When evaluating ∬_R f(x,y) dA using the substitution x = g(u,v), y = h(u,v), what correctly replaces the area element dA?
Adu dv
BJ du dv, where J is the signed Jacobian determinant
C|J| du dv, where |J| is the absolute value of the Jacobian determinant
D(1/J) du dv
The area element transforms as dA = |J| du dv. The absolute value is required because area elements must be positive regardless of whether the transformation preserves or reverses orientation. Using the signed J (option B) would give a negative result for orientation-reversing transformations. Using 1/J (option D) corresponds to the inverse transformation.
Question 2 Multiple Choice
A student correctly sets up a double integral in polar coordinates but forgets to include the Jacobian factor. Her computed answer will be:
ACorrect, because the polar transformation has a Jacobian of 1
BOff by a constant factor equal to the area of the integration region
CWrong because she is effectively computing ∬ f(r,θ) dr dθ instead of ∬ f(r,θ) r dr dθ
DCorrect, because the polar transformation is its own inverse and the errors cancel
The Jacobian for polar coordinates is r (not 1). Forgetting it means integrating without the r factor — the error is not a constant but varies across the region, since r takes different values at different points. The r factor is not a formula to memorize but the Jacobian of the transformation x = r cosθ, y = r sinθ; this is why it appears and why it cannot simply be dropped.
Question 3 True / False
The r that appears in the polar area element r dr dθ is exactly the Jacobian determinant of the polar coordinate transformation.
TTrue
FFalse
Answer: True
True. With x = r cosθ, y = r sinθ, the Jacobian matrix is [[cosθ, −r sinθ], [sinθ, r cosθ]], and its determinant is r cos²θ + r sin²θ = r. The r in r dr dθ is not an ad hoc correction — it is the Jacobian. This unifies a formula that might otherwise seem arbitrary with the general theory.
Question 4 True / False
Any valid substitution x = g(u,v), y = h(u,v) will simplify a double integral, as long as the Jacobian is correctly computed.
TTrue
FFalse
Answer: False
False. A poor substitution can make an integral harder, not easier. The Jacobian correctly accounts for area scaling regardless of the substitution chosen, but if the new region in uv-space is more complicated than R, or if the transformed integrand is messier, the substitution has made the problem worse. The skill is choosing a substitution aligned with the symmetry of the region and integrand — one that turns R into a simpler shape, often a rectangle.
Question 5 Short Answer
Explain why the Jacobian determinant appears in the change-of-variables formula for double integrals. What does it measure, and how does this connect to single-variable substitution?
Think about your answer, then reveal below.
Model answer: In single-variable substitution, the factor |g'(u)| corrects for how much the substitution stretches or compresses the x-axis: a small interval du in u-space corresponds to a length |g'(u)| du in x-space. The Jacobian is the 2D generalization: it measures how much the transformation locally stretches or compresses area. A small rectangle du × dv in uv-space maps to a parallelogram in xy-space with area |J| du dv, so the integral must include |J| to correctly account for that area change.
The geometric interpretation — Jacobian = local area scaling factor — is the key insight. The determinant connection comes from the fact that the columns of the Jacobian matrix are the images of the unit vectors under the linear approximation to the transformation, and the determinant measures the area of the parallelogram they span.