How Are Limits of Integration Derived for Double Integrals Over Complex Regions?

[trap101]

For the following regions in R² express the double integral in terms of iterated integrals in two different ways:

S = the region between the parabolas y = x² and y = 6 - 4x -x²


Solution:

Ok I got everything except one limit of integration in regards to the order dx dy of integration:

for an upper limit of integration the solution said: x = -2 ± (10 - y )^1/2. my question is how do they obtain that limit of integration. It's probably more of an algebraic question than calculus.

Also how do you write integrals with iTex code?

 

[BruceW]

look at the equations you have been given. Can you think of how to rearrange one of them to get the limit?

Also, for more information on latex: https://www.physicsforums.com/showthread.php?t=617567 Or google might help. I think I remember the way to write the integral sign is \int

 

[trap101]

look at the equations you have been given. Can you think of how to rearrange one of them to get the limit?

Also, for more information on latex: https://www.physicsforums.com/showthread.php?t=617567 Or google might help. I think I remember the way to write the integral sign is \int

That's exactly my problem I can't see how to rearrange them to obtain that solution.

 

[HallsofIvy]

Integrating with respect to x should be straightforward. Looking at a graph indicates that, integrating with respect to y you will need to break this into three separate integrals, from y= 0 to y= 1, from y= 1 to y= 9, and from y= 9 to y= 10. Do you see why?

 

[trap101]

Integrating with respect to x should be straightforward. Looking at a graph indicates that, integrating with respect to y you will need to break this into three separate integrals, from y= 0 to y= 1, from y= 1 to y= 9, and from y= 9 to y= 10. Do you see why?

I do somewhat see why, well what I mean by that is I understand exactly the reason for splittiing it up, I think the way I drew the graph doesn't illustrate that. But it is solving for the exact x value that I specified in my first post. I tried to work it backwards and couldn't get anything related to what I have.

 

[BruceW]

You can do it. Get y=6-4x-x² so that x is the subject of the equation. Also, it is important to draw out the graph, and algebraically work out the points of intersection.