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In calculus 2 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.

Thanks.

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  • $\begingroup$ The substitution $y':=y-x$ and $x':=x$ makes the axis of rotation horizontal. $\endgroup$ – Inactive - Objecting Extremism Sep 11 '16 at 10:29
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hint .....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.

$$\binom xy=\frac{1}{\sqrt2}\left(\begin{matrix}1&1\\-1&1\end{matrix}\right)\binom t{t^2}$$

Now consider the volume as $$\pi\int_{t=0}^1 y^2\frac{dx}{dt}dt$$

And it should all work out nicely...

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Another approach using a ''modified'' disk method.

The distance of a point $(x,y)=(x,x^2)$ 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 $$ A=\pi r^2=\pi\frac{(x-x^2)^2}{2} $$ and the volume of a disk of length $\delta x=\sqrt{2}dx$ along the axis of rotation is $$ dV=\frac{\pi}{2} \int_0^1(x-x^2)^2 \delta x=\frac{\pi}{\sqrt{2}} \int_0^1(x-x^2)^2 dx $$ and the volume of the solid of revolution is the integral: $$ V=\frac{\pi}{\sqrt{2}} \int_0^1(x-x^2)^2 dx $$

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  • $\begingroup$ Don't you need to adjust the $dx$ term because the thickness of the element is not $\delta x$ but $\sqrt{2}\delta x$? $\endgroup$ – David Quinn Sep 11 '16 at 14:22
  • $\begingroup$ @DavidQuinn: Yes, thank you ! You are right and I edit my answer, but my result seems different from your.... or I'm wrong? $\endgroup$ – Emilio Novati Sep 11 '16 at 14:53
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The substitution $y':=y-x$ and $x':=x$ makes the axis of rotation horizontal, and the bounds become $$y'=0\qquad\text{ and }\qquad y'=x'^2-x'.$$

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