Let R be the region in the first quadrant that is bounded above by the polar curve r=4cosθ and below by the line θ=1, as shown in the figure above. What is the area of R ?

Answer:

3.99m

Explanation:

The area of R will be equal to 3.999390781

R has a surface area of about 5.74 square units.

The whole area that an object's surface covers is referred to as surface area. In other terms, the area of a three-dimensional object is the total of all of its faces or surfaces. Inches, metres, and other square measurements of surface area are typical.

Given,

The polar curve r = 4cos and the line = 1 together define the region R. We must integrate over the area in terms of polar coordinates to determine the area of R.

We must first determine the boundaries of integration. When cos = 0, which happens at the points = /2 and 3/2, the curve r = 4cos intersects the origin. The curve is where the line = 1 intersects it at = cos1(1/4). As a result, the limits of integration for are cos(1/4/2)/2.

Next, we must translate the curve's equation into terms of r and. We can write the expression r = 4cos as r = 4cos(-/2)/(-/2) = 4sin(-/2). As a result, the curve's polar coordinates equation is r = 4sin(-/2).

The area of R can be calculated as:

A = ∫∫R r dr dθ

where the limits of integration are as described above.

Now, we integrate first with respect to r, from 0 to 4sin(θ-π/2):

A = ∫π/2^arccos(1/4) ∫0^4sin(θ-π/2) r dr dθ

Integrating with respect to r gives:

A = ∫π/2^arccos(1/4) [1/2 r^2]0^4sin(θ-π/2) dθ

Simplifying:

A is equal to 1/2 arccos(1/4) 16sin(-/2) d.

Using the formula sin2(-2/2) = -1/2 cos(2-), we get the following:

A = 1/2 arccos(1/4), (8 - 8cos(2-)), and d

Calculating the integral results in:

A = [4θ - 2sin(2θ-π)]π/2^arccos(1/4)

The restrictions of integration and simplification are replaced to produce:

A is equal to (5/6 - cos(1/4)) sqrt(15)/2= 5.74 sq. units

Therefore, R has a surface area of about 5.74 square units.

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All of these align with the idea of sustainable development EXCEPT

limited use of chemicals
universal education
widely available health care
preservation of resources
use of cheap labor