How can I define the intersection area of an ellipse and a circle?

Question

So I would need to solve how to define the intersection area of a circle and an ellipse. The coordinates for circle and ellipse are known and also the radius of circle and semi-major and -minor axes of an ellipse. Here is plot to help to understand what I mean Intersection area plot. I am not the best with mathematics and integrals etc. and I will appreciate a ton if someone is able to help me with this.

Answer 1

If $(x,y_l)$ and $(x,y_u)$ are the points of intersection of the two curves, then the required area can be calculated through integration w.r.t $y$: $$\int_{y_l}^{y_u} x^+(y)_{\text{ellipse}}-x^-(y)_{\text{circle}} \ dy$$ Here, $x^+$ and $x^-$ mean that you need to take the positive and negative square roots respectively, when solving for $x$ in terms of $y$.

EDIT: After finding the intersection points and integrating, you should get the answer as approximately $\boxed{2003708.58843}$.

Answer 2

Well, we know that the equation of a circle is given by:

$$\left(x-\text{a}\right)^2+\left(\text{y}-\text{b}\right)^2=\text{r}^2\tag1$$

Where $\left(\text{a},\text{b}\right)$ are the center coordinates of the circle and $\text{r}$ is the radius of the circle.

In your case, we have $\text{a}=\text{b}=5000$ and $\text{r}=2000$. So:

$$\left(x-5000\right)^2+\left(\text{y}-5000\right)^2=2000^2\tag2$$

We know that the equation of an ellipse is given by:

$$\left(\frac{x-x_0}{\alpha}\right)^2+\left(\frac{\text{y}-\text{y}_0}{\beta}\right)^2=1\tag3$$

Where $\left(x_0,\text{y}_0\right)$ are the center coordinates of the ellipse and $\alpha$ is the semi-major axis and $\beta$ is the semi-minor axis.

In your case, we have $x_0=2500$, $\text{y}_0=5000$, $\alpha=2000$, and $\beta=1000$. So:

$$\left(\frac{x-2500}{2000}\right)^2+\left(\frac{\text{y}-5000}{1000}\right)^2=1\tag4$$

Now, I used Mathematica to plot this with the following code:

In[1]:=ContourPlot[{(x - 5000)^2 + (y - 5000)^2 == 
   2000^2, ((x - 2500)/2000)^2 + ((y - 5000)/1000)^2 == 1}, {x, 2000, 
  8000}, {y, 2000, 8000}]

And got the following output:

We can solve for the intersection points, using:

In[2]:=FullSimplify[
 Solve[{(x - 5000)^2 + (y - 5000)^2 == 
    2000^2, ((x - 2500)/2000)^2 + ((y - 5000)/1000)^2 == 1, 
   x > 0 && y > 0}, {x, y}]]

Out[2]={{x -> -(500/3) (-35 + 2 Sqrt[61]), 
  y -> -(500/3) (-30 + Sqrt[5 (-25 + 4 Sqrt[61])])}, {x -> -(500/
     3) (-35 + 2 Sqrt[61]), 
  y -> 500/3 (30 + Sqrt[5 (-25 + 4 Sqrt[61])])}}

Using gridlines we can use the following code:

ContourPlot[{(x - 5000)^2 + (y - 5000)^2 == 
   2000^2, ((x - 2500)/2000)^2 + ((y - 5000)/1000)^2 == 1}, {x, 2000, 
  8000}, {y, 2000, 8000}, 
 GridLines -> {{-(500/3)*(2*Sqrt[61] - 35), 3000, 4500}, {}}]

To see:

Now, it is not hard to show that the desired area is given by:

$$\mathcal{A}:=\text{I}_1+\text{I}_2\tag5$$

Where:

I1 = Integrate[
   5000 + Sqrt[-(-7000 + x) (-3000 + x)], {x, 3000, \[Tau]}] - 
  Integrate[5000 - Sqrt[-(-7000 + x) (-3000 + x)], {x, 3000, \[Tau]}]

I2 = Integrate[
   5000 + 1/2 Sqrt[-(-4500 + x) (-500 + x)], {x, \[Tau], 4500}] - 
  Integrate[
   5000 - 1/2 Sqrt[-(-4500 + x) (-500 + x)], {x, \[Tau], 4500}]

Where $\tau=\frac{500}{3}\left(35-2\sqrt{61}\right)$.

So, we get:

$$\mathcal{A}\approx2.00371\cdot10^6\tag8$$

And the exact value is:

250000/3 (-5 Sqrt[5 (-25 + 4 Sqrt[61])] + 
   48 (ArcCsc[2 Sqrt[1/15 (4 + Sqrt[61])]] + 
      2 ArcSec[2 Sqrt[2/65 (-7 + 2 Sqrt[61])]]))