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.

- liizzyliizz

- chestercat

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- amistre64

|dw:1327356271299:dw|

- anonymous

Amistre64, can you hep me once you're done here please?

- amistre64

i can try

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## More answers

- liizzyliizz

o.O lol oh lord.

- amistre64

im assuming by cross section you are trying to find the area when the cross sections are semicircles?

- liizzyliizz

Yes sir.

- 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

that***

- amistre64

y=8 is a horizontal line thru y=8 yes

- amistre64

do you kow how the semicircles are crosssectioned in there? say flat part up hopefully?

- amistre64

|dw:1327359342185:dw|

- liizzyliizz

That last part has me a bit thrown off, can you explain?

- amistre64

im trying to determine how the semicircles are orientated in the shaded area

- amistre64

with any luck the flat end is up and the rounded end down to make life simpler

- liizzyliizz

ooh i see.

- 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

y = e^x - 2
y = 6
would be the new way it would look but it would retain the same content

- amistre64

it might be good to know where these lines cross at so that we can set up boundaries to integrate by

- amistre64

e^x - 2 = 6
e^x = 8
x = ln(8) looks about right to me

- amistre64

|dw:1327360185033:dw|

- 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

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

R = 7 - e^x is our radiuses then

- liizzyliizz

oh my lol. hmm.

- amistre64

:) its easier than it looks, i hope

- 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

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

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

But thank you anyway :D your help is always greatly appreciated !

- amistre64

10? ugh
yeah, wish i could be more helpful :)
good luck with it tho

- liizzyliizz

Thank you very much. and lol yesss -__- cutting 10 little semi circles is not fun :c

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