Problem with setting the region of integration

  • Thread starter Thread starter Amaelle
  • Start date Start date
  • Tags Tags
    Integration
Amaelle
Messages
309
Reaction score
54
Homework Statement
look at the image below!
Relevant Equations
double integrals
Good day !
I have a problem with the solution of the floowing integrals
Indeed i don't understand why they choose such borders for integral
b/a<c
y<c
doesn't mean that y<b/a !
many thanks in advance!

1614189909864.png
 

Attachments

  • 1614189699163.png
    1614189699163.png
    37.4 KB · Views: 157
Physics news on Phys.org
Amaelle said:
doesn't mean that y<b/a !
For ##(x,y)\in\Omega##, yes that is correct. That's why they can set the upper limit on the outer integral to ##b/a## rather than ##c##, even though ##c## may be greater than ##b/a##.

The area of integration is a triangle with vertices (0,0), (0, b/a), (b,0).

Given that c>b/a, the value of c is not used in the calculation. I think they just put it in the question to confuse people!
 
that was an amzing shot! thanks a million!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top