Consider the polar to cartesian coordinate transformation
𝑥
=
𝑟
cos
⁡
𝜃
and
𝑦
=
𝑟
sin
⁡
𝜃
the Jacobian for this would be

1)
𝑟

2)
𝑟
2

3)
𝑟
2
sin
⁡
𝜃

4)
𝑟
sin
⁡
𝜃
cos
⁡
𝜃

Question

Consider the polar to cartesian coordinate transformation 
𝑥
=
𝑟
cos
⁡
𝜃
 and 
𝑦
=
𝑟
sin
⁡
𝜃
 the Jacobian for this would be

1) 
𝑟

2) 
𝑟
2

3) 
𝑟
2
sin
⁡
𝜃

4) 
𝑟
sin
⁡
𝜃
cos
⁡
𝜃

BlackTom AI · Accepted Answer

For the transformation x = r cos θ, y = r sin θ the Jacobian determinant equals r. This follows from computing the 2×2 matrix of partial derivatives and taking its determinant.

Relevant knowledge points:
- The Jacobian for a change of variables (r, θ) → (x, y) is det(∂(x,y)/∂(r,θ)).
- For polar coordinates the area element satisfies dA = |J| dr dθ, and standard result is dA = r dr dθ.

Step-by-step computation:
1. Form the Jacobian matrix of partial derivatives:
   ∂x/∂r = cos θ,     ∂x/∂θ = -r sin θ
   ∂y/∂r = sin θ,     ∂y/∂θ =  r cos θ
   So J = [ [cos θ, -r sin θ], [sin θ, r cos θ] ].

2. Compute the determinant:
   det J = (cos θ)(r cos θ) - (-r sin θ)(sin θ)
         = r cos^2 θ + r sin^2 θ
         = r (cos^2 θ + sin^2 θ).

3. Use the Pythagorean identity cos^2 θ + sin^2 θ = 1:
   det J = r.

Why this is correct:
- The algebra shows the determinant simplifies exactly to r, so the Jacobian factor converting dr dθ to dx dy is r. Geometrically, a small rectangle in (r,θ) of sides dr and dθ maps to an area ≈ r dr dθ because the arc length for angle dθ at radius r is r dθ.

Why other options are incorrect:
- Option 2: r^2 — no factor of r^2 arises; determinant calculation yields a single factor r, not r^2.
- Option 3: r^2 sin θ — introduces an extra r and a sin θ factor not present after simplification; determinant had cos^2+sin^2, which equals 1, not sin θ.
- Option 4: r sin θ cos θ — would result from incorrectly multiplying off-diagonal terms without combining correctly; algebraic simplification gives r, not r sin θ cos θ.

Conclusion: The correct Jacobian is r, matching the provided answer (choice 1).

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