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Question Number 68241 by mathmax by abdo last updated on 07/Sep/19

c a l c u l a t e \int \int_{w} (x^{2} - 2 y^{2}) \sqrt{x^{2} + 3 y^{2}} d x d y

w i t h w = {(x, y) \in R^{2} / 0 ⩽ x ⩽ 1 a n d 1 ⩽ y ⩽ 2}

Commented by mathmax by abdo last updated on 10/Sep/19

w e u s e t h e d i f f e o m o r p h i s m (r, θ) \to φ (r, θ) = (x, y) = (r c o s θ, \frac{r}{\sqrt{3}} s i n θ)

= (φ_{1}, φ_{2}) w e h a v e 0 ⩽ x^{2} ⩽ 1 a n d 1 ⩽ y^{2} ⩽ 4 \Rightarrow 3 ⩽ 3 y^{2} ⩽ 12 \Rightarrow

3 ⩽ x^{2} + 3 y^{2} ⩽ 13 \Rightarrow \sqrt{3} ⩽ \sqrt{x^{2} + 3 y^{2}} ⩽ \sqrt{13} \Rightarrow \sqrt{3} ⩽ r ⩽ \sqrt{13}

M_{j} (φ) = (\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{1}{\sqrt{3}} s i n θ \frac{r}{\sqrt{3}} c o s θ \end{matrix}) \Rightarrow d e t (M_{j} (φ)) = \frac{r}{\sqrt{3}} {c o s}^{2} θ + \frac{r}{\sqrt{3}} {s i n}^{2} θ

= \frac{r}{\sqrt{3}} \Rightarrow \int \int_{w} (x^{2} - 2 y^{2}) \sqrt{x^{2} + 3 y^{2}} d x d y

= \int \int_{\sqrt{3} ⩽ r ⩽ \sqrt{13} a n d 0 ⩽ θ ⩽ \frac{π}{2}} (r^{2} {c o s}^{2} θ - \frac{2}{3} r^{2} {s i n}^{2} θ) r \frac{r}{\sqrt{3}} d r d θ

= \frac{1}{\sqrt{3}} \int_{\sqrt{3}}^{\sqrt{13}} r^{4} d r \int_{0}^{\frac{π}{2}} ({c o s}^{2} θ - \frac{2}{3} {s i n}^{2} θ) d θ w e h a v e

\int_{\sqrt{3}}^{\sqrt{13}} r^{4} d r = {[\frac{r^{5}}{5}]}_{\sqrt{3}}^{\sqrt{13}} = \frac{1}{5} {{(\sqrt{13})}^{5} - {(\sqrt{3})}^{5}

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

= \frac{1}{3} \int_{0}^{\frac{π}{2}} (3 \frac{1 + c o s (2 θ)}{2} - 2 \frac{1 - c o s (2 θ)}{2}) d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (2 θ)) d θ - \frac{1}{3} \int_{0}^{\frac{π}{2}} (1 - c o s (2 θ)) d θ

= \frac{π}{4} + \frac{1}{4} {[s i n (2 θ)]}_{0}^{\frac{π}{2}} - \frac{π}{6} + \frac{1}{6} {[s i n (2 θ)]}_{0}^{\frac{π}{2}}

= \frac{π}{12} \Rightarrow

\int \int_{w} (x^{2} - 2 y^{2}) \sqrt{x^{2} + 3 y^{2}} d x d y = \frac{1}{5 \sqrt{3}} {{(\sqrt{13})}^{5} - {(\sqrt{3})}^{5}} \frac{π}{12}

= \frac{π}{60 \sqrt{3}} {{(\sqrt{13})}^{5} - {(\sqrt{3})}^{5}} .