Calculations for the Jacobian:

• The direct Jacobian of Cartesian coordinates $(x, y)$ with respect to polar coordinates $(r, \theta)$ is:
  $$J = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\[6pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix}$$
  $$J = r\cos^2\theta - (-r\sin^2\theta) = r(\cos^2\theta + \sin^2\theta) = r$$
• By the reciprocal property of Jacobians:
  $$\frac{\partial(r, \theta)}{\partial(x, y)} = \frac{1}{\frac{\partial(x, y)}{\partial(r, \theta)}} = \frac{1}{r}$$

Integrate by parts using $\int u\, v'\, dx = u v - \int u'\, v\, dx$:
Let $u = \tan^{-1} x \implies u' = \frac{1}{1 + x^2}$
Let $v' = x \implies v = \frac{x^2}{2}$

Displacement is the definite integral of velocity over the time interval $[0, 3]$:

Evaluate the double integral over the rectangular region $R = [1, 2] \times [0, 3]$:

• Inner integral with respect to $x$:
  $$\int_1^2 (4xy - y^2)\, dx = \left[ 2x^2 y - y^2 x \right]_{x=1}^2$$
  $$= (2(4)y - 2y^2) - (2(1)y - y^2) = (8y - 2y^2) - (2y - y^2) = 6y - y^2$$
• Outer integral with respect to $y$:
  $$\int_0^3 (6y - y^2)\, dy = \left[ 3y^2 - \frac{y^3}{3} \right]_0^3 = 3(9) - \frac{27}{3} = 27 - 9 = 18$$