Find the surface area of the part of the surface dfrac x²
Find the surface area of the part of the surface $\dfrac{x^2}a+\dfrac{y^2}b=2z$ cut out by the elliptic cylinder $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$.
Explanation of Method
Solving for $z=\frac{x^2}{2a}+\frac{y^2}{2b}$, the surface-area integrand $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+\frac{x^2}{a^2}+\frac{y^2}{b^2}}$ matches the shape of the cutting cylinder exactly, so the elliptic substitution $x=ar\cos\theta,\,y=br\sin\theta$ (Jacobian $abr$) turns the region into the unit disk and reduces the problem to a single elementary radial integral.
- Compute the surface-area integrand.
- Substitute elliptic coordinates matching the base ellipse.
- Evaluate via $u=1+r^2$.
Step 1 · Surface-area integrand
With $z=\dfrac{x^2}{2a}+\dfrac{y^2}{2b}$: $z_x=\dfrac xa,\,z_y=\dfrac yb$, so $1+z_x^2+z_y^2=1+\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$.
Step 2 · Elliptic coordinates
With $x=ar\cos\theta,\,y=br\sin\theta$ ($0\le r\le1,\,0\le\theta\le2\pi$), the Jacobian is $abr$ and $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=r^2$:
$$ S = \int_0^{2\pi}\int_0^1\sqrt{1+r^2}\cdot ab\,r\,dr\,d\theta = 2\pi ab\int_0^1r\sqrt{1+r^2}\,dr. $$
Step 3 · Evaluate
With $u=1+r^2$:
$$ S = \pi ab\int_1^2\sqrt u\,du = \pi ab\left[\frac23u^{3/2}\right]_1^2 = \frac{2\pi ab}3\big(2\sqrt2-1\big). $$
Additional Notes
- This derivation matches the source exactly. Note the surface equation uses $a,b$ (not $a^2,b^2$) in its denominators, unlike the cutting cylinder — this asymmetry is exactly what makes the substitution $x=ar\cos\theta,\,y=br\sin\theta$ work cleanly, since it is $z_x,z_y$ (not $z$ itself) that need to match the cylinder's shape.
- Verified numerically: direct evaluation of the double integral (e.g. $a=1.3,\,b=2.1$) gives $\approx10.454395$, matching $\dfrac{2\pi ab}3(2\sqrt2-1)$ exactly.
Result
The surface area is: