# Finding area enclosed by the polar curve

• Coderhk
In summary, the attempted solution to the homework equation using the formula given yields incorrect results.

## Homework Statement

Question attached in attachments

## Homework Equations

Area enclosed by polar graph is ∫0.5r^2
where r is the radius as a function of angle theta

## The Attempt at a Solution

I attempted to use the formula above and I subtracted the area of the inside from the outside but it seems to yield an incorrect solution.

##0.5∫((2cos(3x))^2)dx-0.5∫4dx##
The limit of integration are from ##0## to ##2π##

#### Attachments

• a.png
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integrating from 0 to ##2\pi## without adjustment is no good. There are at least two problems with that:

1. The three-leaved flower shape is traversed twice as the integration variable travels from 0 to ##2\pi##. So there is double-counting going on.

2. ##r## is negative for half the time, which will give negative integrals.

Instead I suggest, to get the area of the three-leaved shape, you instead integrate over half a leaf, for instance from 0 to ##\pi/6##. If you do that then ##r## is never negative over that range, and every part of the area is covered only once. Then multiply the area you get by six to get the total area of all three identical leaves. Then take that away from the area of the circle.

scottdave and Coderhk
I don't think either of points 1 and 2 of #2 are correct. r is squared so there is no negative. And from 0 to 2π is going round the circle once isn't it?
True you don't have to integrate around the whole circle, you could integrate around a third of it with any lower limit you like, or in the upper limit you like as long as the difference between them is 2π/3 - then multiply by 3.

For your answer you give no indications of what you got wrong. Area enclosed by polar graph is ∫0.5r^2 . For the shaded area you have got the signs wrong. For the area of the whole disk, I think you could be allowed to use a well-known formula! Then compare with your calculation to see you got everything right.

Coderhk
epenguin said:
I don't think either of points 1 and 2 of #2 are correct. r is squared so there is no negative. And from 0 to 2π is going round the circle once isn't it?
True you don't have to integrate around the whole circle, you could integrate around a third of it with any lower limit you like, or in the upper limit you like as long as the difference between them is 2π/3 - then multiply by 3.

For your answer you give no indications of what you got wrong. Area enclosed by polar graph is ∫0.5r^2 For the shaded area you have got the signs wrong. For the area of the whole disk, I think you could be allowed to use a well-known formula! Then compare with your calculation to see you got everything right.
I tried integrating from 0 to #pi/6# and then multiplied it by 6. Afterwards I got the area using pi*R*R and subtracted the white space using the result I got in the first step. I solved it already but thank you very much

## 1. How do you find the area enclosed by a polar curve?

To find the area enclosed by a polar curve, you can use the formula A = 1/2 ∫ab r² dθ, where r is the radius of the curve and θ is the angle of rotation.

## 2. Can you explain the concept of polar coordinates and curves?

Polar coordinates are a way of representing points in a two-dimensional plane using a distance from the origin (r) and an angle of rotation (θ). A polar curve is a graph of points that follow a specific pattern, determined by the equation r = f(θ). This allows for a different way of representing and analyzing curves compared to Cartesian coordinates.

## 3. What is the difference between finding the area enclosed by a polar curve and a Cartesian curve?

The main difference is that polar curves are defined in terms of r and θ, while Cartesian curves are defined in terms of x and y. This means that the formulas for finding the area enclosed by each type of curve are different. Additionally, the way the curves are graphed and analyzed also differ due to the different coordinate systems.

## 4. Can you provide an example of finding the area enclosed by a polar curve?

Sure, let's say we have the polar curve r = 2sin(θ) and we want to find the area enclosed by one loop of the curve. First, we need to determine the limits of integration, which in this case would be 0 and π (since one loop of the curve covers a full rotation of 2π). Then, we plug in the values into the formula A = 1/2 ∫0π (2sin(θ))² dθ. After solving the integral, we get the area to be 2π square units.

## 5. Are there any special cases or restrictions when finding the area enclosed by a polar curve?

Yes, there are a few things to keep in mind. First, the curve must be a closed loop, meaning it starts and ends at the same point. Additionally, the curve must not intersect itself. If it does, you will need to split the integral into multiple parts to find the total area. Lastly, if the curve crosses the origin, the area will need to be calculated in separate parts for each section of the loop.