1. ## Order of integration

Order of integration has never been my forte. I combines the two things I hate the most...shapes...and inequalities .

Ok so I don't understand what I am doing wrong here.

Change the order of integration

$\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\cos(x)}f(x,y)~dy~dx$

So

$\displaystyle 0\leq{x}\leq\frac{\pi}{2}$

and

$\displaystyle 0\leq{y}\leq\cos(x)$

$\displaystyle \Rightarrow\frac{\pi}{2}\leq\arccos(y)\leq{x}$

$\displaystyle \Rightarrow\arccos(y)\leq{x}\leq\frac{\pi}{2}$

and I can see that from the picture that the outer bound should be

$\displaystyle \int_0^{1}$

But could someone give me a quick run through on how to manipulate the inequalities?

I know the first part is wrong and the second right, but if you wouldn't mind doing both.

2. Hello,

$\displaystyle 0 \le x \le \tfrac \pi 2 \implies 0 \le \cos(x) \le 1$

$\displaystyle 0 \le y \le \cos(x) \le 1 \implies 0 \le y \le 1$

3. Originally Posted by Moo
Hello,

$\displaystyle 0 \le x \le \tfrac \pi 2 \implies 0 \le \cos(x) \le 1$

$\displaystyle 0 \le y \le \cos(x) \le 1 \implies 0 \le y \le 1$
Maybe I am being stupid, maybe it's late or I don't know, sorry to bother you but I have no idea what you did

$\displaystyle 0\leq{x}\leq{1}$
$\displaystyle \overbrace{\Rightarrow}^{\text{?}}\cos(0)\leq{\cos (x)}\leq\cos\left(\frac{\pi}{2}\right)$

$\displaystyle \Rightarrow{1\leq\cos(x)\leq{0}}$

Or do you have to switch the direction of the inequalities

and for the second one

I see how you got it (assuming I got the first) part

4. Originally Posted by Mathstud28
Maybe I am being stupid, maybe it's late or I don't know, sorry to bother you but I have no idea what you did

$\displaystyle 0\leq{x}\leq{1}$
$\displaystyle \overbrace{\Rightarrow}^{\text{?}}\cos(0)\leq{\cos (x)}\leq\cos\left(\frac{\pi}{2}\right)$

$\displaystyle \Rightarrow{1\leq\cos(x)\leq{0}}$

Or do you have to switch the direction of the inequalities
Between $\displaystyle 0$ and $\displaystyle \tfrac \pi 2$, the cosine function is decreasing

This is why we reverse the inequality order ! And saying that $\displaystyle 1 \le \dots \le 0$ is just a nonsense

and for the second one

I see how you got it (assuming I got the first) part

Just not taking care of cos(x).

5. Originally Posted by Mathstud28
Order of integration has never been my forte. I combines the two things I hate the most...shapes...and inequalities .

Ok so I don't understand what I am doing wrong here.

Change the order of integration

$\displaystyle \int_0^{\frac{\pi}{2}}\int_{{\color{red}y = }0}^{ {\color{red}y = } \cos(x)}f(x,y)~dy~dx$ Mr F says: Note the red stuff I've added.

[snip]

2. Draw the region defined by the integral terminals.

It should then be evident that after reversing the order of integration you get:

$\displaystyle \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~dx ~dy$.

6. Originally Posted by mr fantastic
Why ?
Doesn't the order of dy dx define the order of integration ?

$\displaystyle \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~{\color{red}dx} ~{\color{red}dy}$

7. Originally Posted by Moo
Between $\displaystyle 0$ and $\displaystyle \tfrac \pi 2$, the cosine function is decreasing

This is why we reverse the inequality order ! And saying that $\displaystyle 1 \le \dots \le 0$ is just a nonsense

Just not taking care of cos(x).
Oh of course! For then we would have to have that if $\displaystyle x_0<x_1$

that $\displaystyle f(x_0){\color{red}<}f(x_1)$

Great that was the missing key!

8. Originally Posted by mr fantastic

2. Draw the region defined by the integral terminals.

It should then be evident that after reversing the order of integration you get:

$\displaystyle \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~dx ~dy$.
I got that, but I am a really analytical person and graphs mean nothing to me =(

9. Originally Posted by Moo
Why ?
Doesn't the order of dy dx define the order of integration ?

$\displaystyle \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~{\color{red}dx} ~{\color{red}dy}$
You'd be amazed at how much easier it is for novices to draw the region defined by the integral terminals when you add the red stuff and hence explicitly state the equations of the curves that bound the region ....... Even the act of doing this forces them to think of the equations of the curves that bound the region.

An expert obviously doesn't need to do this ......

10. Originally Posted by Mathstud28
I got that, but I am a really analytical person and graphs mean nothing to me =(
You have my sympathy. I'll try to avoid future replies that expose your handicap. Perhaps some day gene therapy can cure you.

11. Originally Posted by mr fantastic
You have my sympathy. I'll try to avoid future replies that expose your handicap. Perhaps some day gene therapy can cure you.
Ahh what a glorious day that will be .

Here is my other handicap again

I should have paid more attention in fourth grade

I have been able to solve all the ones I have tried in the last ten minutes (about 65 or so) but this one...and I am not entirely sure why...

$\displaystyle 0<x<1$

$\displaystyle x<y<2-x$