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Calculate the double integral in polar coordinates

Posted 5 years ago
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POSTED BY: ajin mj
9 Replies
In[1]:= 1/\[Pi] Integrate[(x^2 + y^2) Boole[x^2 + y^2 <= 36], {x, 0, 
   6}, {y, -6, 0}]

Out[1]= 162
POSTED BY: Frank Kampas

Define

x = r Cos[t];
y = r Sin[t];

Then

x^2 + y^2 // Simplify

r^2

So your r is equal to 6.

Your constraints are fulfilled if 3Pi/2 <= t <= 2 Pi. Look for example at

Plot[{x /. r -> 1, y /. r -> 1.5}, {t, 0, 2 Pi}]

The Volumeelement in Polar Coordinates (in R2 ) is dV = r dr dt, so your integral is written as

In[68]:= 1 / Pi Integrate[ r^2  r , {r, 0, 6}, {t, 3 Pi/2, 2 Pi}]

Out[68]= 162

Obviously your integral "lives" on a disk with radius r. Then you can equally write y as function of x and your integral is as well

In[77]:= 1/Pi Integrate[ x^2 + y^2, {x, 0, 6}, {y, 0, Sqrt[36 - x^2]}]

Out[77]= 162
POSTED BY: Hans Dolhaine

That's not the correct syntax for entering an integral. Also, I don't understand why you're doing a maximization.

POSTED BY: Frank Kampas

Ok Frank, sorry! I attach the NMaximize.

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Christos, It would be easier to read you code if you post it using the code sample icon (the one after Add Notebook)

POSTED BY: Frank Kampas

Frank, What is the problem with the following?

NMaximize[{\!( *SubsuperscriptBox[([Integral]), (0), (6)]( *SubsuperscriptBox[([Integral]), (-6), (0)](( *SuperscriptBox[(x), (2)] + *SuperscriptBox[(y), (2)]))/[Pi]\ \ [DifferentialD]y [DifferentialD]x)),x^2+y^2<=36,y<=0},{x,y}]

The solution is: {275.02, {x -> 3.59228, y -> -0.0941189}}

Why don't you follow your own suggestion (in the post title) and transform the problem into polar coordinates, which immediately gives:

Integrate[r^3/Pi, {r, 0, 6}] Pi/2
(*  Out:   162   *)

But if one hesitates to do any math one can choose the straightforward approach:

reg = ImplicitRegion[{x^2 + y^2 <= 36, x >= 0, y <= 0}, {x, y}];
Integrate[(x^2 + y^2)/Pi, {x, y} \[Element] reg]
(*  Out:   162   *)
POSTED BY: Henrik Schachner

It is rather unclear what you want to integrate. Below there are three alternatives with three different values: NMaximize[{[Integral]\!( *SubsuperscriptBox[([Integral]), ( *SuperscriptBox[(x), (2)] + *SuperscriptBox[(y), (2)]), (36)]( *FractionBox[(1), ([Pi])] (( *SuperscriptBox[(x), (2)] + *SuperscriptBox[(y), (2)])) [DifferentialD]x [DifferentialD]y\ )), x >= 0, y >= 0, x^2 + y^2 <= 36}, {x, y}]

Out[69]= {25789., {x -> 0., y -> 6.}}

NMaximize[{[Integral][Integral]1/[Pi] (x^2 + y^2) [DifferentialD]x [DifferentialD]y, x >= 0, y >= 0, x^2 + y^2 <= 36}, {x, y}]

Out[70]= {68.7549, {x -> 4.24264, y -> 4.24264}}

NMaximize[{\!( *SubsuperscriptBox[([Integral]), ( *SuperscriptBox[(x), (2)] + *SuperscriptBox[(y), (2)]), (36)]( *SubsuperscriptBox[([Integral]), ( *SuperscriptBox[(x), (2)] + *SuperscriptBox[(y), (2)]), (36)] *FractionBox[(1), ([Pi])] (( *SuperscriptBox[(x), (2)] + *SuperscriptBox[(y), (2)])) [DifferentialD]x [DifferentialD]y\ )), x >= 0, y >= 0, x^2 + y^2 <= 36}, {x, y}]

Out[72]= {-1.1913110^9, {x -> 2.0236310^-9, y -> 5.45121*10^-9}}

use Boole[ ] to implement the constraint

POSTED BY: Frank Kampas
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