Integrate Volumes: Revolve Region Bounded by x=0, x=1, y=0, y=x^5

physicsnewb7
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Homework Statement


Revolve the region bounded by x=0, x=1, y=0 and y=x^5 about the y-axis use shells to find the volume

I know how to set up the integral I just don't know where I'm integrating from. Is it from 0 to 1?
 
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Hi physicsnewb7! :smile:

You can use "horizontal" strips, to give you disc shells, or "vertical" strips, to give you cylindrical shells …

in either case, use the endpoints of the strip. :wink:
 
thank you tiny tim
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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