let-U-x-y-R-2-1-x-2-2y-2-3-calculate-U-x-y-x-2-y-2-dxdxy-

Question Number 51998 by maxmathsup by imad last updated on 01/Jan/19

l e t U = {(x, y) \in R^{2} / 1 ⩽ x^{2} + 2 y^{2} ⩽ 3}

c a l c u l a t e \int \int_{U} \frac{x - y}{x^{2} + y^{2}} d x d x y

Commented by Abdo msup. last updated on 05/Jan/19

l e t c o n s i d e r t h e d i f f e o m o r p h i s m

(r, θ) \to φ (r, θ) = (x, y) w i t h x = r c o s θ a n d y = \frac{r}{\sqrt{2}} s i n θ

1 ⩽ x^{2} + 2 y^{2} ⩽ 3 \Rightarrow 1 ⩽ r^{2} ⩽ 3 \Rightarrow 1 ⩽ r ⩽ \sqrt{3} a n d

0 ⩽ θ ⩽ 2 π \Rightarrow

\int \int_{U} \frac{x - y}{x^{2} + y^{2}} d x d y = \int \int_{1 ⩽ r ⩽ \sqrt{3} a n d 0 ⩽ θ ⩽ 2 π} \frac{r (c o s θ - \frac{s i n θ}{\sqrt{2}})}{r^{2} {c o s}^{2} θ + \frac{r^{2}}{2} {s i n}^{2} θ} r d r d θ

\int_{1}^{\sqrt{3}} d r \int_{0}^{2 π} \frac{\sqrt{2} c o s θ - s i n θ}{2 {c o s}^{2} θ + {s i n}^{2} θ} 2 d θ

= 2 (\sqrt{3} - 1) \int_{0}^{2 π} \frac{\sqrt{2} c o s θ - s i n θ}{1 + {c o s}^{2} θ} d θ b u t

\int_{0}^{2 π} \frac{\sqrt{2} c o s θ - s i n θ}{1 + {c o s}^{2} θ} d θ = \sqrt{2} \int_{0}^{2 π} \frac{c o s θ}{1 + {c o s}^{2} θ} d θ

+ \int_{0}^{2 π} \frac{- s i n θ}{1 + {c o s}^{2} θ} d θ

\int_{0}^{2 π} \frac{- s i n θ}{1 + {c o s}^{2} θ} d θ = {[a r c t a n (c o s θ)]}_{0}^{2 π}

= 0 a l s o

\int_{0}^{2 π} \frac{c o s θ}{1 + {c o s}^{2} θ} d θ = \int_{0}^{π} \frac{c o s θ}{1 + {c o s}^{2} θ} + \int_{π}^{2 π} \frac{c o s θ}{1 + {c o s}^{2} θ} d θ

\int_{π}^{2 π} \frac{c o s θ}{1 + {c o s}^{2} θ} d θ =_{θ = π + t} \int_{0}^{π} \frac{- c o s t}{1 + {c o s}^{2} t} d t \Rightarrow

\int_{0}^{2 π} \frac{c o s θ}{1 + {c o s}^{2} θ} d θ = 0 \Rightarrow

\int \int_{U} \frac{x - y}{x^{2} + y^{2}} d x d y = 0