Find the surface area of the part of the paraboloid
Find the surface area of the part of the paraboloid $z^2=2xy$ (where $z>0$) bounded by the planes $x=0,\,x=a,\,y=0,\,y=b$.
Explanation of Method
Solving for $z=\sqrt{2xy}$ over the rectangle $[0,a]\times[0,b]$, the surface-area integrand $\sqrt{1+z_x^2+z_y^2}$ simplifies (via the identity $1+\frac y{2x}+\frac x{2y}=\frac{(x+y)^2}{2xy}$) to the elementary expression $\frac{x+y}{\sqrt{2xy}}$, which separates into two elementary power-law integrals.
- Compute the surface-area integrand and simplify it to a perfect square under the root.
- Split the integrand into two separable terms.
- Evaluate the resulting elementary integrals.
Step 1 · Surface-area integrand
With $z=\sqrt{2xy}$: $z_x=\sqrt{\dfrac y{2x}},\,z_y=\sqrt{\dfrac x{2y}}$, so:
$$ 1+z_x^2+z_y^2 = 1+\frac y{2x}+\frac x{2y} = \frac{x^2+2xy+y^2}{2xy} = \frac{(x+y)^2}{2xy}, \qquad \sqrt{1+z_x^2+z_y^2}=\frac{x+y}{\sqrt{2xy}}. $$
Step 2 · Set up and separate the integral
$$ S = \int_0^a\int_0^b\frac{x+y}{\sqrt{2xy}}\,dy\,dx = \int_0^a\left[\sqrt{\frac x2}\int_0^by^{-1/2}\,dy + \frac1{\sqrt{2x}}\int_0^by^{1/2}\,dy\right]dx. $$
Step 3 · Evaluate
With $\displaystyle\int_0^by^{-1/2}dy=2\sqrt b$ and $\displaystyle\int_0^by^{1/2}dy=\frac{2b^{3/2}}3$:
$$ S = \int_0^a\left[\sqrt{2b\,x}+\frac{\sqrt2\,b^{3/2}}{3\sqrt x}\right]dx = \sqrt{2b}\cdot\frac{2a^{3/2}}3 + \frac{\sqrt2\,b^{3/2}}3\cdot2a^{1/2} = \frac{2\sqrt2}3\sqrt{ab}\,(a+b). $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct evaluation of the double integral (e.g. $a=1.7,\,b=2.3$) gives $\approx7.4571368$, matching $\dfrac{2\sqrt2}3\sqrt{ab}(a+b)$ exactly.
Result
The surface area is: