liizzyliizz
  • liizzyliizz
Can someone help me with Cross sections please? -Region bounded by y=e^x +1 , y=8 and the y-axis cross sections are semi circles.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
|dw:1327356271299:dw|
anonymous
  • anonymous
Amistre64, can you hep me once you're done here please?
amistre64
  • amistre64
i can try

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liizzyliizz
  • liizzyliizz
o.O lol oh lord.
amistre64
  • amistre64
im assuming by cross section you are trying to find the area when the cross sections are semicircles?
liizzyliizz
  • liizzyliizz
Yes sir.
liizzyliizz
  • liizzyliizz
for the y=8 would hat be considered another line o.o its just that my teacher gave us a piece of paper and the information listed above is all that is written. So i've been afraid to do anything crazy because it is a project lol
liizzyliizz
  • liizzyliizz
that***
amistre64
  • amistre64
y=8 is a horizontal line thru y=8 yes
amistre64
  • amistre64
do you kow how the semicircles are crosssectioned in there? say flat part up hopefully?
amistre64
  • amistre64
|dw:1327359342185:dw|
liizzyliizz
  • liizzyliizz
That last part has me a bit thrown off, can you explain?
amistre64
  • amistre64
im trying to determine how the semicircles are orientated in the shaded area
amistre64
  • amistre64
with any luck the flat end is up and the rounded end down to make life simpler
liizzyliizz
  • liizzyliizz
ooh i see.
amistre64
  • amistre64
we can also move the graph and not change the area; lower it down so it begine at 0,0 - in effect we subtract 2 from our heights
amistre64
  • amistre64
y = e^x - 2 y = 6 would be the new way it would look but it would retain the same content
amistre64
  • amistre64
it might be good to know where these lines cross at so that we can set up boundaries to integrate by
amistre64
  • amistre64
e^x - 2 = 6 e^x = 8 x = ln(8) looks about right to me
amistre64
  • amistre64
|dw:1327360185033:dw|
amistre64
  • amistre64
the radius of each semicircle is then the distance from y=6 to y=e^x-2 or simply: R = 6-(e^x -2) = 6 -e^x +2 = 8 - e^x
amistre64
  • amistre64
got some numbers off in me head; e^x is normal ar y=1 +1 means we are at y=2 drop 2 means we are at e^x -1, not e^x -2 it bites getting old
amistre64
  • amistre64
R = 7 - e^x is our radiuses then
liizzyliizz
  • liizzyliizz
oh my lol. hmm.
amistre64
  • amistre64
:) its easier than it looks, i hope
liizzyliizz
  • liizzyliizz
This is like going over my head, lol I'll try it out, it doesn't look so hard, I guess I just get intimidated by graphs /
amistre64
  • amistre64
the graph is more of a picture to help focus your thought on; gives you something more concrete to play with than an abstract notion of integrations and such
liizzyliizz
  • liizzyliizz
It's just that I have to find someway to make 10 semicircles on my graph. so that has me nervous. lol I think the actually solving the integral is much easier.
liizzyliizz
  • liizzyliizz
But thank you anyway :D your help is always greatly appreciated !
amistre64
  • amistre64
10? ugh yeah, wish i could be more helpful :) good luck with it tho
liizzyliizz
  • liizzyliizz
Thank you very much. and lol yesss -__- cutting 10 little semi circles is not fun :c

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