Calculate the surface area of the part of the (parabolic)
Calculate the surface area of the part of the (parabolic) cylinder $z^2=4x$ that lies in the first octant, cut by the cylinder $y^2=4x$ and the plane $x=1$.
Explanation of Method
In the first octant, $z=2\sqrt x$ is a function of $x$ alone, so $\partial z/\partial y=0$ and the surface-area integrand $\sqrt{1+z_x^2+z_y^2}$ simplifies to a function of $x$ only. The base region is bounded by $y^2=4x$ and $x=1$.
- Compute the surface-area integrand from $z=2\sqrt x$.
- Set up the double integral over the base region $0\leqslant x\leqslant1,\,0\leqslant y\leqslant2\sqrt x$.
- Evaluate.
Step 1 · Surface-area integrand
With $z=2\sqrt x$: $z_x=\dfrac1{\sqrt x},\,z_y=0$, so $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+\dfrac1x}=\dfrac{\sqrt{x+1}}{\sqrt x}$.
Step 2 · Set up the integral
$$ S=\int_0^1\int_0^{2\sqrt x}\frac{\sqrt{x+1}}{\sqrt x}\,dy\,dx = \int_0^1\frac{\sqrt{x+1}}{\sqrt x}\cdot2\sqrt x\,dx = 2\int_0^1\sqrt{x+1}\,dx. $$
Step 3 · Evaluate
$$ 2\int_0^1\sqrt{x+1}\,dx = 2\left[\frac{2(x+1)^{3/2}}3\right]_0^1 = \frac43\big(2\sqrt2-1\big) = \frac{8\sqrt2-4}3. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct evaluation of the double integral gives exactly $\dfrac{8\sqrt2-4}3$.
Result
The surface area is: