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What is the volume when $f(x) = x^2$ is rotated around the line $y = x$?

For each individual $x$, I was considering the difference between $\begin{pmatrix} x\\x \end{pmatrix}$ and the projection of $\begin{pmatrix} x^2\\x \end{pmatrix}$ onto $\begin{pmatrix} x\\x \end{pmatrix}$ (which would give vectors perpendicular to the line $y = x$ and of the correct length, going from a point on $y=x$ to the corresponding point on $y=x^2$).

I would get $\begin{pmatrix} x\\x \end{pmatrix} \cdot \left(1 - \frac{x^3+x^2}{2x^2}\right)$, and I could integrate the length of that squared times $\pi$, going from $0$ to $1$ to find the volume.

I would get $\pi \cdot \int_0^1 2*\left(x - \frac{x^3+x^2}{2x}\right)^2 = \frac{\pi}{60}$, but the correct answer is $\frac{\sqrt{2}\pi}{60}$.

Incidentally, $\frac{d\begin{vmatrix} x\\x \end{vmatrix}}{dx} = \sqrt{2}$.

I'm missing something, and can't figure out exactly how this (rigorously) fits together.

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[Sorry, previous answer was wrong]

You are right in saying that "Incidentally, $\frac { d {\begin{matrix}x\\x\\ \end{matrix}} }{dx} = \sqrt{2}$. In changing the coordinates, from integrating on the x-axis, to integrating on the line y=x, you also need to account for the difference in volume, which is precisely this amount.

For example, if you consider the length of the line on $y=x$ from $x=0$ to $1$, you do not say that it has length $\int_{x=0}^1 1 dx = 1$, but it has length $\int_{x=0}^1 \sqrt{2} dx = \sqrt{2}$. The Jacobian takes care of this.

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