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2013-09-01 01:42 6002139 Anonymous (1375333975079.jpg 1024x686 624kB)
I've been working on this problem for 4 hours.. help me out here PLEASE Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line y = x^3, y = 0, x = 1; rotate about the line x = 2 the textbook answer is 3pi/5 but I can only get that answer if I am rotating it about the line x = 1. that is the radius is y^(1/3). But if you look at the actual radius its more like 2 - y^(1/3)

5 min later 6002146 Anonymous
anyone?

7 min later 6002153 Anonymous
>>6002146 give me a sec

7 min later 6002154 Anonymous
herro?

17 min later 6002165 Anonymous (hmmm.png 687x458 26kB)
Not OP, but wondering if I set this up wrong. >inb4 underageb& cant into calculus

21 min later 6002171 Anonymous
>>6002153 still workin?

24 min later 6002175 Anonymous
>>6002171 no got this >>6002165 but unless your book is lying to you, we are waiting for a charitable anon to come and help both of us

29 min later 6002185 Anonymous
>>6002175 OP here, looks like you set it up right, unless you meant that 1 to not be there? don't really get what you're asking?

36 min later 6002204 Anonymous
Your textbook is correct. The correct formula is: \int_0^1 \pi \left( \left(2 - y^{1 \over 3} \right)^2 -1\right) \mathrm{d}y = \frac{3}{5} \pi Just draw a pic for problems like this.

41 min later 6002218 Anonymous
>>6002204 why is there a -1?

43 min later 6002226 Anonymous
>>6002218 you are subtracting the volume of the inner cylinder, pi(1^2). the pi gets pulled out in front. >>6002204 why are we using the cube root of y and integrating with respect to y? why doesn't the one above work?

44 min later 6002227 Anonymous
>>6002204 wow just figured it out thanks a lot, just wow now i feel stupid

48 min later 6002236 Anonymous
The volume is \int_0^1 A(y) \mathrm{d}y Where A(y) is the area at the cross section at height y.

2.063 0.037