2} ∫{√8}^{4sinθ} cos θ · r · (1/r) dr dθ = ∫_{π/4}^{π/2} (4 sin θ cos θ − √8 cos θ) dθ.
Evaluate: [−cos 2θ − √8 sin θ]_{π/4}^{π/2} = (1 − √8) − (0 − 2) = 3 − √8.
\iint_R f(x,y)\,dA = \int_\alpha^\beta \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\;r\,dr\,d\thetax = r\cos\theta,\quad y = r\sin\theta,\quad x^2 + y^2 = r^2,\quad dA = r\,dr\,d\thetaThe most common error in polar conversion is forgetting the extra factor of r. dA = r dr dθ, not just dr dθ.
When reversing the order, students sometimes swap the limits mechanically without sketching the region. This leads to wrong bounds. Always draw the region first.
x² + y² = 4y is a circle centred at (0, 2) with radius 2, not centred at the origin. Students who assume origin-centred circles get the wrong polar bounds.
The intersection of two polar curves must be found by setting them equal. Do not assume standard angles.
⚠️ Reversing the order of integration appears on nearly every MATH 126 final. The integral is usually chosen so that one order is impossible to evaluate directly.
⚠️ Polar coordinate problems test both the conversion mechanics and your ability to find the correct θ-bounds from circle intersections.
⚠️ Forgetting the r in dA = r dr dθ is worth losing full marks on a polar integral.
Fill in the blank: In polar coordinates, dA = ______ dr dθ. (r)
True or False: x² + y² = 4y in polar coordinates becomes r = 4 sin θ. (True.)
Fill in the blank: If the original limits are 0 ≤ x ≤ 2, x/2 ≤ y ≤ 1, the reversed outer limits for y are ______ to ______. (0 to 1)
Q: Reverse the order of integration and evaluate ∫₀² ∫_{x/2}¹ x/(y³+1) dy dx.
A: Reversed: ∫₀¹ ∫₀^{2y} x/(y³+1) dx dy. Inner integral gives 2y²/(y³+1). Substituting u = y³+1 yields (2/3) ln 2.
Q: Evaluate ∬_R x/(x²+y²) dA where R is inside x²+y² = 4y, outside x²+y² = 8, first quadrant.
A: In polar, integrand simplifies to cos θ. Bounds: π/4 ≤ θ ≤ π/2, √8 ≤ r ≤ 4 sin θ. Answer: 3 − √8.
Double integrals in polar coordinates connect to surface area and surface integrals in Chapter 16. The technique of reversing integration order is a special case of Fubini's theorem, which also underpins triple integrals.
double integral, iterated integral, reversing order of integration, Fubini's theorem, polar coordinates double integral, r dr dtheta, Jacobian polar, circle in polar, MATH 126, Calculus III, UW Math 126