Q. 31 - Calculus

Question

In Exercises 27–32, functions x = x(u, v) and y = y(u, v) are given that determine transformations from an XY-coordinate system to a UV-coordinate system in R2. Use these functions to determine a region in the XY-plane that has the image specified for the given values of u and v, and find the Jacobian of the transformation.

$x = u \sin v and y = u \cos v for 0 \le u \le 2 and 0 \le v \le \pi$

Step-by-Step Solution

Answer

The Jacobian is equal to $J=-u$.

Step 1: Given information

The functions are,

$x = u \sin v and y = u \cos v for 0 \le u \le 2 and 0 \le v \le \pi$

Step 2: Find the Jacobian

The Jacobian is computed as,

$$\begin{gathered} \frac{\partial (x, y)}{\partial (u,v)}=det\left[\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v}\end{array}\right] \\ \frac{\partial (x, y)}{\partial (u,v)}=det\left[\begin{array}{cc}\sin v& cos v\\ u\cos v& -u \sin v\end{array}\right] \\ \frac{\partial (x, y)}{\partial (u,v)}=-u{\sin }^2 v-u {\cos }^{2}v \\ \frac{\partial (x, y)}{\partial (u,v)}=-u\left({\sin }^2 v+ {\cos }^{2}v\right) \\ \frac{\partial (x, y)}{\partial (u,v)}=-u\times 1 \\ \frac{\partial (x, y)}{\partial (u,v)}=-u \end{gathered}$$