Let R be the region of the plane bounded by the curves y = 2^x and 6x − y = 8. The region R is rotated around the y-axis. Use the washer method to find an integral
whose value is the volume of the solid that you get.
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OpenStudy (zarkon):
where are you stuck?
OpenStudy (anonymous):
from the start :(
OpenStudy (zarkon):
can you find where the two curves intersect?
OpenStudy (anonymous):
no i need help
OpenStudy (zarkon):
y = 2^x and 6x − y = 8
6x − 2^x = 8
solve for x
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OpenStudy (anonymous):
x = (y+8)/6 ?
OpenStudy (zarkon):
no
OpenStudy (anonymous):
help i don't know?
OpenStudy (zarkon):
you need to solve \[ 6x − 2^x = 8\] for x
OpenStudy (zarkon):
you should get 2 numbers
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OpenStudy (anonymous):
I don't know how
OpenStudy (zarkon):
you might be able to do it by inspection
OpenStudy (anonymous):
what do i do
OpenStudy (zarkon):
you either use a calculator or a computer...or just look at the equation and notice that the solutions are 'nice' numbers
OpenStudy (anonymous):
so where do i go from 6x-2^x = 8
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OpenStudy (anonymous):
you can't combine them its 2 separate equations........y = 2^x and 6x-y=8
OpenStudy (zarkon):
there is no algebraic way of solving this type of problem...use my privious post
OpenStudy (anonymous):
idk how to even use your previous step
OpenStudy (zarkon):
yes you can...to find where they cross each other
OpenStudy (anonymous):
shouldn't it be x = log base 2 (y)
x = (y+8)/6
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OpenStudy (anonymous):
other than that idk what to do
OpenStudy (zarkon):
I find it easier to see the solution for x first ...then it is easy to get the values of y
OpenStudy (anonymous):
so can you do it for me since i have no idea what im doing?
OpenStudy (zarkon):
notice that x=2 and x=4 satisfy the above equation
OpenStudy (anonymous):
ok where
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OpenStudy (anonymous):
oh yes
OpenStudy (zarkon):
plugging into \(y=2^x\) gives us that
y=4 and 16
OpenStudy (zarkon):
so your integral will go from 4 to 16
OpenStudy (anonymous):
x=4 you mean?
OpenStudy (zarkon):
\(2^2=4\)
\(2^4=16\)
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OpenStudy (anonymous):
yes
OpenStudy (zarkon):
\[2\le x\le 4\Rightarrow 4\le y\le 16\]
OpenStudy (zarkon):
now just set up your integral
OpenStudy (anonymous):
i need your help with that
OpenStudy (zarkon):
you will have something of the form
\[\pi\int\limits_{a}^{b}\left(f(y)^2-g(y)^2\right)dy\]
where \(f(y)\) is the 'outer' function and \(g(y)\) in the inner function
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OpenStudy (anonymous):
why does the integral go from 4 to 16?
OpenStudy (anonymous):
and isnt it 2pi
OpenStudy (zarkon):
that would be for the method of cylindrical shells
OpenStudy (zarkon):
you said to use the 'washer method'
OpenStudy (anonymous):
so whats it gonna look like can you show me
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OpenStudy (zarkon):
you tell me
OpenStudy (anonymous):
with everything plugged in
OpenStudy (zarkon):
I've done all the work here...I'm done...your move
OpenStudy (anonymous):
But i dont know what it is?
OpenStudy (zarkon):
then you might want to look at some examples in your book
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OpenStudy (anonymous):
I already tried that why do you think i came here ?
OpenStudy (zarkon):
I'm here to help...not to teach the entire topic...I teach during the day...not at night
OpenStudy (anonymous):
so F(y) is 16?
OpenStudy (zarkon):
no
OpenStudy (anonymous):
i have no clue
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