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Question Number 27595 by abdo imad last updated on 10/Jan/18

f i n d \int \int_{D} x y \sqrt{x^{2} + y^{2}} d x d y w i t h

D = {(x, y) \in R^{2} / x^{2} + 2 y^{2} ⩽ 1, x ⩾ 0, y ⩾ 0}

Commented by abdo imad last updated on 15/Jan/18

w e u s e t h e p o l a r c o o r d i n a t e l e t u s e t h e c h . x = r c o s θ

a n d y = \frac{1}{\sqrt{2}} s i n θ d u e t o t h e d i f f e o m o r p h i s m e w e m u s t h a v e

0 ⩽ θ ⩽ \frac{π}{2} a n d 0 ⩽ r ⩽ 1

I = \int \int_{D} x y \sqrt{x^{2} + 2 y^{2}} d x d y = \int \int_{w} Φ o f / j_{Φ} / d r d θ

(r, θ) - (f_{1} (r, θ), f_{2} (r, θ)) = (x y) = (r c o s θ, \frac{r}{\sqrt{2}} s i n θ)

M_{j} = (_{\frac{1}{\sqrt{2}} s i n θ}^{c o s θ}_{\frac{r}{\sqrt{2}} c o s θ}^{- r s i n θ})

I = \int \int_{0 ⩽ r ⩽ 1 a n d 0 ⩽ θ ⩽ \frac{π}{2}} r c o s θ . \frac{r}{\sqrt{2}} s i n θ r . \frac{r}{\sqrt{2}} d r d θ

I = \frac{1}{2} \int_{0^{}}^{1} r^{4} \int_{0}^{\frac{π}{2}} c o s θ s i n θ d θ = \frac{1}{4} {[\frac{1}{5} r^{5}]}_{0}^{1} \int_{0}^{\frac{π}{2}} s i n (2 θ) d θ

= \frac{1}{40} {[- c o s (2 θ)]}_{0}^{\frac{π}{2}} = \frac{1}{20} .