Unit Circle Double Integral: Is 2π/3 the Answer?

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SUMMARY

The discussion centers on evaluating the double integral \(\int\int_R \sqrt{x^2+y^2} \, dx \, dy\) over the unit circle \(R\). The initial computation presented by the user resulted in an answer of \(\frac{2\pi}{3}\), which was later corrected to reflect the proper limits of integration, specifically \(\int_{-1}^1\) for the second integral. The correct setup requires ensuring that \(r\) remains non-negative throughout the integration process.

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squenshl
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For the double integral [tex]\int\int_R[/tex] sqrt(x^2+y^2) dx dy where R is the unit circle.
I got[tex]\int_0^\pi\int_1^1[/tex] sqrt(r2) r dr dtheta
Then after the integration I got an answer of 2[tex]pi[/tex]/3 as my final answer.
Is this right.
The bottom of the 2nd integral is -1 not 1
 
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squenshl said:
For the double integral [tex]\int\int_R[/tex] sqrt(x^2+y^2) dx dy where R is the unit circle.
I got[tex]\int_0^\pi\int_1^1[/tex] sqrt(r2) r dr dtheta
Then after the integration I got an answer of 2[tex]pi[/tex]/3 as my final answer.
Is this right.
The bottom of the 2nd integral is -1 not 1


Hi squenshl! :smile:

(have a pi: π and a theta: θ and try using the X2 tag just above the Reply box :wink:)

(and you needed \int_{-1}^1)

erm :redface:you can't have r less than 0! :wink:
 

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