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Mathematics 14 Online
OpenStudy (anonymous):

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.

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

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

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

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

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

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):

\(2^2=4\) \(2^4=16\)

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

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

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

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

OpenStudy (zarkon):

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