How to calculate surface area of the intersection of an elliptic cylinder and plane?

Question

Given the equations $x^2+2y^2 \leq 1$, and $x+y+z=1$, how do I find the surface area of their intersection?

I approached this question by first parameterizing the equation for the elliptic cylinder.

$x=\cos(u)$, $y= \frac{\sqrt2}{2} \sin(u)$, $z = v$

Then, I calculated the tangent vectors $T_u$ and $T_v$.

$T_u = (-\sin(u), \frac{\sqrt2}{2}\cos(u),0)$.

$T_v=(0,0,1)$.

$T_u \times T_v = -\frac{\sqrt2}{2}\cos(u)\cdot i-\sin(v)\cdot j$.

$|T_u \times T_v| = \sqrt{\frac{1}{2}\cdot\cos^2(u)+\sin^2(u)}$.

I could not integrate the above expression. After looking through various resources, they all say to parameterize the elliptic cylinder the way I did above. So now I am looking for either other methods of parametrization or a different approach to this problem overall.

Any help is appreciated, thank you!

Answer 1

You are cutting an elliptical cylinder with a plane, leading to an ellipse. By a simple change of variable ($y=Y/2$) this is the same as cutting a cylinder with a plane. In such a case the area of the section is $\pi R^2 |\sec\theta|$, where $R$ is the radius of the cylinder and $\theta$ the the angle between the cutting plane and a plane containing a circular section of the cylinder. Can you compute $R$, $\cos\theta$ and finish by scaling?

Answer 2

Thanks to hardmath, I was able to figure out the answer to this problem. We parameterize the plane equation $x+y+z=1$.

$x = u$, $y=v$, $z=1-u-v$.

Now our $T_u$ = $(1,0,-1)$ and $T_v=(0,1,-1)$.

$T_u \times T_v = i+j+k$.

$|T_u \times T_v| = \sqrt{3}$.

Our integral is $\iint_s \sqrt{3} \, dx\, dy = \sqrt{3} \cdot \pi ab$, where $s$ is the horizontal cross section of our original elliptic cylinder equation $x^2+2y^2=1$. Now we have$\iint_s dx\, dy = \pi ab$ since the area of $s$ is $\pi ab$, and $a$ and $b$ are the lengths of its semi-major and semi-minor axes. We have $a=1$ and $b= \frac{\sqrt2}{2}$ from $x^2+2y^2=1$.

Thus, the final surface area is $\frac{\pi \cdot \sqrt{6}}{2}$.