Area of the section enclosed by the curve Curve is formed by the intersection of the surface:
$\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
with the plane $2x + 2y + z = 1$
What is the area enclosed by this curve?
Eliminating z from both equations, I get an implicit equation relating $x$ and y, with terms in $x^2, y^2, xy, x,$ and $y$. But I don't know where to go from there.
 A: Two given surfaces are
$ \ \displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \ $ and
$ \ 2x+2y+z=1$
We need to find area enclosed by the intersection curve. The projection of the enclosed area in XY plane is ellipse
$x^2+y^2-\frac{x}{2}-\frac{y}{2}+ \frac{3xy}{2} = 0$.
As the enclosed area is in plane $z = 1- 2x - 2y$, the area element can be expressed in terms of the area element of its projection in $XY$ plane,
$dS = \sqrt{1+2^2+2^2} \ dA \implies dS = 3 \ dA$
Now I will suggest you to plot the ellipse $x^2+y^2-\frac{x}{2}-\frac{y}{2}+ \frac{3xy}{2} = 0$ in a tool and see. It is apparent that to make it easier to work, we either rotate the ellipse clockwise or anticlockwise by $\frac{\pi}{4}$. If we rotate it anticlockwise, we will get an ellipse with major axis parallel to $x-$ axis and minor axis along $y-$axis. A combination of rotation and shifting of origin will convert it into an ellipse centered at the origin and major and minor axes along coordinate axes. We can first rotate by substituting,
$x = \frac{X+Y}{\sqrt2}, y = \frac{Y-X}{\sqrt2}$. So the equation of ellipse becomes,
$\displaystyle \frac{X^2} {(\sqrt{2/7})^2} + \frac{(Y - \sqrt2 / 7)^2}{(\sqrt2 / 7)^2} = 1$
Now you can directly apply the formula for area of ellipse which is $A = \pi ab = \displaystyle \pi \frac{\sqrt2}{\sqrt7} \frac{\sqrt2}{7} = \frac{2 \pi}{7 \sqrt7}$
So the area enclosed by the curve is $\displaystyle \frac{6 \pi}{7 \sqrt7}$.

If you want to find the area by integral, use polar coordinates with the following transformation -
$X = \sqrt{\frac{2}{7}} \ r \cos\theta, Y = \frac{\sqrt2}{7} (1 + r \sin\theta)$
which transforms the ellipse into a unit circle centered at origin.
Now find the Jacobian of transformation and integrate with limits $0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$.
