Find the area of the intersection of the surface
Find the area of the intersection of the surface: $$ x^2+y^2+z^2-xy-xz-yz=a^2 $$ and the plane $x+y+z=0$.
Explanation of Method
Eliminating $z=-x-y$ reduces the surface equation to a quadratic curve in $x,y$ — but this curve is the orthogonal projection of the true intersection onto the $xy$-plane, not the true curve itself, since the plane $x+y+z=0$ is tilted relative to the $xy$-plane. Computing "the area of the ellipse" using $x,y$ as if they were ordinary orthonormal coordinates gives only the projected area; the actual area of the planar curve, as it sits inside the tilted plane in $3$-space, must include the area-distortion (Jacobian) factor between the plane and its $xy$-projection.
- Eliminate $z$ using the plane equation and simplify the quadric to a 2-variable quadratic form.
- Compute the true area using an orthonormal coordinate system within the plane itself (not the $x,y$-projection), or equivalently apply the projection-area correction factor $\sec\theta$, where $\theta$ is the angle between the two planes.
- Evaluate the resulting (circular, as it turns out) area.
Step 1 · Eliminate $z$
With $z=-x-y$: $z^2=x^2+2xy+y^2$, $-xz=x^2+xy$, $-yz=xy+y^2$. Substituting into $x^2+y^2+z^2-xy-xz-yz=a^2$ and collecting terms:
$$ x^2+y^2+(x^2+2xy+y^2)-xy+(x^2+xy)+(xy+y^2) = 3x^2+3y^2+3xy = a^2 \quad\Longrightarrow\quad x^2+y^2+xy=\frac{a^2}3. $$
This is the equation of the curve's projection onto the $xy$-plane (since every point of the plane $x+y+z=0$ has a unique $(x,y)$, dropping the $z$-coordinate is exactly an orthogonal projection onto the $xy$-plane).
Step 2 · Find the true area using an orthonormal basis for the plane
An orthonormal basis for the plane $x+y+z=0$ is $\mathbf e_1=\frac1{\sqrt2}(1,-1,0)$, $\mathbf e_2=\frac1{\sqrt6}(1,1,-2)$. Writing a point of the plane as $(x,y,z)=u\,\mathbf e_1+v\,\mathbf e_2$ (so $u,v$ are true, distance-preserving coordinates within the plane) and substituting into the quadratic form $x^2+y^2+z^2-xy-xz-yz$:
$$ x^2+y^2+z^2-xy-xz-yz = \frac32u^2+\frac32v^2 $$
(the cross term vanishes and the two coefficients are equal — the curve is, remarkably, a circle in the plane's own orthonormal coordinates, even though its projection $x^2+y^2+xy=\frac{a^2}3$ is a genuine ellipse). So the boundary curve is:
$$ \frac32\big(u^2+v^2\big)=a^2 \quad\Longrightarrow\quad u^2+v^2=\frac{2a^2}3, $$
a circle of radius $r=a\sqrt{2/3}$ in the true, orthonormal $(u,v)$ coordinates of the plane.
Step 3 · Compute the area
$$ A = \pi r^2 = \pi\cdot\frac{2a^2}3 = \frac{2\pi a^2}3. $$
Additional Notes
- A note on the source computation: the source correctly derived $x^2+y^2+xy=\frac{a^2}3$, then computed "the area of this ellipse" as $\frac{2\pi a^2\sqrt3}9$ — but this treats $x,y$ as if they were an orthonormal coordinate system for the plane $x+y+z=0$, which they are not: they are the projection onto the $xy$-plane. The relationship between the true area (in the tilted plane) and the projected area (in the $xy$-plane) is $A_{\text{true}}=A_{\text{projected}}/\cos\theta$, where $\theta$ is the angle between the plane $x+y+z=0$ (normal $\frac1{\sqrt3}(1,1,1)$) and the $xy$-plane (normal $(0,0,1)$): $\cos\theta=\frac1{\sqrt3}$, so $A_{\text{true}}=\sqrt3\cdot A_{\text{projected}}=\sqrt3\cdot\frac{2\pi a^2\sqrt3}9=\frac{2\pi a^2}3$. This matches the direct orthonormal-coordinate computation above, and is exactly $\sqrt3$ times the source's answer.
- Verified three independent ways for $a=1.7$: (1) the orthonormal-basis circle-area formula gives $A\approx6.05280$; (2) a direct numerical computation of the planar-polygon area in $3$-space (sampling the ellipse via its eigen-parametrization in $x,y$, lifting to $(x,y,z)$, and applying the vector cross-product polygon-area formula) gives $A\approx6.05280$; (3) $\frac{2\pi a^2}3\approx6.05280$. All three agree to $6$ significant figures.
Result
The true area of the intersection ellipse is: