let-D-x-y-R-2-x-2-y-2-x-lt-0-and-x-2-y-2-y-gt-0-and-y-gt-0-calculate-D-x-y-2-dxdy-

Question Number 36193 by prof Abdo imad last updated on 30/May/18

l e t D = {(x, y) \in R^{2} / x^{2} + y^{2} - x < 0 a n d

x^{2} + y^{2} - y > 0 a n d y > 0}

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

Commented by maxmathsup by imad last updated on 15/Aug/18

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 g i v e

x^{2} + y^{2} - x < 0 \Rightarrow r^{2} - r c o s θ < 0 \Rightarrow r < c o s θ

x^{2} + y^{2} - y > 0 \Rightarrow r^{2} - r s i n θ > 0 \Rightarrow s i n θ < r \Rightarrow s i n θ < r < c o s θ

\int \int_{D} {(x + y)}^{2} d x d y = \int \int_{D} (x^{2} + y^{2} + 2 x y) d x d y

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

= \int_{0}^{2 π} (\int_{s i n θ}^{c o s θ} r^{3} d r + \int_{s i n θ}^{c o s θ} 2 r^{3} s i n θ c o θ d r) d θ

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

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

= 0 + \frac{1}{4} \int_{0}^{2 π} s i n θ {c o s}^{3} θ d θ - \frac{1}{4} \int_{0}^{2 π} c o s θ {s i n}^{3} θ

= - \frac{1}{16} {[{c o s}^{4} θ]}_{0}^{2 π} - \frac{1}{16} {[{s i n}^{4} θ]}_{0}^{2 π} = 0 \Rightarrow I = 0