Question: ![enter image description here](https://i.stack.imgur.com/mSDvh.png)

The volume of the object in the above picture is given by, $$V=\int\_{0}^{4} \pi y^2dx$$

While dealing with calculating surface areas of objects, I was introduced to the integral, $$S=\int\_{0}^{4} 2\pi y\sqrt{1+(y')^2} dx$$ where $\sqrt{1+(y')^2} dx$ is the infinitesimal arc length. 

Is it reasonable to improve the definition of the volume of a solid object to be $V=\int\_{0}^{4} \pi y^2\sqrt{1+(y')^2} dx$?

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![enter image description here](https://i.stack.imgur.com/ZpmIf.png)

Geometrically, the segment length is a better approximation. However, when the interval tends to zero, the arc, segment and $\Delta x$ tend to coincide. So, as the interval goes to zero, I can consider the infinitesimal segment length instead of $dx$, can I not? 

What I would like to know is if $V=\int\_{0}^{4} \pi y^2\sqrt{1+(y')^2} dx$ is valid or not. If not, then why?

Answer: 
This is perhaps best explained in terms of Pappus's Centroid Theorems:

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](https://en.wikipedia.org/wiki/Pappus%27s_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 cases that

$$
S=2\pi\int\_0^4 y\ ds =2\pi\int\_0^4 y\sqrt{1+y'^2} dx\\
V=\pi\int\_0^4 y^2\ dx 
$$

This is just as stated in the OP. Notice that here, the length, $L$ and the volume, $V$ have cancelled out, thus disguising the true nature of the equations.

Now, clearly you can't have

$$V=\pi\int\_0^4 y^2\ dx=\pi\int\_0^4 y^2\sqrt{1+y'^2} ~dx$$

But more to the point, the volume pertains to the area under the curve, whereas, the term $\sqrt{1+y'^2}$ in the surface area equation refers to the area along the arc of $ds$ and it has nothing to do with the volume of revolution.
It is reasonable to improve the definition of the volume of a solid object that can be defined to be

$$V=\int\_{0}^{4} \pi y^2 dx$$

by assembly of infinitesimally thin disks shown, so that the direction of building up of volume is strictly perpendicular to the base circle plane.

( Surface area of the solid is obtained by assembling lateral areas of connected cones).