Csdrhrt

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.

$$begin{align}
color{gray}{y} &color{gray}{=} color{gray}{-x+2x_2}\
y &=x
end{align}$$
$$x=-x+2x_2$$
$$2x=2x_2$$
$$x_2=xtag{2}$$

$$begin{align}
y_2&=f(x_2)\
&=xtag{3}
end{align}$$

$$begin{align}
color{gray}{y} &color{gray}{=} color{gray}{-x_1+2x_2}\
color{gray}{y} &color{gray}{=} color{gray}{-x_1+2x}\
y &={x_1}^2
end{align}$$
$${x_1}^2=-x_1+2x$$
$$0=1{x_1}^2+1x_1+-2x$$
$$begin{align}
x_1&=frac{-1+sqrt{1^2-4(1)(-2x)}}{2(1)}\
&=frac{sqrt{1+8x}-1}{2}tag{4}
end{align}$$

$$begin{align}
y_1&=g(x_1)\
&=bigg(frac{sqrt{1+8x}-1}{2}bigg)^2tag{5}
end{align}$$
Find the variables in the distance formula as functions of $x$ (Eqns. 2-5 with derivations listed above them, respectively).

$$begin{align}
d&=sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\
&=sqrt{Bigg(x-frac{sqrt{1+8x}-1}{2}Bigg)^2+Bigg(x-bigg(frac{sqrt{1+8x}-1}{2}bigg)^2Bigg)^2}tag{6}
end{align}$$

Plug Eqns. 2-5 into the distance formula.

$$begin{align}
d&=sqrt{Bigg(x-frac{sqrt{1+8x}-1}{2}Bigg)^2+Bigg(x-bigg(frac{sqrt{1+8x}-1}{2}bigg)^2Bigg)^2}\
&=sqrt{Bigg(x-frac{sqrt{1+8x}-1}{2}Bigg)^2+Bigg(x-bigg(frac{(1+8x)-2sqrt{1+8x}+1}{4}bigg)Bigg)^2}\
&=sqrt{Bigg(x-frac{sqrt{1+8x}-1}{2}Bigg)^2+Bigg(x-bigg(frac{2+8x-2sqrt{1+8x}}{4}bigg)Bigg)^2}\
&=sqrt{Bigg(frac{2x}{2}-frac{sqrt{1+8x}-1}{2}Bigg)^2+Bigg(frac{2x}{2}-frac{1+4x-sqrt{1+8x}}{2}Bigg)^2}\
&=sqrt{Bigg(frac{2x-sqrt{1+8x}+1}{2}Bigg)^2+Bigg(frac{2x-1-4x+sqrt{1+8x}}{2}Bigg)^2}\
&=sqrt{Bigg(frac{1+2x-sqrt{1+8x}}{2}Bigg)^2+Bigg(frac{-1-2x+sqrt{1+8x}}{2}Bigg)^2}\
&=sqrt{2Bigg(frac{1+2x-sqrt{1+8x}}{2}Bigg)^2}\
&=sqrt{2Bigg(frac{1+4x^2+(1+8x)+4x-2sqrt{1+8x}-4xsqrt{1+8x}}{4}Bigg)}\
&=sqrt{frac{4x^2+12x-(4x+2)sqrt{1+8x}+2}{2}}\
&=sqrt{2x^2+6x-(2x+1)sqrt{1+8x}+1}tag{7}
end{align}$$

Simplify Eqn. 6.

$$begin{align}
r&=sqrt{2bigg(frac{x}{sqrt{2}}bigg)^2+6bigg(frac{x}{sqrt{2}}bigg)-bigg(2bigg(frac{x}{sqrt{2}}bigg)+1bigg)sqrt{1+8bigg(frac{x}{sqrt{2}}bigg)}+1}\
&=sqrt{2bigg(frac{x^2}{2}bigg)+6bigg(frac{sqrt{2}x}{2}bigg)-bigg(2bigg(frac{sqrt{2}x}{2}bigg)+1bigg)sqrt{1+8bigg(frac{sqrt{2}x}{2}bigg)}+1}\
&=sqrt{x^2+3sqrt{2}x-big(sqrt{2}x+1big)sqrt{1+4sqrt{2}x}+1}tag{8}
end{align}$$

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):
$$color{red}{intbig(sqrt{2}x+1big)sqrt{1+4sqrt{2}x} dx}$$
$$
begin{array}{|c|}
hline
mathbf{u=sqrt{2}x+1}, mathbf{dv=sqrt{1+4sqrt{2}x} dx}\
hline
begin{array}{c|c}
frac{du}{dx}=sqrt{2} & int dv=intsqrt{1+4sqrt{2}x} dx\
mathbf{du=sqrt{2} dx} & v=intsqrt{w} frac{dw}{4sqrt{2}}\
& v=frac{1}{4sqrt{2}}intsqrt{w} dw\
& v=frac{1}{4sqrt{2}}timesfrac{w^frac{3}{2}}{frac{3}{2}}\
& v=frac{2}{12sqrt{2}}w^frac{3}{2}\
& mathbf{v=frac{1}{6sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}}\
end{array}\
hline
end{array}
$$

$$begin{align}
&=uv-int v du\
&=big(sqrt{2}x+1big)bigg(frac{1}{6sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}bigg)-int bigg(frac{1}{6sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}bigg)big(sqrt{2} dxbig)\
&=frac{1}{6sqrt{2}}big(sqrt{2}x+1big)big(1+4sqrt{2}xbig)^frac{3}{2}-frac{1}{6}int big(1+4sqrt{2}xbig)^frac{3}{2} dx\
&=frac{1}{6sqrt{2}}big(sqrt{2}x+1big)big(1+4sqrt{2}xbig)^frac{3}{2}-frac{1}{6}int w^frac{3}{2} frac{dw}{4sqrt{2}}\
&=frac{1}{6sqrt{2}}big(sqrt{2}x+1big)big(1+4sqrt{2}xbig)^frac{3}{2}-frac{1}{24sqrt{2}}int w^frac{3}{2} dw\
&=frac{1}{6sqrt{2}}big(sqrt{2}x+1big)big(1+4sqrt{2}xbig)^frac{3}{2}-frac{1}{24sqrt{2}}timesfrac{w^frac{5}{2}}{frac{5}{2}}\
&=frac{1}{6sqrt{2}}big(sqrt{2}x+1big)big(1+4sqrt{2}xbig)^frac{3}{2}-frac{2}{120sqrt{2}}w^frac{5}{2}\
&=frac{1}{6sqrt{2}}big(sqrt{2}x+1big)big(1+4sqrt{2}xbig)^frac{3}{2}-frac{1}{60sqrt{2}}big(1+4sqrt{2}xbig)^frac{5}{2}\
&=frac{1}{60sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}big(10big(sqrt{2}x+1big)-big(1+4sqrt{2}xbig)big)\
&=frac{1}{60sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}big(10sqrt{2}x+10-1-4sqrt{2}xbig)\
&=frac{1}{60sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}big(6sqrt{2}x+9big)\
end{align}$$

Work:
$$begin{align}
V&=int_0^sqrt{2}pisqrt{x^2+3sqrt{2}x-big(sqrt{2}x+1big)sqrt{1+4sqrt{2}x}+1}^2 dx\
&=int_0^sqrt{2}pi x^2+3pi sqrt{2}x-pi big(sqrt{2}x+1big)sqrt{1+4sqrt{2}x}+pi dx\
&=int_0^sqrt{2}pi x^2 dx+int_0^sqrt{2}3pi sqrt{2}x dx-int_0^sqrt{2}pi big(sqrt{2}x+1big)sqrt{1+4sqrt{2}x} dx+int_0^sqrt{2}pi dx\
&=piint_0^sqrt{2}x^2 dx+3pi sqrt{2}int_0^sqrt{2}x dx-pi color{red}{int_0^sqrt{2}big(sqrt{2}x+1big)sqrt{1+4sqrt{2}x} dx}+piint_0^sqrt{2}dx\
&=pibigg[frac{x^3}{3}bigg]_0^sqrt{2}+3pi sqrt{2} bigg[frac{x^2}{2}bigg]_0^sqrt{2}-pi bigg[frac{1}{60sqrt{2}}big(1+4sqrt{2}xbig)^frac{3}{2}big(6sqrt{2}x+9big)bigg]_0^sqrt{2}+pi[x]_0^sqrt{2}\
&=pibigg[frac{2sqrt{2}}{3}-frac{0}{3}bigg]+3pisqrt{2}bigg[frac{2}{2}-frac{0}{2}bigg]-pibigg[frac{(9)^frac{3}{2}(21)}{60sqrt{2}}-frac{(1)^frac{3}{2}(9)}{60sqrt{2}}bigg]+pibig[sqrt{2}-0big]\
&=pibigg[frac{2sqrt{2}}{3}bigg]+3pisqrt{2}[1]-pibigg[frac{558}{60sqrt{2}}bigg]+pibig[sqrt{2}big]\
&=frac{2}{3}pisqrt{2}+3pisqrt{2}-frac{93}{20}pisqrt{2}+pisqrt{2}\
&=pisqrt{2}bigg(frac{40}{60}+frac{180}{60}-frac{279}{60}+frac{60}{60}bigg)\
&=pisqrt{2}bigg(frac{1}{60}bigg)\
&=frac{pisqrt{2}}{60}
end{align}$$

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 $frac{pisqrt{2}}{60}$ is definitely pretty simple. I have to believe that there is a more concise way of solving this problem.

All thoughts/answers welcome, thanks!

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 $fracpi{sqrt2}(x - x^2)^2 dx,$ so we integrate
$$
int_0^1 fracpi{sqrt2}(x - x^2)^2 dx
= fracpi{sqrt2}left[frac{x^5}5 - frac{x^4}2 + frac{x^3}3right]_0^1
= frac{pisqrt2}{60}.
$$

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

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

integral.

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

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

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

How about using the rotation matrix $begin{pmatrix}cosfrac{pi}4&-sinfrac{pi}4\sinfrac{pi}4&cosfrac{pi}4end{pmatrix}$ 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: $y=frac{-(2x+sqrt2)pmsqrt{2+8sqrt2 x}}2=frac{-2x-sqrt2pmsqrt2 sqrt{1+4sqrt2x}}2$.

So now we need to integrate. We need $piint_0^{sqrt2}y^2operatorname dx$ 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.