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I am finding a bounded volume. The question says

Find the volume of the solid obtained by rotating the region bounded by $y=x^{2}$ and $x=y^{2}$. Rotating about $y=1$

I got an intercept of those functions which was (1,1). I tried to use washer method then I got

$\pi\int_0^1[(x)^2-(\sqrt x )^2]dx$

and I took integral of the functions but my volume was not right number. I think my way to solve was not right. Could you post correct way to find the volume?

Thank you !

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up vote 3 down vote accepted

Your formula assumes that you are rotating about the $x$-axis (that is, about $y=0$), but you aren't; you are rotating about $y=1$. Also, you forgot that your other function was $y=x^2$ and not $y=x$ (the $(x)^2$ factor was probably meant to be $(x^2)^2$?).

If you have a thin strip of width $\Delta x$ that starts at the point $x_i$, then you have a washer with outer radius $1-x_i^2$ and inner radius $1-\sqrt{x_i}$ (draw a picture, and remember the axis of rotation). So the volume described by the washer is approximated by $$\pi\left( (1-x_i^2)^2 - (1-\sqrt{x_i})^2\right)\Delta x.$$ Take the Riemann sum, then limits as $\Delta x\to 0$ to turn it into an integral.

Added. In this case, the fact that you are rotating about $y=1$ instead of about $y=0$ turns out not to matter because of the symmetry involved; had you used $(x^2)^2$ instead of $(x)^2$ you would have gotten the right result. But you should nonetheless be careful and make sure you are computing on the basis of what you are doing, and not on the basis of formulas that make assumptions that you may not have in the case at hand.

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