MHB Integration in polar coordinates

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The integral of the function (x^2+y^2)^(7/2) over the disk defined by x^2+y^2≤16 is evaluated using polar coordinates. The region D is described with 0≤θ≤2π and 0≤ρ≤4. The integral is expressed as I=∫_0^(2π)dθ∫_0^4(ρ^2)^(7/2)ρ dρ. The calculation results in I=2π[(ρ^9)/9] evaluated from 0 to 4, yielding a final result of (2^19)π/9. This demonstrates the effectiveness of polar coordinates in simplifying the evaluation of double integrals over circular regions.
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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