Evaluate the following integral over the specified region. Assume (r comma theta )(r,θ) are polar coordinates. ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper R StartFraction dA Over (16 plus x squared plus y squared )squared EndFraction∫∫R dA 16+x2+y22; Upper R equals StartSet (r comma theta ): 1 less than or equals r less than or equals 3 comma 0 less than or equals theta less than or equals pi EndSetR={(r,θ): 1≤r≤3, 0≤θ≤π} M
Accepted Solution
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Answer:Step-by-step explanation:We have to integrate the function [tex]16+x^2+y^2[/tex] over the region [tex]1\leq r\leq 3\\0\leq \theta \leq \pi[/tex]Let us convert into polar coordinatesx=cost : y = sintThen we get dxdy = rdr dtSo we get the integral as[tex]\int \int _R (16+x^2+y^2)^2 dxdy\\= \int\limits^3_0 \int\limits^\pi_0( {16+r^2}^2) (r)drdt\\ =(t) (\frac{(16+r^2)^3}{6}[/tex]Substitute the limits to getAnswer =[tex](\pi-0) [\frac{1}{6} (25-16)\\=1.5 \pi[/tex]