Group Abstract Group Abstract

Message Boards Message Boards

0
|
4K Views
|
5 Replies
|
1 Total Like
View groups...
Share
Share this post:
GROUPS:
Mathematics

Can't compute a double integral by using If command with Mathematica 10.2

Juan Felix Avila, UCR
Posted 10 years ago

I used to do this computation with Mathematica 9.0:

Hold[Simplify[Integrate[Integrate[ If[{x^2/a^2 + y^2/b^2 <= 1}, 1, 0], {y, -Infinity, Plus[Infinity]}], {x, -Infinity, Plus[Infinity]}], {a > 0, b > 0}]]

or alternative notation 

Simplify[\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(-\[Infinity]\), \(+\[Infinity]\)]\(\((
\*SubsuperscriptBox[\(\[Integral]\), \(-\[Infinity]\), \(+\[Infinity]\)]\((If[{x^2/a^2 + y^2/b^2 <= 1}, 1, \.010])\) \[DifferentialD]y)\) \[DifferentialD]x\)\), {a >    0, b > 0}]

and get a b Pi. However when I try to compute the same on Mathematica 10.2, it gets stuck. What's going on? Thank your for your valuable time.
POSTED BY: Juan Felix Avila
5 Replies
Sort By:

Try copying some of the code people posted below. Do you get the same resutls? If not, please run "Quit" to make sure that the kernel is fresh and has no previously defined values.

POSTED BY: Sean Clarke
Bill Simpson
Bill Simpson
Posted 10 years ago

In 10.1

In[1]:= Simplify[Integrate[Integrate[If[{x^2/a^2+y^2/b^2<=1},1,0], {y,-Infinity,Infinity}], {x,-Infinity,Infinity}], {a>0,b>0}]

Out[1]= a b Pi

In[2]:= Simplify[Integrate[Integrate[If[x^2/a^2+y^2/b^2<=1,1,0], {y,-Infinity,Infinity}], {x,-Infinity,Infinity}], {a>0,b>0}]

Out[2]= a b Pi

In[3]:= Simplify[Integrate[Integrate[Boole[x^2/a^2+y^2/b^2<=1], {y,-Infinity,Infinity}], {x,-Infinity,Infinity}], {a>0,b>0}]

Out[3]= a b Pi

In[4]:= Simplify[Integrate[Integrate[Piecewise[{{1,x^2/a^2+y^2/b^2<=1}}, 0], {y,-Infinity,Infinity}], {x,-Infinity,Infinity}], {a>0,b>0}]

Out[4]= a b Pi

In[5]:= Simplify[Integrate[Integrate[Which[x^2/a^2+y^2/b^2<=1,1,True,0], {y,-Infinity,Infinity}], {x,-Infinity,Infinity}], {a>0,b>0}]

Out[5]= a b Pi
POSTED BY: Bill Simpson 
In[2]:= Integrate[
 If[x^2/a^2 + y^2/b^2 <= 1, 1, 0], {y, -Infinity, 
  Infinity}, {x, -Infinity, Infinity}]

Out[2]= \[Pi]/Sqrt[1/(a^2 b^2)]
POSTED BY: Frank Kampas

Thank your for your valuable time. There you go:

Hold[Simplify[Integrate[Integrate[ If[{x^2/a^2 + y^2/b^2 <= 1}, 1, 0], {y, -Infinity, Plus[Infinity]}], {x, -Infinity, Plus[Infinity]}], {a > 0, b > 0}]]

Best regards, Juan F. Avila

Attachments:
POSTED BY: Juan Felix Avila

It looks like you were trying to paste Traditional Math notation.

Please run Hold and InputForm on your integral:

putYourIntegralHere//Hold//InputForm

This will show the actual code that you are using, which you can then post here.

POSTED BY: Sean Clarke
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard