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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