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 :)

enter image description here

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).

enter image description here

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\tag{1}$$ 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.

enter image description here 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 $x$.


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!

  • $\begingroup$ I don't want to sound discouraging, but this looks like more of an essay or an article than a question. $\endgroup$ – mathreadler Dec 17 '18 at 22:17
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    $\begingroup$ @mathreadler Not at all! Generally, I’ve been told on SE to show things that I have already tried. So I figure that, even though this is long, it helps people understand where I am and what I’m thinking. Maybe that’s just me... $\endgroup$ – Shady Puck Dec 17 '18 at 22:36

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 $$ \int_0^1 \frac\pi{\sqrt2}(x - x^2)^2 dx = \frac\pi{\sqrt2}\left[\frac{x^5}5 - \frac{x^4}2 + \frac{x^3}3\right]_0^1 = \frac{\pi\sqrt2}{60}. $$

  • $\begingroup$ I suppose if your starting point did not take into account just how amazing and robust the subject of Calculus is, you should put on a dunce cap! (+1) $\endgroup$ – CopyPasteIt Dec 19 '18 at 12:11
  • $\begingroup$ An incredibly simple and efficient method. Thank you very much! I just applied this method to the problem of the more general case ($y=ax^2$ rotated about $y=bx$). It made finding my final answer a breeze. +1 & accepted! One question: Do you mean the lateral area of the cone instead of the surface area? Because that's what your math seemed to say and what made more sense. $\endgroup$ – Shady Puck Jan 1 at 19:37
  • $\begingroup$ The lateral area is what I meant; another way to describe it is what is left after you remove the circular bottom surface area from a finite right circular cone. $\endgroup$ – David K Jan 1 at 20:16

Yes, using the disk method and you wind up having to contend with some unwieldy calculations. Use cylindrical shells to make your life easier. Here are some equations/algebra that will be needed:

The distance between the line $y = x + c$ and the line $y =x$ is equal to $\frac{|c|}{\sqrt 2}$.

If both $y = x + c$ and $y = x^2$ are true, then

$\tag 1 x^2 -x -c = 0$

Using the quadratic formula,

$$\tag 2 x_0 = \frac{1 - \sqrt{1 + 4c}}{2} \text{ and } x_1 = \frac{1 + \sqrt{1 + 4c}}{2} $$

The distance between $(x_0, x_0+c)$ and $(x_1, x_1+c)$ is given by $\sqrt {2}\,\sqrt {1+4c}$.

Letting $c$ vary, it ranges from $0$ to $-\frac{1}{4}$. Using a change of variable, set $u = -\frac{c}{\sqrt 2}$, so that

$\tag 3 u \text{ varies from } 0 \text{ to } \frac{\sqrt 2}{8}$

You are a few steps away from setting up your

$$\quad \int_{u=0}^{\frac{\sqrt 2}{8}} du$$


I worked it out using Wolfram and the volume is $0.074048\dots$, which is equal to $\frac{\pi\sqrt{2}}{60}$.

Integral Answer (use cursor as a 'spoiler'):

$$\quad 2 \pi \,\sqrt 2 \int_{u=0}^{\frac{\sqrt 2}{8}} u \sqrt{1 + 4\sqrt2 \,u} \;du$$

  • $\begingroup$ This is a great adaptation of the shell method! I found this very useful, and it helped confirm my equation for the general case, $y=ax^2$ rotated about $y=bx$. (+1) Truly a very clever procedure, except that I feel I have to accept the one by @DavidK based on its exceptional simplicity. Still, this was very helpful. Thank you very much! $\endgroup$ – Shady Puck Jan 1 at 19:41
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    $\begingroup$ @ShadyPuck And thank you for the question - the work that you put into it was an inspiration to look at it from other angles. $\endgroup$ – CopyPasteIt Jan 1 at 19:50

How about using the rotation matrix $\begin{pmatrix}\cos\frac{\pi}4&-\sin\frac{\pi}4\\\sin\frac{\pi}4&\cos\frac{\pi}4\end{pmatrix}$ to rotate $(x,y)$, and then you can integrate along $x$.

Then I get the equation $y^2+x^2-\sqrt2 x+\sqrt2 y+2xy=0$.

To solve for y we can use the quadratic formula: $y=\frac{-(2x+\sqrt2)\pm\sqrt{2+8\sqrt2 x}}2=\frac{-2x-\sqrt2\pm\sqrt2 \sqrt{1+4\sqrt2x}}2$.

So now we need to integrate. We need $\pi\int_0^{\sqrt2}y^2\operatorname dx$ and this can be done by integrating by parts, as you noted.

I used an integral calculator (too lazy) to check this and your answer appears to be correct.


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