Volume by double integral

[jade abyss]

Hi, I have a confusion about knowing what to start the integration with, dx or dy? I mean I can tell from the f(x,y) function easily but I can never tell just by looking at the graph, I don't see lines switch, I don't get it, can anyone explain that please
Thank you

[dawn bluff]

same

[stuck lynx]

Always let the outer integral have constant bounds and the the inner integral have non constant bounds.

For example y goes from 0 to x which is a bound that depends on x so you know you have to integrate with respect to dy first.

Notice that in the volume calculation, they leave
$\int_{0}^{1}$ last.

[rain horizon]

AstroGuy

[stuck lynx]

That’s because of what I’ve told you earlier. Outer integral must have constant bounds. If not, then it’s not a defined integral

[jade abyss]

Ya ya I know, the outer integral must have constant bounds, I am asking about the inner one, do I start with dx or dy

[stuck lynx]

You start with dy

Why? Because y has non constant bounds

0<= y <= x

While x:

0<= x <= 1

x is the one that has constant bounds so you integrate with respect to y first

That’s what they did

dy first because its bounds are non constant

If you knew that outer integral must have constant bounds then you would have known that you need to start with dy

[jade abyss]

y does have constant bounds I mean like the bounds are 2 points (0,0) and (1,1) , so in both functions u can get the bounds with respect to y or x , and if u can look at the pic , u can see that he wrote it in both ways , so the idea here is not about the constant bounds .

[stuck lynx]

He changed the region

He defined a different region that’s why he was allowed to change bounds

Well. It describes the same triangle but using a different region

You’re talking about that one right?

For the whole calculation he used the integral at the left

And you can’t just switch a region like that. You would have to redefine everything

y has constant bounds because you decided to change the region and it changes the integral completely

In the first configuration, y has non constant bounds. In THAT configuration. But if you choose to change the region then in THAT new configuration, y has constant bounds

[jade abyss]

Redefining is not a problem, in some problems we redefine, like this one
The idea is how to know if we should start integrating with respect to dy or dx by looking at the graph

[stuck lynx]

By looking at the graph, you choose the order that makes the region simplest to describe.

By should, you mean the most convenient way.

Here the most convenient way would be to integrate with respect to x first because if you look at the boundary curves, they are both already written nicely as x= (something in y)

Look at the boundary curves always

See if their expression are nicer for x or for y

[jade abyss]

Wdym by nicer? Like they are both ugly to me🙂‍↕️

[stuck lynx]

If you keep y=sqrt(2x+6) will have to deal with two branches:

y=-sqrt(2x+6) and y=+sqrt(2x+6)

But if you do it with x:

x=(1/2)y^2-3 and x=y+1

Easy bounds compared to sqrt

[stuck lynx]

Both are ugly but x less..

[jade abyss]

Well I mean in the last question sent , ya I can see if with respect to y , then I have to integrate as u said between the first curve boundries then I'll switch to the 2 curves , so it's pretty easy to tell in second question, but in the first one I sent? Can u put it in the same way?cuz I see if with respect to y then i have to deal with x=0 and x=y , and with respect to x I have to deal with y=1 and y=x, all the time so i don't see it there

[stuck lynx]

Well for first one you sent, You can either deal with y=0 to y=x if you start integrating with dy right?

Or deal with x=y to x=1 if you start with dx

So both configurations look great right?

The problem here is the integrand

e^(y/x)

If you integrate with respect to x first, you will get logarithms

Do you agree?

So in the second picture it was rather about “what boundary curves look nicer, dx or dy first then?”

But first picture it was about: “integrand will be simpler to integrate if i choose dx first or dy first,”