Evaluate $$\int\int_D \sqrt {x^2+y^2}dxdy~~~~D=\{(x,y):x\leq x^2+y^2\leq 2x\} $$ After applying the change of variables i got the integral set up as $$2\int_0^{\pi/2}\int_{\cos \theta}^{2cos \theta}r^2dr d\theta$$ But i want to change the order of integration in the polar coordinates, so i got the integral set up as $$2\left (\int_0^1\int_{\arccos r}^{\arccos r/2}r^2 d\theta dr +\int_1^2\int_{0}^{\arccos r/2}r^2 d\theta dr \right)$$ Is this correct
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1 Answer
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Looks good to me. Here's a plot of your two regions of integration (top half) compared to the original region of integration (bottom half). You can see they match up perfectly.
Made using Mathematica, with the command
RegionPlot[{x<x^2+y^2<2x && y<0,Sqrt[x^2+y^2]<1 && ArcCos[Sqrt[x^2+y^2]]<ArcTan[x,y]<ArcCos[Sqrt[x^2+y^2]/2],1<Sqrt[x^2 + y^2]<2 && 0<ArcTan[x,y]<ArcCos[Sqrt[x^2+y^2]/2]},{x,0,2},{y,-1,1}]
