The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the y-axis, rotated around x=36.

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The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the y-axis, rotated around x=36.

Mathematics
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we should get the same results if im thinking of it correctly
but just in case; think of this as a solid cylinder that we are removing the under part of the sqrt(x)
|dw:1327534530288:dw|

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so, cylindar itself; r=36, h=6; pi 36^2*6 is total area of this cylindar, now we gut out the area under the sqrt(x)
so it's like an upside down bowl
yes, and lets disc out the underside, i think it might integrate easier that way
ok after I integrated I got, 18x^2 - 2/3 * x^(3/2)
|dw:1327534763514:dw|
r = 36-y^2; when y=0, r=36; when y=6, r=0 that works in my head lol
V = int (pi [f(y)]^2) from 0 to 6 V = int (pi(36-y^2)^2) from 0 to 6, im just gonna paste this into wolfram to verify :)
so I couldn't leave the y=sqrt(x), but would have to convert it to y^2 instead?
yes, for disc; maybe shell is easier?
Gotcha! So your r should always equal 0 if a limit is plugged in?
not always, the radius changes as the integration moves across the interval; in this case, with the disc, the radius is determined by 36 over - y^2 back
but lets do shell and see if my thoughts improve lol
haha alright
the shell moves it radius from 0 to 36 as it does, the height is determined by sqrt(x); but i tend to like to move everything to the yaxis to get a clearer picture; so; sqrt(x+36) moves everything to the left by 36 and gets our rotation around the y axis
|dw:1327535297927:dw|
Ahh. and that isn't what we want!
this is another view of what we want; its better to me
\[V=2pi\int_{0}^{-36}x(\sqrt{x+36})dx\] i believe thats right;
sorry, but I gotta get; hope i havent messed it up too bad. good luck tho :)
thank you for your help!

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