calculate-x-2-2y-2-1-x-2-y-2-dxdy-

Question Number 34714 by abdo mathsup 649 cc last updated on 10/May/18

c a l c u l a t e \int \int_{x^{2} + 2 y^{2} ⩽ 1} (x^{2} - y^{2}) d x d y

Commented by math khazana by abdo last updated on 10/May/18

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

(r, θ) \to (x, y) = (r c o s θ, \frac{r}{\sqrt{2}} s i n θ) = (φ_{1} (r, θ), φ_{2} (r, θ))

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} c o s θ - r s i n θ \\ \frac{s i n θ}{\sqrt{2}} \frac{r}{\sqrt{2}} c o s θ \end{matrix}) a n d d e t (M_{j}) =

\frac{r}{\sqrt{2}} {c o s}^{2} θ + \frac{r}{\sqrt{2}} {s i n}^{2} θ = \frac{1}{\sqrt{2}} r

I = \int \int_{0 ⩽ r ⩽ 1, - π ⩽ θ ⩽ π} (r^{2} {c o s}^{2} θ - \frac{r^{2}}{2} {s i n}^{2} θ) \frac{r}{\sqrt{2}} d r d θ

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

= \frac{1}{4 \sqrt{2}} \int_{- π}^{π} {c o s}^{2} θ d θ - \frac{1}{8 \sqrt{2}} \int_{- π}^{π} {s i n}^{2} θ d θ b u t

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

= \int_{0}^{π} (1 + c o s (2 θ)) d θ = π

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

= \int_{0}^{π} (1 - c o s (2 θ)) d θ = π \Rightarrow

I = \frac{π}{4 \sqrt{2}} - \frac{π}{8 \sqrt{2}} = \frac{2 π - π}{8 \sqrt{2}} = \frac{π}{8 \sqrt{2}}

I = \frac{π}{8 \sqrt{2}} .