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!
Best Answer
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 $\frac{\pi \cdot \sqrt{6}}{2}$.