the region enclosed by the graphs of y=e^(x/2), y=1, and x=ln3 is revolved about the x-axis. Find the volume of the solid generated??????? I NEED CALCULUS HELP PLEASE

- anonymous

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- amistre64

yay!! an easy one

- anonymous

easy? lol i would love your help then :)

- amistre64

lets see where these curves meet; and what our boundaries are

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## More answers

- amistre64

e^(x/2) = 1 when x=0 right?

- anonymous

yes

- amistre64

x = ln(3) is just a constant; we can find it on a calculator

- anonymous

yea 1.098

- amistre64

right; 1.09861...

- amistre64

let me draw a pic of my interpretation of the bonds then

- anonymous

i think they intersect at 0 and ln3 but i dont know how to integrate it

- amistre64

like this?

##### 1 Attachment

- amistre64

theres only two things to integrate here; from 0 to ln(3) right?

- anonymous

yes

- amistre64

lets find the volume of the top curve; then we will cut out the section below y=1ok

- anonymous

yes that would work but im suppose to do something with (pie)

- amistre64

pi {S} [e^(1/2(x))]^2 dx ; [0,ln(3)]

- amistre64

what is e^(x/2)^2 =? e^x right?

- anonymous

yes because they would cancel

- amistre64

the key here is to realize that we are adding up all the areas of circles that have been sliced;
the area of a circle = pi r^r it just so happens the r = f(x)

- amistre64

r^2 that is lol

- amistre64

so; [f(x)]^2 = e^(x/2)^2 = e^x
pi {S} e^x dx ; [0,ln(3)]
whats the integral of e^x?

- anonymous

the antiderivetive? its justs e^x

- amistre64

e^x is its own derivative; so its its own integral :) yes

- amistre64

pi [ e^(ln(3)) e^(0) ] = ?

- amistre64

typoed that

- amistre64

pi [ e^(ln(3)) - e^(0) ] = ?

- anonymous

an example my teacher did shows it being {S} (pi (e^x)-1) and then taking the antiderivetive of that

- amistre64

lets stay on one track here; well jump the track when we get to the end

- anonymous

ok i understand that i just dont understand where the minus 1 came from?

- amistre64

right now we are making a solid piece; we will then cut out the senter

- amistre64

pi [e^(ln3) - e^0] = ??

- anonymous

o so we r taking the whole volume minus the part we dont need which is the y=1

- amistre64

exaclty :) we can do it in parts or all togther; makes no difference

- anonymous

gotcha and i got 2 for that answer

- amistre64

2 or 2pi?

- anonymous

2 pi

- amistre64

good :)
now let integrate y = 1 from 0 to ln(3); or do you know a shortcut for this one?

- amistre64

pi {S} 1 dx ; [0,ln3] becomes?

- anonymous

pi (x)= pi (ln3)

- amistre64

good :)
2pi - ln(3) pi is your answer then

- amistre64

pi [ 2 - ln(3)] even lol

- anonymous

and that is what the back of the book says...you are good lol

- amistre64

toldja it was easy lol

- amistre64

adding up areas of circles; thats all

- anonymous

yea i just suck at it when my teacher tries to explain it..i may need more help in a few

- amistre64

im sure well be around

- anonymous

ok thank you :)

- anonymous

u still there? i have a quick question?

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