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anonymous
 4 years ago
The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the yaxis, rotated around x=36.
anonymous
 4 years ago
The region bounded above by the line y=6, below by the curve y=sqrt(x) and to the left by the yaxis, rotated around x=36.

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0we should get the same results if im thinking of it correctly

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0but just in case; think of this as a solid cylinder that we are removing the under part of the sqrt(x)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327534530288:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0so, 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so it's like an upside down bowl

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0yes, and lets disc out the underside, i think it might integrate easier that way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok after I integrated I got, 18x^2  2/3 * x^(3/2)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327534763514:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0r = 36y^2; when y=0, r=36; when y=6, r=0 that works in my head lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0V = int (pi [f(y)]^2) from 0 to 6 V = int (pi(36y^2)^2) from 0 to 6, im just gonna paste this into wolfram to verify :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so I couldn't leave the y=sqrt(x), but would have to convert it to y^2 instead?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0yes, for disc; maybe shell is easier?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Gotcha! So your r should always equal 0 if a limit is plugged in?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0not 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0but lets do shell and see if my thoughts improve lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0the 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327535297927:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ahh. and that isn't what we want!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0this is another view of what we want; its better to me

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[V=2pi\int_{0}^{36}x(\sqrt{x+36})dx\] i believe thats right;

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, but I gotta get; hope i havent messed it up too bad. good luck tho :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you for your help!
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