Difficulty: Advanced | Prerequisites: Single-variable integration, u-substitution, polar coordinates basics, geometric series, and familiarity with the basic Taylor series for 1/(1−x), eˣ, sin x, and cos x.
This covers the second half of MATH 126: double integrals (including reversing the order of integration), applications of integration in polar coordinates (mass, centre of mass), Taylor polynomials with error bounds, and Taylor series manipulation. These topics are heavily weighted on the final exam and require both computational skill and conceptual understanding. If you are behind, make sure you are comfortable with single-variable integration techniques and the basic series formulas before diving in.
Double integrals compute accumulated quantities (area, volume, mass) over 2D regions; when the integrand is hard to evaluate in one order, reverse the order of integration by re-describing the region. Polar coordinates simplify integrals over circular regions by replacing dA with r dr dθ. Taylor polynomials approximate functions near a base point, with error bounds given by the Taylor remainder theorem. Taylor series are built by substituting into known basic series, and their interval of convergence comes from the underlying geometric series condition.
Double integral
An integral over a 2D region, written ∬_D f(x, y) dA. Think of it as adding up tiny contributions f(x, y) · dA over every small patch of the region D.
Reversing the order of integration
Swapping the dx dy integral to dy dx (or vice versa) by re-describing the region's bounds. In simple terms, you are slicing the region in the other direction to make the integral easier to evaluate.
Polar coordinates
A coordinate system where each point is described by distance from the origin (r) and angle from the positive x-axis (θ). x = r cos θ, y = r sin θ, and the area element dA = r dr dθ.
Taylor polynomial T_n(x)
The degree-n polynomial that best approximates a function f(x) near a base point b, built from f and its derivatives at b. Think of it as a polynomial that matches the function's value, slope, curvature, and higher behaviour at that single point.
Taylor remainder (error bound)
The maximum possible error |f(x) − T_n(x)| on an interval, bounded by (M / (n+1)!) |x − b|^(n+1), where M is the maximum of |f^(n+1)| on the interval.
Taylor series
The infinite series version of a Taylor polynomial. For many functions, the Taylor series converges to the original function within some interval (the interval of convergence).
Interval of convergence
The set of x values for which a Taylor (or power) series converges. Found by applying the ratio test or by tracking the convergence condition through substitutions.
If the inner integral cannot be evaluated in closed form (e.g. ∫ x²/(1 + x⁵) dx has no elementary antiderivative), try reversing the order.
Sketch the region first. Identify the curves that bound it and determine the new limits.
Draw the region described by the original limits.
Re-describe the region with the other variable as the outer integral.
Example: ∫₁⁵ ∫_{√(y−1)}^{2} x²/(1 + x⁵) dx dy. The region is: 1 ≤ y ≤ 5, √(y − 1) ≤ x ≤ 2. Sketch this: x ranges from 0 to 2, and for each x, y ranges from 1 to x² + 1. Reversed: ∫₀² ∫₁^{x²+1} x²/(1 + x⁵) dy dx.
The inner integral in y is now trivial (just y), leaving ∫₀² x⁴/(1 + x⁵) dx, which is a standard u-substitution (u = 1 + x⁵).
Answer: (1/5) ln(33).
For the volume enclosed by surfaces, identify the height function (top minus bottom) and the region of integration in the xy-plane.
Example: Volume under z = 4 − y, above z = 0, inside the cylinder y = x². The region in xy is −2 ≤ x ≤ 2, x² ≤ y ≤ 4. Height = 4 − y. Volume = ∫{−2}^{2} ∫{x²}^{4} (4 − y) dy dx.
Replace x = r cos θ, y = r sin θ, and dA = r dr dθ.
The extra factor of r in dA is essential and commonly forgotten.
Mass = ∬_D ρ(x, y) dA, where ρ is the density function.
Express ρ in polar coordinates: for example, ρ(x, y) = x/√(x² + y²) becomes r cos θ / r = cos θ.
Example: Region D is in the first quadrant, inside x² + y² = 4, outside x² + (y − 2)² = 4. The outer circle is r = 2. The inner circle x² + (y − 2)² = 4 becomes r = 4 sin θ in polar. They intersect when 4 sin θ = 2, giving θ = π/6.
Mass = ∫₀^{π/6} ∫_{4 sin θ}^{2} cos θ · r dr dθ. Evaluating gives Mass = 2/3.
The y-coordinate of the centre of mass is ȳ = M_x / m, where M_x = ∬_D y · ρ(x, y) dA.
In polar: y = r sin θ, so the integrand becomes r sin θ · cos θ · r = r² sin θ cos θ.
From the example: M_x = ∫₀^{π/6} ∫_{4 sin θ}^{2} r² sin θ cos θ dr dθ. After u-substitution (u = sin θ), M_x = 1/5. So ȳ = (1/5)/(2/3) = 3/10.
The nth-degree Taylor polynomial for f(x) centred at b is: T_n(x) = f(b) + f′(b)(x − b) + f″(b)(x − b)²/2! + ⋯ + f⁽ⁿ⁾(b)(x − b)ⁿ/n!.
Compute f, f′, f″, ... up to the nth derivative, all evaluated at b.
Example: f(x) = sin(x − 1) + x² + eˣ, b = 1. f(1) = 1 + e, f′(1) = 3 + e, f″(1) = 2 + e. So T₂(x) = (1 + e) + (3 + e)(x − 1) + (1/2)(2 + e)(x − 1)².
|f(x) − T_n(x)| ≤ (M / (n + 1)!) |x − b|^(n+1), where M = max of |f⁽ⁿ⁺¹⁾(c)| on the interval.
To find M, you need to bound the (n+1)th derivative on the given interval.
Example: For f(x) = sin(x − 1) + x² + eˣ and T₂ on [1/2, 3/2], the third derivative is f‴(x) = −cos(x − 1) + eˣ. On [1/2, 3/2]: |cos(x − 1)| ≤ 1 and eˣ ≤ e^(3/2). So M = 1 + e^(3/2).
Error ≤ (1/(3!)) · (1 + e^(3/2)) · |x − 1|³ ≤ (1/6)(1 + e^(3/2))(1/2)³ = (1/48)(1 + e^(3/2)).
Start from the basic series: 1/(1 − x) = Σ xᵏ, arctan(x) = Σ (−1)ᵏ x^(2k+1)/(2k+1), etc.
Substitute to build new series. To find f(x) = x² arctan(−3x):
Start with arctan(x) = Σ_{k=0}^∞ (−1)ᵏ x^(2k+1)/(2k+1).
Replace x with −3x: arctan(−3x) = Σ (−1)ᵏ (−3x)^(2k+1)/(2k+1) = Σ (−1)^(k+1) 3^(2k+1) x^(2k+1)/(2k+1).
Multiply by x²: f(x) = Σ (−1)^(k+1) 3^(2k+1) x^(2k+3)/(2k+1).
The base series 1/(1 − x) converges for |x| < 1.
After substituting x → −x², convergence requires |−x²| < 1, so |x| < 1.
After substituting x → −3x, convergence requires |−3x| < 1, so |x| < 1/3.
Interval of convergence: (−1/3, 1/3).
The coefficient of xⁿ in a Taylor series based at 0 equals f⁽ⁿ⁾(0)/n!.
To find f⁽²⁰²³⁾(0), locate the term with x²⁰²³ in the series.
In the example, x^(2k+3) = x²⁰²³ when 2k + 3 = 2023, so k = 1010.
The coefficient is (−1)^(1011) · 3^(2021) / 2021 = −3²⁰²¹/2021.
So f⁽²⁰²³⁾(0)/2023! = −3²⁰²¹/2021, giving f⁽²⁰²³⁾(0) = −3²⁰²¹ · 2023!/2021.
Basic Taylor series (memorise these)
1/(1 − x) = Σ_{k=0}^∞ xᵏ, |x| < 1
eˣ = Σ_{k=0}^∞ xᵏ/k!
sin x = Σ_{k=0}^∞ (−1)ᵏ x^(2k+1)/(2k+1)!
cos x = Σ_{k=0}^∞ (−1)ᵏ x^(2k)/(2k)!
arctan x = Σ_{k=0}^∞ (−1)ᵏ x^(2k+1)/(2k+1), |x| ≤ 1
Taylor polynomial
T_n(x) = Σ_{k=0}^n f⁽ᵏ⁾(b)(x − b)ᵏ / k!
Taylor remainder bound
|f(x) − T_n(x)| ≤ M|x − b|^(n+1) / (n+1)!
Polar area element
dA = r dr dθ
Mass and centre of mass
m = ∬ ρ dA, ȳ = (1/m) ∬ y ρ dA, x̄ = (1/m) ∬ x ρ dA
Reversing the order of integration is used in probability theory when computing expected values over joint distributions. Polar integrals for mass and centre of mass are directly applicable in mechanical engineering for computing moments of inertia of circular components. Taylor series are the foundation of numerical methods: calculators and computers evaluate sin, cos, and exp by summing a few terms of their Taylor series.
Students frequently forget the extra r factor when converting to polar coordinates. The area element is r dr dθ, not just dr dθ.
When reversing the order of integration, students sometimes swap the limits without re-deriving them from the region. Always sketch the region and find the new bounds from the sketch.
For Taylor error bounds, students sometimes use |f⁽ⁿ⁺¹⁾(b)| instead of the maximum of |f⁽ⁿ⁺¹⁾| on the entire interval. You need the worst-case value, not the value at the centre.
When building a Taylor series by substitution, students often lose track of signs. Be especially careful with (−3x)^(2k+1): the odd power preserves the negative sign.
⚠️ Reversing the order of integration is a classic exam question. If the inner integral looks impossible, that is your cue to reverse.
⚠️ Polar coordinate problems almost always involve the r = 4 sin θ or r = 2 cos θ families of circles. Know how to convert these and find their intersection angles.
⚠️ For Taylor error bounds, the exam will specify the interval. You must show how you found M (the bound on the next derivative) to receive full credit.
⚠️ The "find f⁽ⁿ⁾(0)" question is a favourite. The trick is always: write the series, identify the xⁿ term, and equate its coefficient to f⁽ⁿ⁾(0)/n!.
Fill in the blank: When converting to polar coordinates, dA = ______ dr dθ. (r)
True or False: The Taylor series for eˣ converges for all x. (True)
Fill in the blank: The interval of convergence for the geometric series Σ xᵏ is (______, ______). (−1, 1)
True or False: To find f⁽ⁿ⁾(0), you multiply the coefficient of xⁿ in the Taylor series by n!. (True)
True or False: When reversing the order of integration, the integrand f(x, y) changes. (False, only the limits change)
Q: Compute ∫₁⁵ ∫_{√(y−1)}^{2} x²/(1 + x⁵) dx dy by reversing the order of integration.
A: Sketch the region: 0 ≤ x ≤ 2, 1 ≤ y ≤ x² + 1. Reversed: ∫₀² ∫₁^{x²+1} x²/(1 + x⁵) dy dx = ∫₀² x⁴/(1 + x⁵) dx. Let u = 1 + x⁵. Answer: (1/5) ln(33).
Q: Set up (do not evaluate) the integral for the volume enclosed by y = x², y + z = 4, and z = 0.
A: Region: −2 ≤ x ≤ 2, x² ≤ y ≤ 4. Height: 4 − y. Volume = ∫{−2}^{2} ∫{x²}^{4} (4 − y) dy dx.
Q: Find the mass of the lamina in the first quadrant inside x² + y² = 4, outside x² + (y−2)² = 4, with density ρ = x/√(x² + y²).
A: In polar: ρ = cos θ, r ranges from 4 sin θ to 2, θ from 0 to π/6. Mass = ∫₀^{π/6} ∫_{4 sin θ}^{2} cos θ · r dr dθ = 2/3.
Q: Find T₂(x) for f(x) = sin(x − 1) + x² + eˣ at b = 1.
A: f(1) = 1 + e, f′(1) = 3 + e, f″(1) = 2 + e. T₂(x) = (1 + e) + (3 + e)(x − 1) + (1/2)(2 + e)(x − 1)².
Q: Find the Taylor series for x² arctan(−3x) centred at 0 and its interval of convergence.
A: Σ_{k=0}^∞ (−1)^(k+1) 3^(2k+1) x^(2k+3) / (2k+1). Converges on (−1/3, 1/3).
Q: Find f⁽²⁰²³⁾(0) for f(x) = x² arctan(−3x).
A: The x²⁰²³ term has k = 1010. Coefficient = −3²⁰²¹/2021. So f⁽²⁰²³⁾(0) = −3²⁰²¹ · 2023! / 2021.
Double integrals extend to triple integrals in MATH 324, where you add a third dimension and work in cylindrical or spherical coordinates. Taylor series connect to Fourier series, which decompose periodic functions into sines and cosines. The error-bound techniques here are the foundation of numerical analysis, where you need to know how much accuracy you lose by truncating an infinite process.
Double integral, iterated integral, reversing order of integration, Fubini's theorem, polar coordinates, r dr dθ, mass, centre of mass, density function, moment, Taylor polynomial, Taylor series, Maclaurin series, Taylor remainder theorem, error bound, interval of convergence, ratio test, power series, substitution in series, f⁽ⁿ⁾(0), higher derivatives from series, arctan series, geometric series, MATH 126, Calculus III, UW