A student sets up the integral of f over a disk of radius 3 as ∫₀^{2π} ∫₀^3 f(r cosθ, r sinθ) dr dθ. What error have they made?
AThe limits for θ should be 0 to π, not 0 to 2π
BThey should not substitute x = r cosθ and y = r sinθ into the integrand
CThey omitted the Jacobian factor r, writing dr dθ instead of r dr dθ
DThe r integral should be on the outside and θ on the inside
The area element in polar coordinates is r dr dθ, not dr dθ. The extra factor of r is the Jacobian of the polar transformation, accounting for the fact that polar 'cells' grow in size as r increases. Without this factor, the integral treats all radii equally, systematically undercounting regions far from the origin. This is the single most common error in polar integration.
Question 2 Multiple Choice
Why does the integral ∬_{ℝ²} e^{−(x²+y²)} dA become tractable in polar coordinates when it resists direct computation in Cartesian coordinates?
AIn Cartesian form, e^{−x²} and e^{−y²} cannot be separated into a product of single-variable integrals
BThe region ℝ² cannot be expressed with finite bounds in Cartesian coordinates
CIn polar form, x²+y² = r², giving e^{−r²}, and the area element r dr dθ allows the integral to factor into a tractable product of separate integrals in r and θ
DPolar coordinates introduce an algebraic simplification specific to exponential functions
In Cartesian form, ∫∫ e^{−x²−y²} dx dy = (∫e^{−x²}dx)(∫e^{−y²}dy), which requires evaluating ∫e^{−x²}dx — an integral with no elementary closed form when taken alone. In polar form, x²+y² = r² and dA = r dr dθ, giving ∫₀^{2π}dθ · ∫₀^∞ e^{−r²} r dr. The inner integral ∫r e^{−r²}dr has elementary antiderivative −½e^{−r²}, so the full computation yields π. The key is that the Jacobian factor r makes the r-integral tractable.
Question 3 True / False
The area element in polar coordinates is dr dθ, analogous to dx dy in Cartesian coordinates.
TTrue
FFalse
Answer: False
The polar area element is r dr dθ, not dr dθ. This is because a 'polar rectangle' between r and r+dr, and between θ and θ+dθ, is a curved wedge whose arc length along its outer edge is r·dθ, giving area ≈ r dr dθ. The factor r is the Jacobian of the transformation from Cartesian to polar coordinates. Near the origin (small r), polar cells are tiny; far away (large r), they are large. Without the r factor, all radial positions are weighted equally, producing a systematically wrong answer.
Question 4 True / False
Polar coordinates are most useful for double integrals when the region of integration involves x² + y² or has circular symmetry.
TTrue
FFalse
Answer: True
The core benefit of polar coordinates is the conversion x²+y² → r², which simplifies both circular regions (disks, annuli, sectors) and integrands containing x²+y² into clean expressions in r. A disk of radius R becomes simply 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π — a rectangle in (r,θ) space — whereas in Cartesian coordinates it requires messy square-root bounds. The Gaussian integral e^{−(x²+y²)} is the canonical example of an integrand that becomes elementary after the substitution.
Question 5 Short Answer
Why does the polar area element include an extra factor of r, and what goes wrong if you forget it?
Think about your answer, then reveal below.
Model answer: A polar 'cell' between r and r+dr, and between θ and θ+dθ, is a wedge shape. Its width in the angular direction is the arc length r·dθ (not just dθ), so its area is approximately r dr dθ. This factor r is the Jacobian of the coordinate transformation: it corrects for the fact that equal increments of (r, θ) do not correspond to equal areas — cells near the origin are tiny while cells far from the origin are large. Forgetting r means treating all radial positions as equally weighted, which undercounts the contribution from large r and gives an incorrect integral value.
The Jacobian appears whenever you change variables in a multiple integral: the area element transforms as dA = |∂(x,y)/∂(r,θ)| dr dθ, and computing this determinant yields exactly r. This is not a coincidence or a convention to memorize — it is the precise correction factor that makes the area calculation accurate under the coordinate change.