never_lose
New member
- Joined
- Jul 9, 2011
- Messages
- 7
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?
\(\displaystyle \int_{1}^{4}\int_{x}^{2x}f(x,y) dy dx\)
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?