Question

double integral

Answer 1

What do you need answered??

Answer 2

trying to use equation to write question but not working

Answer 3

yeah, they blame it on user error lol

Answer 4

hmm i click and nothing

Answer 5

{SS} dx.dy ; [a,b] [c,d] works

Answer 6

hey amister!!!

Answer 7

hi ;)

Answer 8

i saw a cute pic and figured... why not help lol

Answer 9

when you switch from 1 to 2 for dx and 0 to lnx for dy
you end up with 0 to ln2 and e^y to 2

Answer 10

it's impossible to do this without equation lol

Answer 11

is it dx.dy order ; or dy.dx order?

Answer 12

but shouldn't dy switch to 0 to 2?

Answer 13

going from dydx to dxdy

Answer 14

so your re adjusting the intervals right?

Answer 15

answer book has it 0 to ln2

Answer 16

switching the order

Answer 17

{S{Sdy}dx} ; x interval = [1,2] ; y interval = [0, ln(x)] originally right?

Answer 18

yeah

Answer 19

then teh easiest thing to do is graph it; with the lines y=0, y=ln(x), x=1, x=2

Answer 20

when you switch shouldn't it be [0, 2] and [e^y, 0]

Answer 21

i mean [0, 2] and [e^y, 2]

Answer 22

yeah i drew the pic but can't figure out the limits

Answer 23

this is the area we want, you agree?

Answer 24

very nice yes

Answer 25

then we determine the switch this way right?

Answer 26

yeah

Answer 27

{S{S} dx} dy} ; new y = [0,e^2] then, we want it a straigh line ...

Answer 28

ln(2) = y

and new x = [e^y, 2]

Answer 29

new y = [0,ln(2)]

new x = [e^y,2] ; you said the book has an answer?

Answer 30

yeah

Answer 31

y = ln(2) at the top right?

so new y = [0,ln(2)]

Answer 32

that makes sense but when i look at another problem
[0, 1] [4x, 4] dydx switches to [0, 4] [0, y/4] dxdy

Answer 33

whoops ignore that
it's [4, 0 ][0, sqrt x]dydx to [0, 2][4, y^2]dxdy

Answer 34

[0, 4][0, sqrt x]dydx to [0, 2][y^2, 4]]dxdy

Answer 35

^^ that's correct one lol

Answer 36

gonna make me draw another pic aintcha lol

Answer 37

if i do it like the way you did it then should it new become [0, sqrt x][y^2, 4]dxdy?

Answer 38

lol you dont have to

Answer 39

but I did lol

Answer 40

yeah

Answer 41

this is so confusing

Answer 42

your just switching out the new for the old; it takes practice, but clearly define the old and new

Answer 43

but the new limits would be [0, sqrt x][y^2, 4]dxdy

Answer 44

but answer says [0, 2][y^2, 4]dxdy

Answer 45

can you tell where i'm getting confused?

Answer 46

look at the picture; the new limit goes to 0,2 becasue its no longer a part of the curve...

Answer 47

sqrt(x) is an old limit for an old y; that curve is no longer a part of the new y

Answer 48

just like old x = 0 is no longer a part of new x interval;

Answer 49

new y matches old x in that it is a limit that is defined from a point to point....

new x matches new y in that it is a limit from a curve to a point.....

Answer 50

its like reading a map backwards to get back home .... you have to reorient it to make sense

Answer 51

i think im going to have to look at all the pictures again very slowly by myself lol

Answer 52

thanks for the help! need to start from top again

Answer 53

good luck ;)