let-D-x-y-R-2-x-gt-0-y-gt-0-x-y-lt-1-1-calculate-D-xy-x-2-y-2-dxdy-2-let-a-gt-0-b-gt-0-calculate-D-a-x-b-y-dxdy-

Question Number 36191 by prof Abdo imad last updated on 30/May/18

l e t D = {(x, y) \in R^{2} / x > 0, y > 0, x + y < 1}

1) c a l c u l a t e \int \int_{D} \frac{x y}{x^{2} + y^{2}} d x d y

2) l e t a > 0, b > 0 c a l c u l a t e \int \int_{D} a^{x} b^{y} d x d y

Commented by maxmathsup by imad last updated on 26/Aug/18

p o l a r c o o r d i n a t e s c h a n g e m e n t x = r c o s θ a n d y = r s i n θ

x > 0 a n d y > 0 \Rightarrow 0 < θ < \frac{π}{2} a n d x + y < 1 \Rightarrow r (c o s θ + s i n θ) < 1 \Rightarrow

0 < r < \frac{1}{c o s θ + s i n θ}

\int \int_{D} \frac{x y}{x^{2} + y^{2}} d x d y = \int_{0}^{\frac{π}{2}} (\int_{0}^{\frac{1}{c o s θ + s i n θ}} \frac{r^{2} c o s θ s i n θ}{r^{2}} r d r) d θ

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

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

I = \int_{0}^{\frac{π}{2}} \frac{c o s θ s i n θ}{2 (1 + 2 s i n θ c o s θ)} d θ = \frac{1}{4} \int_{0}^{\frac{π}{2}} \frac{s i n (2 θ)}{1 + s i n (2 θ)} d θ

=_{2 θ = t} \frac{1}{4} \int_{0}^{π} \frac{s i n t}{1 + s i n t} \frac{d t}{2} = \frac{1}{8} \int_{0}^{π} \frac{1 + s i n t - 1}{1 + s i n t} d t

= \frac{π}{8} - \frac{1}{8} \int_{0}^{π} \frac{d t}{1 + s i n t} c h a n g . t a n (\frac{t}{2}) = u g i v e

\int_{0}^{π} \frac{d t}{1 + s i n t} = \int_{0}^{\infty} \frac{1}{1 + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{\infty} \frac{d u}{1 + u^{2} + 2 u}

= 2 \int_{0}^{\infty} \frac{d u}{{(u + 1)}^{2}} = 2 {[- \frac{1}{u + 1}]}_{0}^{+ \infty} = 2 \Rightarrow I = \frac{π}{8} - \frac{1}{4} .