Inverting integration order for int (int y sin(x) dx) dy

[xoninhas]

I'm sorry I've been bombarding this forum with questions... but you've been a big help to me in this study alone... Thank you!

$\displaystyle \int\limits_{0}^{1} \int\limits_{0}^{arcsin(y)} y*sin(x)dxdy$

Well I did my work and I was pretty convinced that it should be:

$\displaystyle \int\limits_{0}^{\pi /2} \int\limits_{0}^{sin(x)} y*sin(x)dydx$

But I tried in Mathematica and in fact it's not the same...

My thought was that the bound of integration in the x axis will go from 0 to $\displaystyle \pi /2$, end of the arcsin(y) function. As for y it should go from 0 till sin(x) which is the inverse of arcsin(y).

So why is this wrong?! :S

 

[soroban]

Re: Inverting integration order

Hello, xoninhas!

Did you make a sketch?

$\displaystyle \int^1_0\int^{\text{arcsin}(y)}_0 y\sin(x)\,dx\,dy$

Well I did my work and I was pretty convinced that it should be:

$\displaystyle \int^{\frac{\pi}{2}}_0 \int ^{\sin(x)}_0 y\sin(x)\,dy\,dx$

\(\displaystyle \text{The region is defined by: }\;\begin{Bmatrix}x \:= \:0 & \to & x \:= \:\arcsin(y) \\ \\[-3mm] y \:=\: 0 & \to & y\:=\:1 \end{Bmatrix}\)

It looks like this:

        |
       1+ - - - - *
        |:::::*   :   *
        |::*      :      *
        |*        :        *
    - - * - - - - + - - - - * - -
        0       pi/2

\(\displaystyle \text{Reversing the limits, we have: }\;\begin{Bmatrix}y \,= \,\sin x & \to &y \,=\,1 \\ \\[-3mm] x \,= \,0 & \to & x \,= \,\frac{\pi}{2} \end{Bmatrix}\)

$\displaystyle \text{Therefore: }\;\int^{\frac{\pi}{2}}_0\int^1_{\sin x} y\sin x\,dy\,dx$

 

[xoninhas]

Re: Inverting integration order

I was doing the lower part!! that is why I couldn't get it right, cause my drawing is the same as yours... just wrong area chosen... How did you find out which area to integrate? why is it not the under area?

 

[xoninhas]

Re: Inverting integration order

Forget it... my bad!!

$\displaystyle \int\limits_{0}^{1} \int\limits_{sin^{-1}(y)}^{\pi/2} y*sin(x)dxdy$

This would be the area under the arcsin(y) right?

Cool!

 

[xoninhas]

This post is just to thank everyone here because I passed my exam... next challenge complex numbers and differential equations! :S

Once again thank you!

 

[galactus]

This post is just to thank everyone here because I passed my exam... next challenge complex numbers and differential equations! :S

Once again thank you!

That's great. Good to hear. Bring on the DE's. I am taking partial and ordinary DE's this fall for a grad course. Bring 'em on.