∫Calc Practice

Jacobians and change of variables

Problem 11.210 · easy

Find the Jacobian \( \displaystyle \dfrac{\partial(x, y)}{\partial(u, v)} \) of the transformation \( \displaystyle x = e^{u} \cos{\left(v \right)} \), \( \displaystyle y = e^{u} \sin{\left(v \right)} \).
  1. \[ \left[\begin{matrix}\frac{\partial}{\partial u} e^{u} \cos{\left(v \right)} & \frac{\partial}{\partial v} e^{u} \cos{\left(v \right)}\\\frac{\partial}{\partial u} e^{u} \sin{\left(v \right)} & \frac{\partial}{\partial v} e^{u} \sin{\left(v \right)}\end{matrix}\right] = \left[\begin{matrix}e^{u} \cos{\left(v \right)} & - e^{u} \sin{\left(v \right)}\\e^{u} \sin{\left(v \right)} & e^{u} \cos{\left(v \right)}\end{matrix}\right] \]
    The matrix of partial derivatives.✓ Proved
  2. \[ e^{2 u} \sin^{2}{\left(v \right)} + e^{2 u} \cos^{2}{\left(v \right)} = e^{2 u} \]
    Its determinant: x_u·y_v − x_v·y_u.✓ Proved
  3. |J| is the factor by which the transformation stretches small areas: dA = |J| du dv.
    Reviewed
Answer \( e^{2 u} \)

Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the image of a tiny square measured by the shoelace formula

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-10-04
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: fail (misleading) 2026-10-04 — The problem asks for the Jacobian determinant, but the solution concludes with a statement about the absolute value |J| and area scaling. While |J| = J in this specific case (since e^{2u} > 0), defining the Jacobian as the area scaling factor is conceptually incorrect; the Jacobian is the signed determinant, and the area scaling factor is its absolute value. This distinction is crucial for change of variables in integrals where orientation matters or when J could be negative.
  • gpt-oss:20b: pass 2026-10-04

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/jacobian, checked 2026-10-04 with SymPy 1.14.0.