- #1

Saracen Rue

- 149

- 10

- TL;DR Summary
- How do you evaluate the area bound by an implicit curve when you can't rearrange for either variable or convert to a polar equation?

The implicit curve in question is ##y=\operatorname{arccoth}\left(\sec\left(x\right)+xy\right)##; a portion of the equations graph can be seen below:

In particular, I'm interested in the area bound by the curve, the ##x##-axis and the ##y##-axis. As such, we can restrict the domain to ##[0, \frac{\pi}{2}]## and the range to ##[0,\infty)##

From what I can tell, there's no way to solve this equation for either ##y## or ##x##, so standard integration is out of the system. My next thought was to convert it to a poloar equation, which got me ##r\sin\left(\theta\right)=\operatorname{arccoth}\left(\sec\left(r\cos\left(\theta\right)\right)+r^{2}\cos\left(\theta\right)\sin\left(\theta\right)\right)##. However, this has the same issue of not being able to solve the equation for ##r##. After this, I did some Googling and I managed to find some reference to double integral methods on Maths Stack Exchange (the reference I used can be found here; in the answer that was marked as correct). However, I found the whole thing quite confusing and I'm not very confident I did it correctly. Regardless, I made an attempt: Firstly, I defined ##f(x,y)## like so:

$$f(x,y)=

\begin{cases}

1 & \text{if }\operatorname{arccoth}\left(\sec\left(x\right)+xy\right)<y \\

0 & \text{otherwise} \\

\end{cases}

$$

I then proceeded to evaluate the double integeral like so and recieved the following result:

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}f\left(x,y\right)dy \text{ }dx \approx 1.46403068591$$

Okay so I have an actual way to get an answer now, but there are a few issues still.

In particular, I'm interested in the area bound by the curve, the ##x##-axis and the ##y##-axis. As such, we can restrict the domain to ##[0, \frac{\pi}{2}]## and the range to ##[0,\infty)##

From what I can tell, there's no way to solve this equation for either ##y## or ##x##, so standard integration is out of the system. My next thought was to convert it to a poloar equation, which got me ##r\sin\left(\theta\right)=\operatorname{arccoth}\left(\sec\left(r\cos\left(\theta\right)\right)+r^{2}\cos\left(\theta\right)\sin\left(\theta\right)\right)##. However, this has the same issue of not being able to solve the equation for ##r##. After this, I did some Googling and I managed to find some reference to double integral methods on Maths Stack Exchange (the reference I used can be found here; in the answer that was marked as correct). However, I found the whole thing quite confusing and I'm not very confident I did it correctly. Regardless, I made an attempt: Firstly, I defined ##f(x,y)## like so:

$$f(x,y)=

\begin{cases}

1 & \text{if }\operatorname{arccoth}\left(\sec\left(x\right)+xy\right)<y \\

0 & \text{otherwise} \\

\end{cases}

$$

I then proceeded to evaluate the double integeral like so and recieved the following result:

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}f\left(x,y\right)dy \text{ }dx \approx 1.46403068591$$

Okay so I have an actual way to get an answer now, but there are a few issues still.

- I have no idea if I set my piecewise function up correctly; specifically for the domain on value for 1 - I just copied the general layout from another implicit curve area example and did my best to fit my equation too it.
- I have no idea if my integration bounds are correct either - I assume they are but I also have a sinking suscipcsion that I have missed a step in determining the values to use here.
- And finally, even if I
*have*done everything else correctly, apparently this double integral method only generates approximate results and may not even be accurate.