The given double integral is in Cartesian coordinates (x, y). To change the coordinate system to polar coordinates (r, θ), we need to find the new limits of integration.

The given limits for x are from 4 to 2, and for y are from x to x^2.

In polar coordinates, x = rcos(θ) and y = rsin(θ).

Substituting x = 4, we get rcos(θ) = 4. Solving for r, we get r = 4sec(θ).

Substituting x = 2, we get rcos(θ) = 2. Solving for r, we get r = 2sec(θ).

For y = x, we get rsin(θ) = rcos(θ), which simplifies to tan(θ) = 1. Solving for θ, we get θ = π/4.

For y = x^2, we get rsin(θ) = (rcos(θ))^2. This is a more complicated equation, but it simplifies to tan(θ) = cos(θ), which has solutions at θ = 0 and θ = π/2.

So, the new limits of integration in polar coordinates are r: 2sec(θ) to 4sec(θ), and θ: 0 to π/4.

Question

The given double integral is in Cartesian coordinates (x, y). To change the coordinate system to polar coordinates (r, θ), we need to find the new limits of integration.

The given limits for x are from 4 to 2, and for y are from x to x^2.

In polar coordinates, x = rcos(θ) and y = rsin(θ).

Substituting x = 4, we get rcos(θ) = 4. Solving for r, we get r = 4sec(θ).

Substituting x = 2, we get rcos(θ) = 2. Solving for r, we get r = 2sec(θ).

For y = x, we get rsin(θ) = rcos(θ), which simplifies to tan(θ) = 1. Solving for θ, we get θ = π/4.

For y = x^2, we get rsin(θ) = (rcos(θ))^2. This is a more complicated equation, but it simplifies to tan(θ) = cos(θ), which has solutions at θ = 0 and θ = π/2.

So, the new limits of integration in polar coordinates are r: 2sec(θ) to 4sec(θ), and θ: 0 to π/4.

Knowee AI · Accepted Answer

The given double integral is in Cartesian coordinates (x, y). To change the coordinate system to polar coordinates (r, θ), we need to find the new limits of integration.

The given limits for x are from 4 to 2, and for y are from x to x^2.

In polar coordinates, x = rcos(θ) and y = rsin(θ).

Substituting x = 4, we get rcos(θ) = 4. Solving for r, we get r = 4sec(θ).

Substituting x = 2, we get rcos(θ) = 2. Solving for r, we get r = 2sec(θ).

For y = x, we get rsin(θ) = rcos(θ), which simplifies to tan(θ) = 1. Solving for θ, we get θ = π/4.

For y = x^2, we get rsin(θ) = (rcos(θ))^2. This is a more complicated equation, but it simplifies to tan(θ) = cos(θ), which has solutions at θ = 0 and θ = π/2.

So, the new limits of integration in polar coordinates are r: 2sec(θ) to 4sec(θ), and θ: 0 to π/4.

Question

Solution

Similar Questions

Upgrade your grade with Knowee