calculate ∫∫_(x^2 +y^2 −2x≤0) xdxdy.


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Question Number 31085 by abdo imad last updated on 02/Mar/18

$c a l c u l a t e \int \int_{x^{2} + y^{2} - 2 x ⩽ 0} x d x d y .$
Commented by abdo imad last updated on 11/Mar/18

$l e t u s e t h e o l a r c o o r d i n a t e s x = r c o s θ a n d y = r s i n θ$ $x^{2} + y^{2} - 2 x ⩽ 0 \Leftrightarrow r^{2} - 2 r c o s θ ⩽ 0 \Leftrightarrow 0 < r ⩽ 2 c o θ d u e t o$ $d i f f e o m o r p h i s m e \frac{- π}{2} ⩽ θ ⩽ \frac{π}{2} \Rightarrow$ $I = \int \int_{- \frac{π}{2} ⩽ θ ⩽ \frac{π}{2} a n d 0 < r ⩽ 2 c o s θ} r c o s θ r d r d θ$ $= \int_{- \frac{π}{2}}^{\frac{π}{2}} (\int_{0}^{2 c o s θ} r^{2} d r) c o s θ d θ b u t$ $\int_{0}^{2 c o s θ} r^{2} d r = {[\frac{1}{3} r^{3}]}_{0}^{2 c o s θ} = \frac{8}{3} {c o s}^{3} θ \Rightarrow$ $I = \frac{8}{3} \int_{- \frac{π}{2}}^{\frac{π}{2}} {c o s}^{4} d θ = \frac{16}{3} \int_{0}^{\frac{π}{2}} \frac{{(1 + c o s (2 θ))}^{2}}{4} d θ$ $= \frac{4}{3} \int_{0}^{\frac{π}{2}} (1 + 2 c o s (2 θ) + \frac{1 + c o s (4 θ)}{2}) d θ$ $= \frac{4}{3} \frac{π}{2} + \frac{8}{3} \int_{0}^{\frac{π}{2}} c o s (2 θ) d θ + \frac{2}{3} \frac{π}{2} + \frac{2}{3} \int_{0}^{\frac{π}{2}} c o s (4 θ) d θ$ $= \frac{2 π}{3} + 0 + \frac{π}{3} + 0 = π \Rightarrow I = π .$
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