Double Integral Doubts: Understanding Regions and Order

Click For Summary
SUMMARY

The discussion centers on the evaluation of a double integral defined by the region \(\int_{-1}^{2}\int_{-\sqrt{4-x^2}}^{1-x^2}f(x,y)dydx\). Participants confirm that the integral represents the area of the region on the left minus the area on the right, leading to potential confusion regarding the sign of the volume generated under the graph of \(f(x,y)\). The equality \(\int_{-3}^{1}\int_{arctg(x)}^{e^x}f(x,y)dydx=\int_{arctg(x)}^{e^x}\int_{-3}^{1}f(x,y)dxdy\) is also discussed, emphasizing the importance of understanding the order of integration in double integrals.

PREREQUISITES
  • Understanding of double integrals and their geometric interpretations
  • Familiarity with the functions \(arctg(x)\) and \(e^x\)
  • Knowledge of the Cartesian coordinate system and area calculations
  • Basic concepts of volume under surfaces in multivariable calculus
NEXT STEPS
  • Study the geometric interpretation of double integrals in multivariable calculus
  • Learn about changing the order of integration in double integrals
  • Explore the implications of negative volumes in integrals
  • Investigate the properties of the functions \(arctg(x)\) and \(e^x\) in integration
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and double integrals, as well as anyone looking to deepen their understanding of integration techniques and geometric interpretations.

Telemachus
Messages
820
Reaction score
30

Homework Statement


Hi there, I've got this doubt about a double integral. I have this region: [tex]\displaystyle\int_{-1}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{1-x^2}f(x,y)dydx[/tex]

And the thing is, how this region would look like? Would it look like this?:
attachment.php?attachmentid=29758&stc=1&d=1289403019.png

The thing is that after the cut between the two curves the order changes, so I think that region would have an opposite sign than the region before.

What you say?

Bye, thanks for posting!
 

Attachments

  • para.PNG
    para.PNG
    2 KB · Views: 478
Physics news on Phys.org
Yes, that's correct. The integral will be equal to the area of the region on the left minus the area of the region on the right.
 
Thanks HallsofIvy. It turns confusing since that area will generate a volume under the graph f(x,y). I don't know if it really have any sense that region planted that way.

Mmm now I think that the volume for the last part would be negative (or positive depending on f) and then it would have some sense.

By the way, is this equality right? [tex]\displaystyle\int_{-3}^{1}\displaystyle\int_{arctg(x)}^{e^x}f(x,y)dydx=\displaystyle\int_{arctg(x)}^{e^x}\displaystyle\int_{-3}^{1}f(x,y)dxdy[/tex]
 
Last edited:

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
29
Views
5K
Replies
9
Views
2K
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K