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I'd like to know how to setup the integral so that I can find the volume of the solid generated by revolving the region between: $y=x$ & $y=\sqrt{x}$ about the line $y=1$ (P.S. No need for a full answer. I just need the proper setup of the integral where I need every part of the integral identified for me to follow what's going on).

Part of the difficulty lies in my poor imagination so if the problem solver can link me to an online utility that can help me see this solid of revolution very clearly then I will be very thankful :)

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What are the boundaries on $x$? –  Henning Makholm Oct 9 '12 at 21:19
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the volume you want to compute doesn't change, if you shift everything one unit of length down and your rotation is executed around the x-axis. If you do this, it looks like the foolowing picture:

enter image description here

Let the blue function be $h(x)$ and let the pink funktion be $g(x)$.

As we did the shift, we get $g(x)=x-1$ and $h(x)= \sqrt{x}-1$.

Now use the formula for computing the volume of a rotation: $V=\pi\cdot\int_a^b (f(x))^2dx$:

Here, our volume is "the volume that we get, if we let blue curve rotate" minus "the volume that we get, if we let pink curve rotate" (and our boundaries are 0 and 1, as these are the solutions of the equation $h(x)=g(x)$):

$V=\pi\cdot\int_0^1 (g(x))^2dx-\pi\cdot\int_0^1 (h(x))^2dx$.

This is the volume you get, when you let the yellow area rotate around the x-axis, which is equal to the volume you are asking for, as previously mentioned.

I hope, I could help.

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