Volumes of solid objects 
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 $V=\int_{0}^{4} \pi y^2\sqrt{1+(y')^2} dx$?


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? 
 A: 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). 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.
A: 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).
