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Question Number 62197 by maxmathsup by imad last updated on 17/Jun/19

c a l c u l a t e \int \int_{{[0, 1]}^{2}} \sqrt{x^{2} + y^{2}} s i n (\sqrt{x^{2} + y^{2}}) d x d y

Commented by mathmax by abdo last updated on 21/Jun/19

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

0 ⩽ x ⩽ 1 a n d 0 ⩽ y ⩽ 1 \Rightarrow 0 ⩽ x^{2} + y^{2} ⩽ 2 \Rightarrow 0 ⩽ r^{2} ⩽ 2 \Rightarrow 0 ⩽ r ⩽ \sqrt{2}

A = \int \int_{{[0, 1]}^{2}} \sqrt{x^{2} + y^{2}} s i n (\sqrt{x^{2} + y^{2}}) d x d y = \int \int_{0 ⩽ r ⩽ \sqrt{2} a n d 0 ⩽ θ ⩽ \frac{π}{2}} r s i n (r) r d r d θ

= \int_{0}^{\sqrt{2}} r^{2} s i n (r) d r \int_{0}^{\frac{π}{2}} d θ = \frac{π}{2} \int_{0}^{\sqrt{2}} r^{2} s i n (r) d r b y p a r t s u = r^{2} a n d v^{^{'}} = s i n r

\int_{0}^{\sqrt{2}} r^{2} s i n (r) d r = {[- r^{2} c o s (r)]}_{0}^{\sqrt{2}} + \int_{0}^{\sqrt{2}} 2 r c o s r d r

= - 2 c o s (\sqrt{2}) + 2 {{[r s i n (r)]}_{0}^{\sqrt{2}} - \int_{0}^{\sqrt{2}} s i n r d r}

= - 2 c o s (\sqrt{2}) + 2 {\sqrt{2} s i n (\sqrt{2}) + {[c o s r]}_{0}^{\sqrt{2}}}

= - 2 c o s (\sqrt{2}) + 2 \sqrt{2} s i n (\sqrt{2}) + 2 (c o s (\sqrt{2}) - 1) \Rightarrow

A = \frac{π}{2} {- 2 c o s (\sqrt{2}) + 2 \sqrt{2} s i n (\sqrt{2}) + 2 c o s (\sqrt{2}) - 2} = π {\sqrt{2} s i n (\sqrt{2}) - 1} .