calculate-W-2x-2-3y-2-x-y-dxdy-with-W-x-y-R-2-0-lt-x-lt-1-and-0-lt-y-lt-1-

Question Number 60506 by prof Abdo imad last updated on 21/May/19

c a l c u l a t e \int \int_{W} \frac{\sqrt{2 x^{2} + 3 y^{2}}}{x + y} d x d y

w i t h W = {(x, y) \in R^{2} / 0 < x < 1 a n d 0 < y < 1 .

Commented by Mr X pcx last updated on 23/May/19

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

x = \frac{r}{\sqrt{2}} c o s θ a n d y = \frac{r}{\sqrt{3}} s i n θ

0 < x^{2} + y^{2} < 2 \Rightarrow 0 < \frac{r^{2}}{2} + \frac{r^{2}}{3} < 2 \Rightarrow

0 < \frac{5}{6} r^{2} < 2 \Rightarrow 0 < r^{2} < \frac{12}{5} \Rightarrow 0 < r < \frac{2 \sqrt{3}}{\sqrt{5}}

M_{j} = (\begin{matrix} \frac{\partial φ_{1}}{\partial r} \frac{\partial φ_{1}}{\partial θ} \\ \frac{\partial φ_{2}}{\partial r} \frac{\partial φ_{2}}{\partial θ} \end{matrix})

= (\begin{matrix} \frac{1}{\sqrt{2}} c o s θ - \frac{r}{\sqrt{2}} s i n θ \\ \frac{1}{\sqrt{3}} s i n θ \frac{r}{\sqrt{3}} c o s θ \end{matrix})

d e t (M_{j}) = \frac{r}{\sqrt{2} \sqrt{3}} \Rightarrow

I = \int \int_{0 < r < \frac{2 \sqrt{3}}{\sqrt{5}} a n d 0 < θ < \frac{π}{2}} \frac{r}{\frac{r}{\sqrt{2}} c o s θ + \frac{r}{\sqrt{3}} s i n θ} \frac{1}{\sqrt{6}} r d r d θ

= \int_{0}^{\frac{2 \sqrt{3}}{\sqrt{5}}} r d r \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{3} c o s θ + \sqrt{2} s i n θ}

\int_{0}^{\frac{2 \sqrt{3}}{\sqrt{5}}} r d r = {[\frac{r^{2}}{2}]}_{0}^{\frac{2 \sqrt{3}}{\sqrt{5}}} = \frac{1}{2} (\frac{4.3}{5}) = \frac{6}{5}

\int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{3} c o s θ + \sqrt{2} s i n θ} =_{t a n (\frac{θ}{2}) = x}

= \int_{0}^{1} \frac{2 d x}{(1 + x^{2}) {\sqrt{3} \frac{1 - x^{2}}{1 + x^{2}} + \sqrt{2} \frac{2 x}{1 + x^{2}}}}

= \int_{0}^{1} \frac{2 d x}{\sqrt{3} - \sqrt{3} x^{2} + 2 \sqrt{2} x}

= \int_{0}^{1} \frac{- 2 d x}{\sqrt{3} x^{2} - 2 \sqrt{2} x - \sqrt{3}}

l e t d r c o m p o s e F (x) = \frac{- 2}{\sqrt{3} x^{2} - 2 \sqrt{2} x - \sqrt{3}}

Δ^{^{'}} = 2 + 3 = 5 \Rightarrow x_{1} = \frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}}

\underset{2}{x} = \frac{\sqrt{2} - \sqrt{5}}{\sqrt{3}} \Rightarrow

F (x) = \frac{- 2}{\sqrt{3} (x - x_{1}) (x - x_{2})}

= \frac{- 2}{\sqrt{3} (x_{1} - x_{2})} {\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}}

= \frac{- 2}{\sqrt{3} (\frac{2 \sqrt{5}}{\sqrt{3}})} {\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}}

= - \frac{1}{\sqrt{5}} {\dots .} \Rightarrow

\int_{0}^{1} F (x) d x ? = \frac{1}{\sqrt{5}} [l n ∣ x - x_{2} ∣ - l n ∣ x - x_{1} ∣]

= \frac{1}{\sqrt{5}} {[l n ∣ \frac{x - x_{2}}{x - x_{1}} ∣]}_{0}^{1}

= \frac{1}{\sqrt{5}} (l n (\frac{1 - \frac{\sqrt{2} - \sqrt{5}}{\sqrt{3}}}{1 - \frac{\sqrt{2} + \sqrt{5}}{\sqrt{3}}}) - l n ∣ \frac{\sqrt{2} - \sqrt{5}}{\sqrt{2} + \sqrt{5}} ∣)

= \frac{1}{\sqrt{5}} (l n ∣ \frac{\sqrt{3} - \sqrt{2} + \sqrt{5}}{\sqrt{3} - \sqrt{2} + \sqrt{5}} ∣ - l n (\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}})

s o t h e v s l u e o f I i s k n o w n .