integrationproof-verification
Say we wanted to derive the formula for the volume of a cube with integration. Each "slice" of the cube has area $x^2$, with "width" $dx$. Integrate from $0$ to $x$, and I believe you would get the following:
$$\int_0^x x^2dx$$
And this equals $\frac 13x^3$, however the volume of a cube is given by $x^3$!
What is wrong with this process?
For $1 \le \alpha \le 2$, it is much easier to visualize the integral by subtracting instead of adding pieces of the volume.
As shown in the picture below, when $1 \le \alpha \le 2$, the volume of cube section below intersection of the plane $x + y + z = \alpha$ is the difference of the volume of one big tetrahedron with width/height/depth $= \alpha$ with three smaller ones with width/height/depth $= \alpha-1$. So the volume above the intersection becomes
$$1 - \left( \frac16 \alpha^3 - 3 ( \frac16 (\alpha-1)^3 )\right) = 1 - \frac16 \alpha^3 + \frac12 (\alpha-1)^3$$
Update
About the question whether this argument can be extended to higher dimension, the answer is yes. Let's look at the 3-dimension $2 \le \alpha \le 3$ case first.
As one increases $\alpha$ beyond $2$, the three tetrahedron in first figure start overlap. As shown in second figure, the intersection of the three tetrahedra are now three even smaller tetrahedra of width/height/depth = $\alpha -2$.
Previous way to compute the "volume" of cube section below the plane $x + y + z = \alpha$ now subtract too much from this three even smaller tetrahedron. One need to add them back. As a result, the volume above the plane becomes:
$$\begin{align}&1 - \left( \frac16 \alpha^3 - 3(\frac16 (\alpha-1)^3 + 3(\frac16 (\alpha-2)^3 \right)\\ = & 1 - \frac16 \alpha^3 + \frac12 (\alpha-1)^3 - \frac12 (\alpha-2)^3\\ = & \frac{(3-\alpha)^3}{6} \end{align}$$
Let us switch to the $k$-dimension case. To compute the "volume" of the hypercube section above the hyperplane $x_1 + \ldots + x_k = \alpha$, the first step is to subtract the volume of a $k$-simplex of size $\alpha$ from 1.
Repeat these arguments and notice in the middle of the process, we need to either add or subtract the volumes of $\binom{k}{i}$ $\;k$-simplices of size $\alpha-i$. The "volume" of interest finally becomes:
$$1 -\sum_{i=0}^{\lfloor \alpha \rfloor} (-1)^i \binom{k}{i} \frac{(\alpha-i)^k}{k!}$$
One way to gain the intuition behind this is to look at what happens in 2 dimensions. Here, rather than surface area and volume, we look at arc length and area under the curve. When we want to find the area under the curve, we estimate using rectangles. This is sufficient to get the area in a limit; one way to see why this is so is that both the error and the estimate are 2-dimensional, and so we aren't missing any extra information.
However, the analogous approach to approximating arc length is obviously bad: this would amount to approximating the curve by a sequence of constant steps (i.e. the top of the rectangles in a Riemann sum) and the length of this approximation is always just the length of the domain. Essentially, we are using a 1-dimensional approximation (i.e. only depending on $x$) for a 2-dimensional object (the curve), and so our approximation isn't taking into account the extra length coming from the other dimension. This is why the arc length is computed using a polygonal approximation by secants to the curve; this approximation incorporates both change in $x$ and change in $y$.
Why is this relevant to solids of revolution? Well, in essence, the volume and surface area formulae are obtained by simply rotating the corresponding 2-dimensional approximation rotated around an axis, and taking a limit. If it wouldn't work in 2 dimensions, it certainly won't work in 3 dimensions.
Best Answer
The problem is that with $\int_0^x x^2dx$ you are actually calculating the volume of a kind of inverted pyramid. As @Fermat suggested, you have to maintain the side fixed (through $a^2$), otherwise it changes as the third axis does.