Solving a Complex Integral: Finding the 4

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SUMMARY

The discussion centers on evaluating the double integral \(\int\int_D |x| dA\) over the domain defined by \(D = x^2 + y^2 \leq a^2\) where \(a > 0\). The user initially computes the integral using polar coordinates, resulting in \(\frac{a^3}{3}\) after evaluating the integral from \(0\) to \(\frac{\pi}{2}\). However, the correct answer is \(4 \frac{a^3}{3}\), which accounts for the entire circle rather than just the first quadrant. The missing factor of 4 arises from the symmetry of the integral across all four quadrants.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with polar coordinates and transformations
  • Knowledge of symmetry in integrals
  • Basic proficiency in evaluating definite integrals
NEXT STEPS
  • Study the properties of double integrals in polar coordinates
  • Learn about the symmetry of functions in multiple dimensions
  • Practice evaluating double integrals over different geometric shapes
  • Explore the concept of area and volume in polar coordinates
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable calculus and integral evaluation, as well as educators seeking to clarify concepts related to double integrals and polar coordinates.

yoleven
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Homework Statement


[tex]\int[/tex][tex]\int[/tex]D[tex]\left|x\right|[/tex]dA
D= X2+y2<=a2 where a>0


Homework Equations


[tex]\int[/tex][tex]\stackrel{\Pi/2}{0}[/tex][tex]\int[/tex][tex]\stackrel{a}{0}[/tex] r cos [tex]\Theta[/tex] r dr d[tex]\Theta[/tex]

I hope that's clear...

I evaluate this to [tex]\frac{a^3}{3}[/tex] sin [tex]\Theta[/tex]


sin [tex]\Pi[/tex]/2 = 1

so I get [tex]\frac{a^3}{3}[/tex]

the book says the answer is 4 [tex]\frac{a^3}{3}[/tex]



I can't see where the 4 is coming from.

can some one explain this to me?
 
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Your domain of [0,pi/2] only includes a quarter of the entire circle.
 
Of course. Thanks.
 

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