- #1
Ral
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Homework Statement
Rewrite by converting to polar coordinates, carefully drawing R.
[tex]\int^{2}_{0}\int^{\sqrt{2x-x^2}}_{0}\sqrt{x^2+y^2}dydx[/tex]
Homework Equations
The Attempt at a Solution
I believe I have the inside part of it right. What I did was replace the x^2 and y^2 in [tex]\sqrt{x^2+y^2}[/tex] with [tex]r^2cos^2\vartheta[/tex] and [tex]r^2sin^2\vartheta[/tex], factored out the r^2, which left me with [tex]cos^2\vartheta+sin^2\vartheta[/tex], which equals 1. So then I would be left with [tex]\sqrt{r^2}rdrd\vartheta[/tex]
What's confusing me is what my limits on the integrals should be. What I have right now is
[tex]\int^{\pi/2}_{0}\int^{2}}_{0}\sqrt{r^2}rdrd\vartheta[/tex]
but I'm still unsure if that's correct or not.
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