Integration in polar coordinates

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SUMMARY

The discussion focuses on evaluating the integral of the function \( (x^2+y^2)^{7/2} \) over the disk defined by \( D: x^2+y^2 \le 16 \) using polar coordinates. The integral is denoted as \( I=\iint_{D}(x^2+y^2)^{7/2}dxdy \). By converting to polar coordinates, the limits are set as \( 0 \le \theta \le 2\pi \) and \( 0 \le \rho \le 4 \). The final evaluation yields \( I = \frac{2^{19}\pi}{9} \).

PREREQUISITES
  • Understanding of polar coordinates in calculus
  • Familiarity with double integrals
  • Knowledge of integration techniques
  • Basic proficiency in evaluating limits of integration
NEXT STEPS
  • Study the conversion of Cartesian coordinates to polar coordinates
  • Learn about double integrals in polar coordinates
  • Explore advanced integration techniques, including substitution methods
  • Investigate applications of polar coordinates in physics and engineering
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Students and professionals in mathematics, particularly those studying calculus and integral evaluation, as well as educators looking for examples of polar coordinate applications.

Fernando Revilla
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I quote a question from Yahoo! Answers

By changing to polar coordinates, evaluate the integral.
(Integrand)(integrand)[(x^2+y^2)^(7/2)… where D is the disk x^2+y^2<=16.

I have given a link to the topic there so the OP can see my response.
 
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Denote $I=\displaystyle\iint_{D}(x^2+y^2)^{7/2}dxdy$ wiith $D\equiv x^2+y^2\le 16$. We have: $D\equiv \left \{ \begin{matrix}0\le \theta\le 2\pi\\0\le \rho \le 4\end{matrix}\right.$, so $$I=\int_0^{2\pi}d\theta\int_0^4(\rho^2)^{7/2}\rho d\rho=2\pi \left[\frac{\rho^9}{9}\right]_0^4=\frac{2^{19}\pi}{9}$$
 

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