MasterChief
  • MasterChief
Volume of a graph (Washer)
Mathematics
schrodinger
  • schrodinger
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MasterChief
  • MasterChief
Find the volume of \[y=\sqrt{x}\] about x=4
MasterChief
  • MasterChief
I started with \[x=y^2\] so would that be the outer radius?
SolomonZelman
  • SolomonZelman
|dw:1441595115874:dw|roughly

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SolomonZelman
  • SolomonZelman
x=y² is not the same as y=√x because it will have a twice larger volume
MasterChief
  • MasterChief
I'm sorry I forgot to mention it is bounded by y=0 so just quadrant 1
SolomonZelman
  • SolomonZelman
oh, and then if it is rotated about x=4, then I will assume that this is where the √x region ends at?
MasterChief
  • MasterChief
Yes
SolomonZelman
  • SolomonZelman
Oh, ok, so you know that your limits of integration are from 0 to 2. \(\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)dy }\)
SolomonZelman
  • SolomonZelman
So, you can tell that your radius is y², and it is from y=0 to y=2. THis is what I would do.
SolomonZelman
  • SolomonZelman
I missed it should e y^4
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)^2dy }\) because radius squared. Sorry
SolomonZelman
  • SolomonZelman
So that is just an integral of \(\pi\)y\(^4\) from y=0 to y=2.
MasterChief
  • MasterChief
Wow I tried this the first time and I guess I integrated incorrectly so I was so confused!
MasterChief
  • MasterChief
Thank you
SolomonZelman
  • SolomonZelman
you can put integral into wolfram. what is important is to get a good practice of making a setup of the integral for volume. integration you know already....
MasterChief
  • MasterChief
Absolutely
SolomonZelman
  • SolomonZelman
32π/5 is what i got. (want to know how to make a π • ÷ × √ with no latex?)
MasterChief
  • MasterChief
Sure!
DanJS
  • DanJS
here are some probs with the solns, pg3 has a nice little summary
MasterChief
  • MasterChief
That will definitely be useful, thank you
SolomonZelman
  • SolomonZelman
I am glitching a bit
SolomonZelman
  • SolomonZelman
Ok, so here is going to be a short guide. The algorithm is: \(1)\) Click ALT and hold it \(2)\) Click the "Number Code" (on the numberpad on the right of the keyboard- if you got one) \(3)\) Release the ALT ------------------------------ Number code Result 0, 2, 1, 5 × 2, 5, 1 √ 7 • 2, 4, 6 ÷ 2, 2, 7 π
SolomonZelman
  • SolomonZelman
there are some others too.... but these are useful examples
MasterChief
  • MasterChief
So if it was x=6 would the outer radius be 2+y^2 and the inner radius be 2?
SolomonZelman
  • SolomonZelman
if it was x=6, |dw:1441596092489:dw|
MasterChief
  • MasterChief
about x=6 I meant
SolomonZelman
  • SolomonZelman
Do you mean, if it was a region of y^2 bound by y=0 and x=4, but rotated about x=6?
MasterChief
  • MasterChief
Yes
SolomonZelman
  • SolomonZelman
Yes, then your radius is y²+2
SolomonZelman
  • SolomonZelman
and if you rotated in a same case about y=4+c then your radius is y²+c
MasterChief
  • MasterChief
I thought I had to subtract the inner area?
SolomonZelman
  • SolomonZelman
yeah, my bad, I am overheating let me think
MasterChief
  • MasterChief
|dw:1441596258832:dw|
MasterChief
  • MasterChief
Haha no problem
SolomonZelman
  • SolomonZelman
|dw:1441596426856:dw|
MasterChief
  • MasterChief
For this problem I can only use washer
SolomonZelman
  • SolomonZelman
Oh, you can do it with respect to x, and do f(x)-g(x)
SolomonZelman
  • SolomonZelman
But, shell is also good.
SolomonZelman
  • SolomonZelman
it is a matter of preference
MasterChief
  • MasterChief
Would f(x) be y^2+2 and g(x) be 2?
MasterChief
  • MasterChief
\[\int\limits_{0}^{2} (2+y^2)^2-(2)^2\] ?
SolomonZelman
  • SolomonZelman
lets rvw the washer with x's
SolomonZelman
  • SolomonZelman
|dw:1441596741353:dw|
SolomonZelman
  • SolomonZelman
of that integral should say π INTEGRAL f(x)²-g(x)² dx
SolomonZelman
  • SolomonZelman
from a to b
MasterChief
  • MasterChief
Yes I get that
MasterChief
  • MasterChief
I just don't understand what g(x) would be in this situation
SolomonZelman
  • SolomonZelman
and with y, you get the same thing|dw:1441596949137:dw|
SolomonZelman
  • SolomonZelman
the area of the cylinder with h=2, r=2 is 8π
SolomonZelman
  • SolomonZelman
Or you can say that g is 2.
SolomonZelman
  • SolomonZelman
if instead of 4, you had some z(x) boundary for the region, then the radius for g would be 6-z(x) (of course from y=0 to y=2)
MasterChief
  • MasterChief
Okay I think I understand this better now thank you
SolomonZelman
  • SolomonZelman
Anytime... thank you for refreshing me on these rotations:)
SolomonZelman
  • SolomonZelman
I mean, I really should tell you that Shell method rocks in so many cases, so try to use that as well. in any case, good luck!
MasterChief
  • MasterChief
Actually, did I plug this in incorrectly? http://www.wolframalpha.com/input/?i=integrate+from+0+to+2+pi%28%282%2By%5E2%29%5E2-4%29
MasterChief
  • MasterChief
The answer is 192π/5
SolomonZelman
  • SolomonZelman
for the problem where you want to know about the redion of y62 bound by y=0 and x=4, rotated about x=6....?
SolomonZelman
  • SolomonZelman
region*
SolomonZelman
  • SolomonZelman
y^2 (not y62)
MasterChief
  • MasterChief
yes
SolomonZelman
  • SolomonZelman
|dw:1441597714083:dw|
SolomonZelman
  • SolomonZelman
I actually do not get why the answer isn't what you got in wolfram.
MasterChief
  • MasterChief
hmmm
SolomonZelman
  • SolomonZelman
you can take the whole volume of radius y^2+2 with limits of y=0 to y=2, and subtract 8π cylinder in the middle, and you get precisely the same.
SolomonZelman
  • SolomonZelman
I have to go, it is almost 12am in my location. maybe I was answering a wrong question idk, but for what I asked, it should be 256π/15
SolomonZelman
  • SolomonZelman
Maybe my brain just shot down xD
MasterChief
  • MasterChief
Hahaha
MasterChief
  • MasterChief
But I really appreciated your time so thank you
SolomonZelman
  • SolomonZelman
I will look at it when I have time. Whatever I can do with my little knowledge:) gtg c(u)
MasterChief
  • MasterChief
Bye!

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