1. A region $$ R $$, in the Cartesian plane is bounded by $$ x^{2}+y^{2}=a^{2}, x^{2}+y^{2}=b^{2}, x \geq 0 $$, $$ y \geq 0 $$, where $$ 00 $$ and $$ 0 \leq heta

Question

1. A region $$ R $$, in the Cartesian plane is bounded by $$ x^{2}+y^{2}=a^{2}, x^{2}+y^{2}=b^{2}, x \geq 0 $$, $$ y \geq 0 $$, where $$ 00 $$ and $$ 0 \leq 	heta<2 \pi $$. Sketch the regions $$ R $$ and $$ R^{\prime} $$ on two different $$ \mathbb{R}^{2} $$ planes. (b) Compute the Jacobian matrix $$ \frac{\partial(x, y)}{\partial(r, 	heta)} $$. (c) Compute the determinant of the Jacobian matrix $$ \left|\frac{\partial(x, y)}{\partial(r, 	heta)}ight| $$. What does the value of this determinant mean with respect to the transformation? (d) What happens if $$ a=0 $$ ?

UpStudy AI · Accepted Answer

Let's solve the problem step by step.

### (a) Determine the image $$ R^{\prime} $$

The region $$ R $$ is bounded by the circles defined by $$ x^{2}+y^{2}=a^{2} $$ and $$ x^{2}+y^{2}=b^{2} $$ in the first quadrant (since $$ x \geq 0 $$ and $$ y \geq 0 $$).

Under the transformation $$ x = r \cos \theta $$ and $$ y = r \sin \theta $$, the equations of the circles become:

1. $$ r^2 \cos^2 \theta + r^2 \sin^2 \theta = a^2 $$ simplifies to $$ r^2 = a^2 $$ or $$ r = a $$.
2. $$ r^2 \cos^2 \theta + r^2 \sin^2 \theta = b^2 $$ simplifies to $$ r^2 = b^2 $$ or $$ r = b $$.

Thus, the image $$ R^{\prime} $$ in polar coordinates is bounded by $$ r = a $$ and $$ r = b $$ for $$ 0 \leq \theta < \frac{\pi}{2} $$ (the first quadrant).

### Sketch of the Regions

- **Region $$ R $$**: This is the area between two circles of radius $$ a $$ and $$ b $$ in the first quadrant.
- **Region $$ R^{\prime} $$**: This is the area between two radial lines at angles $$ 0 $$ and $$ \frac{\pi}{2} $$ from the origin, extending from radius $$ a $$ to radius $$ b $$.

### (b) Compute the Jacobian matrix $$ \frac{\partial(x, y)}{\partial(r, \theta)} $$

The transformation equations are:
- $$ x = r \cos \theta $$
- $$ y = r \sin \theta $$

The Jacobian matrix is given by:

$$
J = \begin{bmatrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}
\end{bmatrix}
$$

Calculating the partial derivatives:

1. $$ \frac{\partial x}{\partial r} = \cos \theta $$
2. $$ \frac{\partial x}{\partial \theta} = -r \sin \theta $$
3. $$ \frac{\partial y}{\partial r} = \sin \theta $$
4. $$ \frac{\partial y}{\partial \theta} = r \cos \theta $$

Thus, the Jacobian matrix is:

$$
J = \begin{bmatrix}
\cos \theta & -r \sin \theta \
\sin \theta & r \cos \theta
\end{bmatrix}
$$

### (c) Compute the determinant of the Jacobian matrix

The determinant of the Jacobian matrix $$ J $$ is calculated as follows:

$$
\text{det}(J) = \left(\cos \theta \cdot r \cos \theta\right) - \left(-r \sin \theta \cdot \sin \theta\right)
$$

This simplifies to:

$$
\text{det}(J) = r \cos^2 \theta + r \sin^2 \theta = r(\cos^2 \theta + \sin^2 \theta) = r
$$

The value of this determinant $$ |J| = r $$ indicates the scaling factor of area when transforming from Cartesian coordinates to polar coordinates. It shows that the area element $$ dx \, dy $$ transforms to $$ r \, dr \, d\theta $$.

### (d) What happens if $$ a=0 $$?

If $$ a = 0 $$, the inner circle collapses to a point at the origin. The region $$ R $$ would then be bounded only by the outer circle $$ x^2 + y^2 = b^2 $$ in the first quadrant. The image $$ R^{\prime} $$ would still be the area between the radial lines at angles $$ 0 $$ and $$ \frac{\pi}{2} $$ extending from the origin to radius $$ b $$.

In this case, the region $$ R $$ would no longer have an inner boundary, effectively becoming a sector of a circle.
(a) The image $$ R^{\prime} $$ is the area between $$ r = a $$ and $$ r = b $$ in the first quadrant. (b) The Jacobian matrix is: $$ J = \begin{bmatrix} \cos \theta & -r \sin \theta \ \sin \theta & r \cos \theta \end{bmatrix} $$ (c) The determinant of the Jacobian is $$ |J| = r $$. This means the area scales by a factor of $$ r $$ during the transformation. (d) If $$ a = 0 $$, the region $$ R $$ becomes a sector bounded by $$ r = 0 $$ and $$ r = b $$ in the first quadrant.

Upstudy AI Solution

Answer

Solution

(a) Determine the image

Sketch of the Regions

(b) Compute the Jacobian matrix

© Compute the determinant of the Jacobian matrix

(d) What happens if ?

Extra Insights

(b) The Jacobian matrix of the transformation is computed as follows:
[
\frac{\partial(x, y)}{\partial(r, \theta)} =

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