Sketch the region then change order of integration.

[never_lose]

Sketch the region $\displaystyle \Omega$ and change the order of integration.

$\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx$

This is my sketch, except there should be vertical asymptotes at x = 1 and x = 4. $\displaystyle \Omega$ would be the region between x = 1 and x = 4 and the two lines.

When I switch, y should be in terms of x... so I solved for x from these two equations: $\displaystyle y = 2x$ and $\displaystyle y = x$ (which are the equations for the red and blue lines)

$\displaystyle x = \frac{y}{2}$ $\displaystyle x = y$.

x should be expressed in terms of y so I use $\displaystyle x = \frac{1}{2}$ $\displaystyle x = y$.

I then get $\displaystyle \int_{1}^{4}\int_{y}^{y/2}f(x,y) dx dy$.

But the answer should be $\displaystyle \int_{1}^{2}\int_{1}^{y}f(x,y) dx dy \; + \; \int_{2}^{4}\int_{y/2}^{y}f(x,y) dx dy \; + \; \int_{4}^{8}\int_{y/2}^{4}f(x,y) dx dy$.

I'm not getting what is happening. How are the integrals getting split up into separate parts?

 

[galactus]

What is f(x,y)?. Is it given?. I see no reason why there should be asymptotes with two lines.

Did you inadvertently leave out f(x,y)?.

As it is, the area between y=2x and y=x from 1 to 4 would be 15/2.

I see no reason to use three separate integrals. There must be something I am missing.

$\displaystyle \int_{1}^{4}xdx=\frac{15}{2}$

$\displaystyle \int_{1}^{4}\int_{x}^{2x}dydx=\frac{15}{2}$

 

[never_lose]

f(x,y) isn't given. It's just simply some function f(x,y).

I see no reason why there should be asymptotes with two lines.

In the original function $\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx$, x is going from 1 to 4, so $\displaystyle \Omega$ would be the region between x = 1, x = 4, and the lines y = x and y = 2x I believe.

 

[pka]

f(x,y) isn't given. It's just simply some function f(x,y). In the original function $\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx$, x is going from 1 to 4, so $\displaystyle \Omega$ would be the region between x = 1, x = 4, and the lines y = x and y = 2x I believe.

Look at the three regions. As we change the order of integration, we see where the new limits come from.

 

[never_lose]

How do you know that it's separated into three regions like that? Is there some algebra work done to see this?

 

[pka]

How do you know that it's separated into three regions like that? Is there some algebra work done to see this?

You need to change from $\displaystyle dydx$ to $\displaystyle dxdy$.
So you have the 'outside' limits on $\displaystyle y$ as $\displaystyle [1,2]\cup[2,4]\cup[4,8]$
On $\displaystyle [1,2]~~dx:1\text{ to }y$,
on $\displaystyle [2,4]~~dx:y/2\text{ to }y$,
on $\displaystyle [4,8]~~dx:y/2\text{ to }4$