Rotate an area around a diagonal line.

Question

I know how to find the volume of the figure formed when you rotate a $2$-dimensional area around a horizontal or vertical line, but what if it were a diagonal line instead?

For example:

Rotate the area between the curve $y=x^2$ and $y=x$ around the line $y=x$.

That should create a diagonal "football" shape, starting at the origin.

Anyways, the way I thought to go about the problem was to rotate the equation $45°$ to the right, and then solve the problem as if it were rotated around the x-axis. So I decided to convert $y=x^2$ to a polar equation, add $\frac{pi}{4}$ to $\theta$, and then turn it back into a rectangular equation. That worked, and it gave me this equation, which gives a diagonal parabola opening toward the first quadrant:

$\frac{\sqrt{2}}{2}(y+x) = \frac{1}{2}xy(x^2+y^2)$

Only problem is that I need to isolate $y$, such that I can solve this problem like an ordinary rotation problem, and that seems impossible. What should I do to solve the problem? I don't know much linear algebra, but could that be used to solve this instead?

Answer 1

Convert $y=x^2$ to a polar equation

$$r\sin\theta = r^2\cos^2 \theta$$ $$r\sin\theta = r^2\cos^2 \theta \rightarrow \hbox{ (add }\frac{\pi}{4}\hbox{ to }\theta\hbox{)}$$ $$r\sin (\theta + \frac{\pi}{4}) = r^2 \cos^2(\theta + \frac{\pi}{4})$$ $$\frac{\sqrt{2}}{2}r\left(\sin \theta + \cos \theta\right) = \frac{1}{2} \left(r\cos\theta -r\sin \theta\right)^2 \rightarrow\hbox{ (convert back) }$$ $$\frac{\sqrt{2}}{2}(y+x) = \frac{1}{2} (x-y)^2$$ which is not $\frac{\sqrt{2}}{2}(x+y)=\frac{1}{2}xy(x^2+y^2)$ anyway.
Thus, it must be a mistake somewhere :)
$\rho((x^2,x),y-x=0)$ can be done much simpler: $\rho=|x^2-x|\cdot \frac{\sqrt{2}}{2}$, so you just have to $\pi\int \rho^2(t) dt$ where $t=\frac{\sqrt{2}}{2}(x+y)=\frac{\sqrt{2}}{2}(x^2+x)$.
Edit:$\frac{\sqrt{2}}{2}(x+y)=\frac{1}{2}xy(x^2+y^2)$ itself leads to an irreducible cubic.

Answer 2

Since the centroid of the region is given by $(\bar{x}, \bar{y})=(\frac{1}{2},\frac{2}{5})$,

and the distance from $(\frac{1}{2},\frac{2}{5})$ to the line $y=x$ is given by $\rho=\frac{\lvert\frac{1}{2}-\frac{2}{5}\lvert}{\sqrt{2}}=\frac{1}{10\sqrt{2}}$,

$\displaystyle V=A(2\pi\rho)=\frac{1}{6}\bigg(2\pi\cdot\frac{1}{10\sqrt{2}}\bigg)=\frac{\pi\sqrt{2}}{60}$ using Pappus's theorem.

Answer 3

The line through $(t,t)$ which is perpendicular to $y=x$ is given by $y=-x+2t$,

so it intersects $y=x^2$ where $\displaystyle x=\frac{\sqrt{1+8t}-1}{2}$ and $\displaystyle y=-x+2t=\frac{4t+1-\sqrt{1+8t}}{2}$.

The distance from $(x,y)$ to the line $y=x$ is given by $\displaystyle R=\frac{|y-x|}{\sqrt{2}}=\frac{|2t+1-\sqrt{1+8t}|}{\sqrt{2}}$,

so $\displaystyle V=\int_0^1\pi R^2 ds=\int_0^1\pi\big(2t^2+6t+1-(2t+1)\sqrt{1+8t}\big)\sqrt{2}dt$

$\hspace{.3 in}\displaystyle=\pi\sqrt{2}\bigg(\left[\frac{2}{3}t^3+3t^2+t\right]_0^1-\int_0^1(2t+1)\sqrt{1+8t}dt\bigg)$

$\hspace{.3 in}\displaystyle=\pi\sqrt{2}\bigg(\frac{14}{3}-\int_1^3\big(\frac{1}{4}(u^2-1)+1\big)u\cdot\frac{1}{4}udu\bigg)=\pi\sqrt{2}\bigg(\frac{14}{3}-\frac{1}{16}\big[\frac{u^5}{5}+u^3\big]_1^3\bigg)$

$\hspace{.3 in}\displaystyle=\pi\sqrt{2}\bigg(\frac{14}{3}-\frac{1}{16}\cdot\frac{372}{5}\bigg)=\frac{\pi\sqrt{2}}{60}$