# Find the area enclosed by the curve y = x csch(x+y)

• I
• Saracen Rue
In summary, the conversation discusses the difficulty in accurately determining the area enclosed by the curve ##y=x \text{ csch}(x+y)## and the ##x##-axis. The equation is represented as a nested function, making it challenging to find a definitive solution. The use of Desmos Graphing calculator was able to provide an estimate of the enclosed area, but it may not be entirely accurate. Alternative methods, such as using WolframAlpha's nest function, may be explored for a more precise estimation.
Saracen Rue
TL;DR Summary
Determine the area enclosed by the curve ##y=x \text{ csch}(x+y)## and the positive ##x##-axis
The title and summary pretty much say it all. I was wondering if it's possible to accurately determine the area enclosed by the curve ## y=x \text{ csch}(x+y)## and the ##x##-axis?

I first tried solving for ##y## and then ##x##, however it doesn't appear possible to solve for either variable. I then recognised that the equation can be represented as a nested function, such that: $$f(x)=x \text{ csch}(x+x\text{ csch}(x+x\text{ csch}(x+x\text{ csch}(x+...))))$$

Through this I was able to use Desmos Graphing calculator to help me estimate the enclosed area to by integrating ##f(x)## from ##0## to ##\infty##. However, the imbeded nature of the function made it hard to properly find a definitive solution and I could only gather that the area appears to be approaching ##1.5## square units, although this might be inaccurate and it could be simply approaching a number near ##1.5##.

How would I go about accurately determining this area?
Any help is greatly appreciated.

Last edited by a moderator:
Saracen Rue said:
Summary:: Determine the area enclosed by the curve ##y=x \text{ csch}(x+y)## and the positive ##x##-axis

The title and summary pretty much say it all. I was wondering if it's possible to accurately determine the area enclosed by the curve ## y=x \text{ csch}(x+y)## and the ##x##-axis?

I first tried solving for ##y## and then ##x##, however it doesn't appear possible to solve for either variable. I then recognised that the equation can be represented as a nested function, such that: $$f(x)=x \text{ csch}(x+x\text{ csch}(x+x\text{ csch}(x+x\text{ csch}(x+...))))$$

Through this I was able to use Desmos Graphing calculator to help me estimate the enclosed area to by integrating ##f(x)## from ##0## to ##\infty##. However, the imbeded nature of the function made it hard to properly find a definitive solution and I could only gather that the area appears to be approaching ##1.5## square units, although this might be inaccurate and it could be simply approaching a number near ##1.5##.

How would I go about accurately determining this area?
Any help is greatly appreciated.
Your equation is equivalent to ##F(x, y) = 0##, with ##F(x, y) = x - xcsch(x + y)##. As such, the graph of F is a surface in ##\mathbb R^3##, and the equation you're working with is the cross-section of the surface in the x-y plane, i.e., where z = 0.
As far as how you would graph F(x, y) = 0, nothing comes to mind, except possibly calculating function values at a bunch of points. There are going to be discontinuities at points where ##x + y = k\pi##, k an integer.

Mark44 said:
As far as how you would graph F(x, y) = 0, nothing comes to mind
I think that was already done. Wolfram plots:
Re: points where x+y=kπ ? it's hyperbolic: $$y = {x\over \sinh(x+y)}$$

BvU said:
I think that was already done. Wolfram plots:
View attachment 261445Re: points where x+y=kπ ? it's hyperbolic: $$y = {x\over \sinh(x+y)}$$
Sorry for not being clear in the OP. You're right, I do already have the graph of ##y=x \text{ csch}(x+y)## and I'm just trying to find the area enclosed by the cruve and the ##x##-axis

Edit: Do you think it would be possible to use WolframAlpha's nest function to create and integrate a heavily nested form of ##y=x\text{ csch}(x+x\text{ csch}(x+...))## to allow for a better estimation of the area? I tried doing it myself but I'm not the most familiar with the language and could only seem to get it to nest every variable ##x## with ##y(x)##

Last edited:

## 1. What is the formula for finding the area enclosed by a curve?

The formula for finding the area enclosed by a curve is to integrate the function representing the curve over the desired interval. In this case, the function is y = x csch(x+y) and the interval would be the x-values where the curve starts and ends.

## 2. How do I find the x-values where the curve starts and ends?

To find the x-values where the curve starts and ends, set the function equal to zero and solve for x. This will give you the x-values where the curve intersects the x-axis. These values will be the starting and ending points of the interval for finding the area enclosed by the curve.

## 3. Can I use any method of integration to find the area?

Yes, you can use any method of integration such as integration by parts, substitution, or partial fractions to find the area enclosed by the curve. However, the choice of method may depend on the complexity of the function and the desired level of accuracy.

## 4. How do I know if the area enclosed by the curve is positive or negative?

The sign of the area enclosed by the curve depends on the orientation of the curve. If the curve is above the x-axis, the area will be positive. If the curve is below the x-axis, the area will be negative. You can determine the orientation of the curve by plotting it or by looking at the sign of the function values at different x-values.

## 5. Is there a way to check if my calculated area is correct?

Yes, there are a few ways to check if your calculated area is correct. One way is to use a graphing calculator or software to graph the function and use the "integral" function to find the area. Another way is to use the Fundamental Theorem of Calculus to take the derivative of your integral and see if it matches the original function. You can also use geometric methods, such as finding the area of a rectangle with the same base and height as the curve, to estimate the area and compare it to your calculated value.

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