Which paraboloid would have a larger surface area? Or would they be the same?

Question

The enclosed area of these two paraboloids should be seen as a cross section of a 3D image. It is a water storage tank. I am trying to find out strengths and weaknesses of each model. Which one would have a smaller surface area (thus cost less), or would they be the same?

Answer 1

Note that the surface are of the paraboloid $z=a(x^2+y^2)$ of height $h$ is given by $$S=\iint_D\sqrt{1+4a^2(x^2+y^2)}dxdy$$ where $D$ is the disc $x^2+y^2\leq r^2=h/a$. Hence, in polar coordinates we get $$S=2\pi\int_0^{\sqrt{h/a}}\sqrt{1+4a^2\rho^2}\rho d\rho=\frac{\pi ((1+4ah)^{3/2}-1)}{6a^2}.$$ The last formula can be obtained as a surface of revolution of the graph of the function $x=f(z)=\sqrt{\frac{z}{a}}$ around the interval $[0,h]$ along the $z$-axis: $$S=2\pi\int_0^hf(z)\sqrt{1+f'(z)^2}dz =2\pi\int_0^h\sqrt{\frac{z}{a}}\sqrt{1+\left(\frac{1}{2\sqrt{az}}\right)^2}dz\\ =\frac{\pi}{a}\int_0^h\sqrt{1+4az}\,dz=\frac{\pi ((1+4ah)^{3/2}-1)}{6a^2}.$$

Answer 2

There is a simpler way to do this problem using Pappus's Centroid Theorems. I will show you how to get both the surface area and the volume of the paraboloid. (Because why would you want to minimize the surface area without considering how much water it will hold?)

Pappus's $(1^{st})$ Centroid Theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid (Pappus's centroid theorem). Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by

$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$

Pappus's $(2^{nd})$ Centroid Theorem says the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$. The centroid of a volume is given by

$$\mathbf{R}=\frac{\int_A \mathbf{r}dA}{\int_A dA}=\frac{1}{A} \int_A \mathbf{r}dA$$

Thus we can say for your case, where the rotation is about the $y$-axis, that

$$ S=2\pi\int x\ ds \\ V=\pi\int xy\ dx $$

Notice that here, the length, $L$ and the volume, $V$ have cancelled out, thus disguising the true nature of the equations.

Now for the details, surface area first (noting that $ds=\sqrt{1+(dy/dx)^2}dx$)

$$ \begin{align} S & =2\pi\int_0^X x\sqrt{1+(dy/dx)^2}dx\\ & =\frac{2\pi}{8a^2}\int_0^{4a^2X^2}\sqrt{1+t}\ dt,\quad t=(dy/dx)^2=(2ax)^2\\ & =\frac{\pi}{6a^2}\left[(1+4a^2X^2)^{3/2}-1 \right] \end{align} $$

where $X=\sqrt{h/a}$ is the width of the paraboloid of height $h$. The volume is given by

$$ \begin{align} V & =2\pi \int_0^X \int_0^{ax^2} xdydx\\ & =2\pi a\int_0^X x^3 dx\\ & =\frac{\pi a X^4}{2} \end{align} $$

These equations should allow you to find the surface area and volume for any paraboloid width, $X$ or height $h$ and parameter $a$. I have verified these equations numerically, as well.