Find the volume obtained by totating the area formed by $y=x$ and $y=\sqrt{x}$ about $y=1$

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The questions asks us to find the volume of solid formed when the area between $y=x$ and $y=\sqrt{x}$ is rotated about the line $y=1$.

I understand that a cone is formed. Now, to find the volume, I integrate from $0$ to $2$, $f(y)=\pi(y^2-y^4)$. (I understand that a ring is formed when we take a cross section, therefore, area of outer radius minus inner radius).

Where am I going wrong? I am getting a wrong answer.

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Using the washer method, you would integrate in terms of $x$ from 0 to 1, with the larger radius being $1-x$ and the smaller radius being $1-\sqrt x$. This gives the equation $$ \int_0^1 \pi ((1-x)^2-(1-\sqrt x)^2)dx $$