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Question Number 65401 by mathmax by abdo last updated on 29/Jul/19

l e t f (x, y) = (x + y) \sqrt{x + y - 1}

c a l c u l a t e \int \int_{D} f (x, y) d x d y w i t h

D = {(x, y) \in R^{2} / 1 ⩽ x ⩽ 2 a n d 1 ⩽ y ⩽ \sqrt{3}}

Commented by mathmax by abdo last updated on 01/Aug/19

\int \int_{D} f (x, y) d x d y = \int_{1}^{\sqrt{3}} (\int_{1}^{2} (x + y) \sqrt{x + y - 1} d x) d y = \int_{1}^{\sqrt{3}} A (y) d y

c h a n g e m e n t \sqrt{x + y - 1} = t g i v e x + y - 1 = t^{2} \Rightarrow x = t^{2} - y + 1 \Rightarrow

A (y) = \int_{\sqrt{y}}^{\sqrt{y + 1}} (t^{2} - y + 1 + y) t (2 t) d t

= 2 \int_{\sqrt{y}}^{\sqrt{y + 1}} (t^{4} + t^{2}) d t = 2 {[\frac{t^{5}}{5} + \frac{t^{3}}{3}]}_{\sqrt{y}}^{\sqrt{y + 1}} = 2 {\frac{{(\sqrt{y + 1})}^{5}}{5} + \frac{{(\sqrt{y + 1})}^{3}}{3}

- \frac{{(\sqrt{y})}^{5}}{5} - \frac{{(\sqrt{y})}^{3}}{3} = 2 {\frac{{(y + 1)}^{2} \sqrt{y + 1}}{5} + \frac{(y + 1) \sqrt{y + 1}}{3} - \frac{y^{2} \sqrt{y}}{5} - \frac{y \sqrt{y}}{3}} \Rightarrow

\int \int_{D} f (x, y) d x d y = \frac{2}{5} \int_{1}^{\sqrt{3}} {(y + 1)}^{2} \sqrt{y + 1} d y + \frac{2}{3} \int_{1}^{\sqrt{3}} (y + 1) \sqrt{y + 1} d y

- \frac{2}{5} \int_{1}^{\sqrt{3}} y^{2} \sqrt{y} d y - \frac{2}{3} \int_{1}^{\sqrt{3}} y \sqrt{y} d y

c h a n g e m e n t \sqrt{y + 1} = u g i v e y + 1 = u^{2} \Rightarrow

\int_{1}^{\sqrt{3}} {(y + 1)}^{2} \sqrt{y + 1} d y = \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{4} u (2 u) d u = 2 \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{6} d u

= \frac{2}{7} {[u^{7}]}_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} = \frac{2}{7} {{(\sqrt{\sqrt{3} + 1})}^{7} - {(\sqrt{2})}^{7}}

\int_{1}^{\sqrt{3}} (y + 1) \sqrt{y + 1} d y = \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{2} u (2 u d u) = 2 \int_{\sqrt{2}}^{\sqrt{\sqrt{3} + 1}} u^{4} d u

= \frac{2}{5} {{(\sqrt{\sqrt{3} + 1})}^{5} - {(\sqrt{2})}^{5}} a l s o c h a n g e m e n t \sqrt{y} = u g i v e y = u^{2}

\Rightarrow \int_{1}^{\sqrt{3}} y^{2} \sqrt{y} d y = \int_{1}^{\sqrt{\sqrt{3}}} u^{4} (u) (2 u d u) = 2 \int_{1}^{\sqrt{\sqrt{3}}} u^{6} d u = \frac{2}{7} [{(\sqrt{\sqrt{3}})}^{7} - 1)

\int_{1}^{\sqrt{3}} y \sqrt{y} d y = \int_{1}^{\sqrt{\sqrt{3}}} u^{2} u (2 u) d u = 2 \int_{1}^{\sqrt{\sqrt{3}}} u^{4} x u = \frac{2}{5} {{(\sqrt{\sqrt{3}})}^{5} - 1}

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