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Question Number 31090 by abdo imad last updated on 02/Mar/18
find ∫∫_(1≤x^2  +y^2 ≤4 and y≥0)    ((dxdy)/( (√(x^2  +y^2 )))) .
$${find}\:\int\int_{\mathrm{1}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{4}\:{and}\:{y}\geqslant\mathrm{0}} \:\:\:\frac{{dxdy}}{\:\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}\:. \\ $$
Commented by abdo imad last updated on 06/Mar/18
let use the polar coordinates x=rcosθ and y=rsinθ  due to diffeomorphisme we must have 1≤r≤2  and  0≤θ≤π ⇒ I= ∫_0 ^π  (∫_1 ^2  ((rdr)/( (√r^2 ))))dθ=∫_0 ^π  (∫_1 ^2 dr)dθ =∫_0 ^π dθ=π .
$${let}\:{use}\:{the}\:{polar}\:{coordinates}\:{x}={rcos}\theta\:{and}\:{y}={rsin}\theta \\ $$$${due}\:{to}\:{diffeomorphisme}\:{we}\:{must}\:{have}\:\mathrm{1}\leqslant{r}\leqslant\mathrm{2}\:\:{and} \\ $$$$\mathrm{0}\leqslant\theta\leqslant\pi\:\Rightarrow\:{I}=\:\int_{\mathrm{0}} ^{\pi} \:\left(\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{rdr}}{\:\sqrt{{r}^{\mathrm{2}} }}\right){d}\theta=\int_{\mathrm{0}} ^{\pi} \:\left(\int_{\mathrm{1}} ^{\mathrm{2}} {dr}\right){d}\theta\:=\int_{\mathrm{0}} ^{\pi} {d}\theta=\pi\:. \\ $$

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