calculate-D-xy-x-2-2y-2-dxdy-D-x-y-0-x-1-and-0-y-1-x-2-

Question Number 78284 by msup trace by abdo last updated on 15/Jan/20

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

D = {(x, y) / 0 ⩽ x ⩽ 1 a n d 0 ⩽ y ⩽ \sqrt{1 - x^{2}}}

Commented by abdomathmax last updated on 17/Jan/20

I = \int \int_{D} x y \sqrt{x^{2} + 2 y^{2}} d x d y \Rightarrow

I = \int_{0}^{1} (\int_{0}^{\sqrt{1 - x^{2}}} y \sqrt{x^{2} + 2 y^{2}} d y) x d x b u t

\int_{0}^{\sqrt{1 - x^{2}}} y \sqrt{x^{2} + 2 y^{2}} d y = {[\frac{1}{6} {(x^{2} + 2 y^{2})}^{\frac{3}{2}}]}_{0}^{\sqrt{1 - x^{2}}}

= \frac{1}{6} {(x^{2} + {(x^{2} + 2 - 2 x^{2})}^{\frac{3}{2}} - x^{3}}

= \frac{1}{6} {(x^{2} - x^{3} + {(2 - x^{2})}^{\frac{3}{2}}} \Rightarrow

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

= \int_{0}^{1} (x^{3} - x^{4}) d x + \int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x

= {[\frac{x^{4}}{4} - \frac{x^{5}}{5}]}_{0}^{1} + \int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x

= \frac{1}{20} + \int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x w e h a v e

\int_{0}^{1} x {(2 - x^{2})}^{\frac{3}{2}} d x =_{x = \sqrt{2} s i n t} \int_{0}^{\frac{π}{4}} \sqrt{2} s i n t {(2 - 2 {s i n}^{2} t)}^{\frac{3}{2}} \sqrt{2} c o s t d t

= 2 \times 2^{\frac{3}{2}} \int_{0}^{\frac{π}{4}} s i n t {c o s}^{4} t d t

= 2^{\frac{5}{2}} {[- \frac{1}{5} {c o s}^{5} t]}_{0}^{\frac{π}{4}} = - \frac{4 \sqrt{2}}{5} ({(\frac{1}{\sqrt{2}})}^{5} - 1)

= - \frac{4 \sqrt{2}}{5} (\frac{1}{4 \sqrt{2}} - 1) = - \frac{1}{5} + \frac{4 \sqrt{2}}{5} \Rightarrow

6 I = \frac{1}{20} - \frac{1}{5} + \frac{4 \sqrt{2}}{5} = \frac{- 3}{20} + \frac{4 \sqrt{2}}{5} \Rightarrow

I = - \frac{1}{40} + \frac{2 \sqrt{2}}{15}