Question
I would like to know the simplest way to find the volume of the solid of revolution created by rotating the parabola $y=x^2$ around the line $y=x$ (the shape shown in blue below). I am currently taking AP BC Calculus as a junior in high school, so a method that uses those concepts would be ideal, but if it is far simpler to use some higher math, I will look into it 🙂
The following is what I have tried using a variation of the disk method. I believe that is correct, but, as the reader can see, it is very complex.
My Method
To employ the disk method, first, derive a function for the radius of the solid as a function of $x$ along $y=x$. Then, square it and multiply by $\pi$. Lastly, integrate on the interval $[0,\sqrt{2}]$.
Begin by constructing a line perpendicular to $y=x$ that intersects $y=x$ (occationally $f(x)$) and $y=x^2$ (occationally $g(x)$) at $(x_2,y_2)$ and $(x_1,y_1)$, respectively (as shown below).
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\tag{1}$$
Use the distance formula to find the distance between these points.
$$\begin{align}
\color{gray}{y} &\color{gray}{=} \color{gray}{-x+2x_2}\\
y &=x
\end{align}$$
$$x=-x+2x_2$$
$$2x=2x_2$$
$$x_2=x\tag{2}$$
$$\begin{align}
y_2&=f(x_2)\\
&=x\tag{3}
\end{align}$$
$$\begin{align}
\color{gray}{y} &\color{gray}{=} \color{gray}{-x_1+2x_2}\\
\color{gray}{y} &\color{gray}{=} \color{gray}{-x_1+2x}\\
y &={x_1}^2
\end{align}$$
$${x_1}^2=-x_1+2x$$
$$0=1{x_1}^2+1x_1+-2x$$
$$\begin{align}
x_1&=\frac{-1+\sqrt{1^2-4(1)(-2x)}}{2(1)}\\
&=\frac{\sqrt{1+8x}-1}{2}\tag{4}
\end{align}$$
$$\begin{align}
y_1&=g(x_1)\\
&=\bigg(\frac{\sqrt{1+8x}-1}{2}\bigg)^2\tag{5}
\end{align}$$
Find the variables in the distance formula as functions of $x$ (Eqns. 2-5 with derivations listed above them, respectively).
$$\begin{align}
d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\
&=\sqrt{\Bigg(x-\frac{\sqrt{1+8x}-1}{2}\Bigg)^2+\Bigg(x-\bigg(\frac{\sqrt{1+8x}-1}{2}\bigg)^2\Bigg)^2}\tag{6}
\end{align}$$
Plug Eqns. 2-5 into the distance formula.
$$\begin{align}
d&=\sqrt{\Bigg(x-\frac{\sqrt{1+8x}-1}{2}\Bigg)^2+\Bigg(x-\bigg(\frac{\sqrt{1+8x}-1}{2}\bigg)^2\Bigg)^2}\\
&=\sqrt{\Bigg(x-\frac{\sqrt{1+8x}-1}{2}\Bigg)^2+\Bigg(x-\bigg(\frac{(1+8x)-2\sqrt{1+8x}+1}{4}\bigg)\Bigg)^2}\\
&=\sqrt{\Bigg(x-\frac{\sqrt{1+8x}-1}{2}\Bigg)^2+\Bigg(x-\bigg(\frac{2+8x-2\sqrt{1+8x}}{4}\bigg)\Bigg)^2}\\
&=\sqrt{\Bigg(\frac{2x}{2}-\frac{\sqrt{1+8x}-1}{2}\Bigg)^2+\Bigg(\frac{2x}{2}-\frac{1+4x-\sqrt{1+8x}}{2}\Bigg)^2}\\
&=\sqrt{\Bigg(\frac{2x-\sqrt{1+8x}+1}{2}\Bigg)^2+\Bigg(\frac{2x-1-4x+\sqrt{1+8x}}{2}\Bigg)^2}\\
&=\sqrt{\Bigg(\frac{1+2x-\sqrt{1+8x}}{2}\Bigg)^2+\Bigg(\frac{-1-2x+\sqrt{1+8x}}{2}\Bigg)^2}\\
&=\sqrt{2\Bigg(\frac{1+2x-\sqrt{1+8x}}{2}\Bigg)^2}\\
&=\sqrt{2\Bigg(\frac{1+4x^2+(1+8x)+4x-2\sqrt{1+8x}-4x\sqrt{1+8x}}{4}\Bigg)}\\
&=\sqrt{\frac{4x^2+12x-(4x+2)\sqrt{1+8x}+2}{2}}\\
&=\sqrt{2x^2+6x-(2x+1)\sqrt{1+8x}+1}\tag{7}
\end{align}$$
Simplify Eqn. 6.
$$\begin{align}
r&=\sqrt{2\bigg(\frac{x}{\sqrt{2}}\bigg)^2+6\bigg(\frac{x}{\sqrt{2}}\bigg)-\bigg(2\bigg(\frac{x}{\sqrt{2}}\bigg)+1\bigg)\sqrt{1+8\bigg(\frac{x}{\sqrt{2}}\bigg)}+1}\\
&=\sqrt{2\bigg(\frac{x^2}{2}\bigg)+6\bigg(\frac{\sqrt{2}x}{2}\bigg)-\bigg(2\bigg(\frac{\sqrt{2}x}{2}\bigg)+1\bigg)\sqrt{1+8\bigg(\frac{\sqrt{2}x}{2}\bigg)}+1}\\
&=\sqrt{x^2+3\sqrt{2}x-\big(\sqrt{2}x+1\big)\sqrt{1+4\sqrt{2}x}+1}\tag{8}
\end{align}$$
Dilate Eqn. 7 by $\sqrt{2}$ in the x-direction to make the distance between the functions x-intercepts equal to the distance between the two intercepts of $f(x)$ and $g(x)$. Simplify to give Eqn. 8. Note that the graph of Eqn. 8 from $[0,\sqrt{2}]$ (below in green) compares to the reflection over the x-axis of the final equation for a parabola rotated 45 degrees given by Ennar (below in red), as it should.
Graph from Desmos.
Integration by parts work (for below):
$$\color{red}{\int\big(\sqrt{2}x+1\big)\sqrt{1+4\sqrt{2}x} \ dx}$$
$$
\begin{array}{|c|}
\hline
\mathbf{u=\sqrt{2}x+1},\ \mathbf{dv=\sqrt{1+4\sqrt{2}x} \ dx}\\
\hline
\begin{array}{c|c}
\frac{du}{dx}=\sqrt{2} & \int dv=\int\sqrt{1+4\sqrt{2}x}\ dx\\
\mathbf{du=\sqrt{2}\ dx} & v=\int\sqrt{w}\ \frac{dw}{4\sqrt{2}}\\
& v=\frac{1}{4\sqrt{2}}\int\sqrt{w}\ dw\\
& v=\frac{1}{4\sqrt{2}}\times\frac{w^\frac{3}{2}}{\frac{3}{2}}\\
& v=\frac{2}{12\sqrt{2}}w^\frac{3}{2}\\
& \mathbf{v=\frac{1}{6\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}}\\
\end{array}\\
\hline
\end{array}
$$
$$\begin{align}
&=uv-\int v \ du\\
&=\big(\sqrt{2}x+1\big)\bigg(\frac{1}{6\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}\bigg)-\int \bigg(\frac{1}{6\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}\bigg)\big(\sqrt{2}\ dx\big)\\
&=\frac{1}{6\sqrt{2}}\big(\sqrt{2}x+1\big)\big(1+4\sqrt{2}x\big)^\frac{3}{2}-\frac{1}{6}\int \big(1+4\sqrt{2}x\big)^\frac{3}{2}\ dx\\
&=\frac{1}{6\sqrt{2}}\big(\sqrt{2}x+1\big)\big(1+4\sqrt{2}x\big)^\frac{3}{2}-\frac{1}{6}\int w^\frac{3}{2}\ \frac{dw}{4\sqrt{2}}\\
&=\frac{1}{6\sqrt{2}}\big(\sqrt{2}x+1\big)\big(1+4\sqrt{2}x\big)^\frac{3}{2}-\frac{1}{24\sqrt{2}}\int w^\frac{3}{2}\ dw\\
&=\frac{1}{6\sqrt{2}}\big(\sqrt{2}x+1\big)\big(1+4\sqrt{2}x\big)^\frac{3}{2}-\frac{1}{24\sqrt{2}}\times\frac{w^\frac{5}{2}}{\frac{5}{2}}\\
&=\frac{1}{6\sqrt{2}}\big(\sqrt{2}x+1\big)\big(1+4\sqrt{2}x\big)^\frac{3}{2}-\frac{2}{120\sqrt{2}}w^\frac{5}{2}\\
&=\frac{1}{6\sqrt{2}}\big(\sqrt{2}x+1\big)\big(1+4\sqrt{2}x\big)^\frac{3}{2}-\frac{1}{60\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{5}{2}\\
&=\frac{1}{60\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}\big(10\big(\sqrt{2}x+1\big)-\big(1+4\sqrt{2}x\big)\big)\\
&=\frac{1}{60\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}\big(10\sqrt{2}x+10-1-4\sqrt{2}x\big)\\
&=\frac{1}{60\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}\big(6\sqrt{2}x+9\big)\\
\end{align}$$
Work:
$$\begin{align}
V&=\int_0^\sqrt{2}\pi\sqrt{x^2+3\sqrt{2}x-\big(\sqrt{2}x+1\big)\sqrt{1+4\sqrt{2}x}+1}^2 \ dx\\
&=\int_0^\sqrt{2}\pi x^2+3\pi \sqrt{2}x-\pi \big(\sqrt{2}x+1\big)\sqrt{1+4\sqrt{2}x}+\pi \ dx\\
&=\int_0^\sqrt{2}\pi x^2 \ dx+\int_0^\sqrt{2}3\pi \sqrt{2}x \ dx-\int_0^\sqrt{2}\pi \big(\sqrt{2}x+1\big)\sqrt{1+4\sqrt{2}x} \ dx+\int_0^\sqrt{2}\pi \ dx\\
&=\pi\int_0^\sqrt{2}x^2 \ dx+3\pi \sqrt{2}\int_0^\sqrt{2}x \ dx-\pi \color{red}{\int_0^\sqrt{2}\big(\sqrt{2}x+1\big)\sqrt{1+4\sqrt{2}x} \ dx}+\pi\int_0^\sqrt{2}dx\\
&=\pi\bigg[\frac{x^3}{3}\bigg]_0^\sqrt{2}+3\pi \sqrt{2} \bigg[\frac{x^2}{2}\bigg]_0^\sqrt{2}-\pi \bigg[\frac{1}{60\sqrt{2}}\big(1+4\sqrt{2}x\big)^\frac{3}{2}\big(6\sqrt{2}x+9\big)\bigg]_0^\sqrt{2}+\pi[x]_0^\sqrt{2}\\
&=\pi\bigg[\frac{2\sqrt{2}}{3}-\frac{0}{3}\bigg]+3\pi\sqrt{2}\bigg[\frac{2}{2}-\frac{0}{2}\bigg]-\pi\bigg[\frac{(9)^\frac{3}{2}(21)}{60\sqrt{2}}-\frac{(1)^\frac{3}{2}(9)}{60\sqrt{2}}\bigg]+\pi\big[\sqrt{2}-0\big]\\
&=\pi\bigg[\frac{2\sqrt{2}}{3}\bigg]+3\pi\sqrt{2}[1]-\pi\bigg[\frac{558}{60\sqrt{2}}\bigg]+\pi\big[\sqrt{2}\big]\\
&=\frac{2}{3}\pi\sqrt{2}+3\pi\sqrt{2}-\frac{93}{20}\pi\sqrt{2}+\pi\sqrt{2}\\
&=\pi\sqrt{2}\bigg(\frac{40}{60}+\frac{180}{60}-\frac{279}{60}+\frac{60}{60}\bigg)\\
&=\pi\sqrt{2}\bigg(\frac{1}{60}\bigg)\\
&=\frac{\pi\sqrt{2}}{60}
\end{align}$$
Using the disk method, integrate $\pi r^2$ from $[0,\sqrt{2}]$ with Eqn. 8 plugged in for $r$ with respect to $x$.
TL;DR
Frankly, the question does not seem that complicated, and the answer of $\frac{\pi\sqrt{2}}{60}$ is definitely pretty simple. I have to believe that there is a more concise way of solving this problem.
All thoughts/answers welcome, thanks!
Best Answer
For $0 < x < 1,$ consider the line segment from $(x,x^2)$ to $(x,x).$ Rotated around the line $y = x,$ this produces a finite conical "hat" with slant height $x - x^2$ and base radius $(x - x^2)/\sqrt2,$ so it has surface area $\pi(x - x^2)^2/\sqrt2.$
The solid is composed of a nested stack of these conical "hats." The volume element between the "hat" at $x$ and the "hat" at $x + dx$ is $\frac\pi{\sqrt2}(x - x^2)^2 dx,$ so we integrate $$ \int_0^1 \frac\pi{\sqrt2}(x - x^2)^2 dx = \frac\pi{\sqrt2}\left[\frac{x^5}5 - \frac{x^4}2 + \frac{x^3}3\right]_0^1 = \frac{\pi\sqrt2}{60}. $$