# Find the volume of the region bounded by y=x and y=x^2, but rotated about the equation of y=x.

Volume of Solid of Revolution about an equation
I learned about the disk and shell method for finding a volume of a solid of revolution. For my question I don't think I can use either method directly. Here is my question:
Find the volume of the region bounded by $y=x$ and $y={x}^{2}$, but rotated about the equation of $y=x$.
Here the axis of revolution is not a vertical or horizontal line, but rather the equation $y=x$. One idea I had was to convert this diagonal line into a horizontal or vertical one, but I can't see to do this. Maybe polar coordinates might be useful, but I'm not sure.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

emarisidie6
Step 1
Another approach using a ''modified'' disk method.
The distance of a point $\left(x,y\right)=\left(x,{x}^{2}\right)$ from the line $y=x$ is:
$r=\frac{|x-{x}^{2}|}{\sqrt{2}}$
so the area of a circle orthogonal to the axis of rotation is
$A=\pi {r}^{2}=\pi \frac{\left(x-{x}^{2}{\right)}^{2}}{2}$
Step 2
And the volume of a disk of length $\delta x=\sqrt{2}dx$ along the axis of rotation is
$dV=\frac{\pi }{2}{\int }_{0}^{1}\left(x-{x}^{2}{\right)}^{2}\delta x=\frac{\pi }{\sqrt{2}}{\int }_{0}^{1}\left(x-{x}^{2}{\right)}^{2}dx$
And the volume of the solid of revolution is the integral:
$V=\frac{\pi }{\sqrt{2}}{\int }_{0}^{1}\left(x-{x}^{2}{\right)}^{2}dx$
###### Did you like this example?
rialsv
Step 1
Start by writing the curve in parametric form, so $x=t,y={t}^{2}$. Now use a matrix to rotate the curve by 45 degrees clockwise so that the new curve is also given parametrically.
$\left(\genfrac{}{}{0}{}{x}{y}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)\left(\genfrac{}{}{0}{}{t}{{t}^{2}}\right)$
Step 2
Now consider the volume as
$\pi {\int }_{t=0}^{1}{y}^{2}\frac{dx}{dt}dt$