I've got a problem in which I'm trying to find the area of an ellipse which is given by the intersection of an elliptic cylinder with a plane. Nothing here is parallel to the coordinate axes, which is making it kind of annoying to work with.

The plane is given by the equation $x+ay+a^2z=0$, and the cylinder is given by $(x-a^2z)^2-a(x-a^2z)(y-az)+a^2(y-az)^2=L^2$.

I can think of some complicated ways to do it with integrals, and I'm wondering if there's something simpler that I'm missing. If I could transform into some coordinates in the plane I'm trying to work with, and I knew that the coordinate transformation would preserve areas, then I would be good, because I know how to get the area of an ellipse in the form $Ax^2+Bxy+Cy^2=1$. I'm really not sure how to do this transformation, though, and whether this is even the best way to proceed.

Maybe I should be using Langrange multipliers with two constraints to just obtain the lengths of the semi-major and semi-minor axes? That sounds like a pain, but doable. At least we're centered at the origin.

Thanks in advance for any assistance.

Edit: If I take $u=(x-a^2z)$ and $v=(y-az)$, then my cylinder becomes $u^2-auv+a^2v^2=L^2$, which is pretty great, but then I've basically cut the cylinder with a plane parallel to $z=0$. In that case, my given plane becomes $u+av+3a^2z=0$, and I'm not sure how that's helpful.

  • $\begingroup$ This looks suspiciously like you are already very close to your required form, where you get $$\frac 1{L^2}u^2-\frac a{L^2}uv+\frac {a^2}{L^2}v^2=1$$ Perhaps this $u,v$ setup can take over in the planar equation as well? $\endgroup$ – abiessu Feb 9 '15 at 21:52
  • $\begingroup$ By graphic view ( not established ) the plane is cut normally by a right circular cylinder for special case a =1. In this case only direction cosines of cylinder ( equally inclined to x,y,z axes) axis need be considered. $\endgroup$ – Narasimham Feb 10 '15 at 10:25
  • $\begingroup$ Is the axis of the cylinder $x=ay=a^2z$, and is this axis perpendicular to the cutting plane? $\endgroup$ – David K Feb 11 '15 at 15:56
  • $\begingroup$ That is the axis of the cylinder, but it is not perpendicular to the cutting plane. A line normal to the plane would be given by $a^2x=ay=z$ instead. $\endgroup$ – G Tony Jacobs Feb 15 '15 at 1:08

Expanding the equation of the cylinder, we get $$ x^2-axy+a^2y^2-a^2xz-a^3yz+a^4z^2 = L^2 $$ Writing this as a quadratic form: $$ \begin{bmatrix}x\\y\\z\end{bmatrix}^T \begin{bmatrix}1 & -\frac{a}{2} & -\frac{a^2}{2} \\ -\frac{a}{2} & a^2 & -\frac{a^3}{2} \\ -\frac{a^2}{2} & -\frac{a^3}{2} & a^4 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = L^2 $$ Diagonalizing this matrix, we find that the eigenvalue $0$ is associated with the eigenvector $\vec{v} = (a^2,a,1)^T$, giving the axis of the cylinder. The other two eigenvalues $\lambda_1$ and $\lambda_2$ are related to the semi-major axes of the ellipse by $s_{1,2} = L/\sqrt{ \lambda_{1,2}}$. The other eigenvalues are: $$ \lambda_{1,2} = \frac{1}{2}\left(1+a^2+a^4\pm\sqrt{(a^2-1)^2(a^4+a^2+1)} \right)$$

The area of the cross section of this cylinder is $$A_0=\pi s_1s_2 = \frac{\pi L^2}{\sqrt{\lambda_1 \lambda_2}} = \frac{2\pi L^2}{a\sqrt{3(1+a^2+a^4)}}$$

The plane cuts the cylinder at an angle, and the area of the slanted cut is $A = A_0/\cos \theta$ where $\theta$ is the angle between the plane's normal axis and the cylinder axis. The plane's normal vector is $\vec{n} = (1,a,a^2)^T$. Using a dot product, $$\cos\theta = \frac{\vec{n}\cdot\vec{v}}{\left\lVert \vec{n} \right\rVert \left\lVert \vec{v} \right\rVert} = \frac{3a^2}{1+a^2+a^4}$$ So then, $$ A = \frac{2\pi L^2}{a\sqrt{3(1+a^2+a^4)}} \frac{1+a^2+a^4}{3a^2} = \frac{2\pi L^2}{3a^3\sqrt{3}} \sqrt{1+a^2+a^4}$$

Ignore this stuff: As OP pointed out, projecting out $x$ doesn't make any sense.

Substituting away $x$ using the plane equation $x=-ay-a^2 z$ into the cylinder equation gives $$ a^2(y+2az)^2+a^2(y+2az)(y-az)+a^2(y-az)^2=L^2 $$ $$ (y+2az)^2+(y+2az)(y-az)+(y-az)^2=\left(\frac{L}{a}\right)^2 $$ Let $u=y+2az$ and $v=y-az$, so $$ u^2+uv+v^2 = \left(\frac{L}{a}\right)^2 $$ The Jacobian of this transformation is $-3a$. The area of the ellipse in $u,v$ coordinates can be found using the second formula here. To obtain the area in the original $x,y,z$ coordinates, multiply by the Jacobian of the transformation.

  • 1
    $\begingroup$ If you substitute away the $x$, then you get an equation that relates the $y$- and $z$-coordinates of each point on the ellipse, i.e., an equation of its projection onto the $yz$-plane. Once I have an equation in two variables, I'm well able to calculate the area. However, that's not the ellipse I'm looking for, or am I missing something? $\endgroup$ – G Tony Jacobs Feb 10 '15 at 21:33
  • $\begingroup$ @GTonyJacobs: Indeed, you are correct. I have edited my answer with a full solution. Let me know if you want clarification on any steps. $\endgroup$ – Victor Liu Feb 10 '15 at 23:31
  • $\begingroup$ This is excellent; thank you. The only thing that's not 100% clear is that I don't know where the equation $s_{1,2}=L/\sqrt{\lambda_{1,2}}$ comes from. This must be a general fact about ellipses and quadratic forms? $\endgroup$ – G Tony Jacobs Feb 15 '15 at 2:08
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    $\begingroup$ @GTonyJacobs: Indeed. The diagonal form of the quadratic form $(x,y)^T diag(a,b) (x,y) = L^2$ leads to $ax^2 + by^2 = L^2$, which you compare to an ellipse with semimajor axes $x^2/s_1^2 + y^2/s_2^2 = L^2$. $\endgroup$ – Victor Liu Feb 15 '15 at 17:52

I think this is less difficult than it appears. If you take any figure with area $S$ in a plane and projected it orthogonally onto another plane, the area of the image of your figure will be $S \cos \theta$, where $\theta$ is the angle between the planes. (Equivalently, $\theta$ is the angle your direction of projection makes with a vector normal to the plane you projected from.)

More generally, if you project the figure obliquely onto another plane, then the image has area $S \cos\theta\sec\phi$ where $\phi$ is the angle between the direction of projection and a vector normal to the plane you projected to.

Now consider the intersection of the cylinder with the $x,y$-plane. This is an ellipse whose formula you can derive simply by setting $z = 0$ in the formula of the cylinder. You know how to find the area $S$ of that ellipse. A projection parallel to the axis of the cylinder takes that ellipse onto your desired ellipse. I believe that projection is an orthogonal projection onto the plane $x+ay+a^2z=0$, in which case all you need to do is find the angle $\theta$ between the axis of the cylinder and the $z$ axis, and then the area of the ellipse you want is $S \cos\theta$.

If I'm wrong about the axis of the cylinder being orthogonal to the plane $x+ay+a^2z=0$, then you also have to find the angle $\phi$ between the axis and that plane, and then the answer is $S \cos\theta\sec\phi$.


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