double integral

- M

double integral

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- jamiebookeater

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- anonymous

What do you need answered??

- M

trying to use equation to write question but not working

- amistre64

yeah, they blame it on user error lol

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## More answers

- M

hmm i click and nothing

- amistre64

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

- anonymous

hey amister!!!

- amistre64

hi ;)

- amistre64

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

- M

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

- M

it's impossible to do this without equation lol

- amistre64

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

- M

but shouldn't dy switch to 0 to 2?

- M

going from dydx to dxdy

- amistre64

so your re adjusting the intervals right?

- M

answer book has it 0 to ln2

- M

switching the order

- amistre64

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

- M

yeah

- amistre64

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

- M

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

- M

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

- amistre64

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- M

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

- amistre64

this is the area we want, you agree?

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- M

very nice yes

- amistre64

then we determine the switch this way right?

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- M

yeah

- amistre64

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

- amistre64

ln(2) = y
and new x = [e^y, 2]

- amistre64

new y = [0,ln(2)]
new x = [e^y,2] ; you said the book has an answer?

- M

yeah

- amistre64

y = ln(2) at the top right?
so new y = [0,ln(2)]

- M

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

- amistre64

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- M

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

- M

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

- M

^^ that's correct one lol

- amistre64

gonna make me draw another pic aintcha lol

- M

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

- M

lol you dont have to

- amistre64

but I did lol

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- M

yeah

- M

this is so confusing

- amistre64

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- amistre64

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

- M

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

- M

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

- M

can you tell where i'm getting confused?

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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.....

- amistre64

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

- M

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

- M

thanks for the help! need to start from top again

- amistre64

good luck ;)

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