Question

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

Answer 1

yay!! an easy one

Answer 2

easy? lol i would love your help then :)

Answer 3

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

Answer 4

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

Answer 5

yes

Answer 6

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

Answer 7

yea 1.098

Answer 8

right; 1.09861...

Answer 9

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

Answer 10

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

Answer 11

like this?

Answer 12

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

Answer 13

yes

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

yes because they would cancel

Answer 19

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)

Answer 20

r^2 that is lol

Answer 21

so; [f(x)]^2 = e^(x/2)^2 = e^x

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

whats the integral of e^x?

Answer 22

the antiderivetive? its justs e^x

Answer 23

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

Answer 24

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

Answer 25

typoed that

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

gotcha and i got 2 for that answer

Answer 35

2 or 2pi?

Answer 36

2 pi

Answer 37

good :)

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

Answer 38

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

Answer 39

pi (x)= pi (ln3)

Answer 40

good :)

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

Answer 41

pi [ 2 - ln(3)] even lol

Answer 42

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

Answer 43

toldja it was easy lol

Answer 44

adding up areas of circles; thats all

Answer 45

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

Answer 46

im sure well be around

Answer 47

ok thank you :)

Answer 48

u still there? i have a quick question?