Evaluate-A-x-y-2-dxdy-over-the-area-bounded-by-the-ellipse-x-2-a-2-y-2-b-2-1-Anybody-

Question Number 190052 by Mastermind last updated on 26/Mar/23

Evaluate \int \int_{A} {(x + y)}^{2} dxdy over the

area bounded by the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

Anybody ?

Answered by CElcedricjunior last updated on 26/Mar/23

\int \int_{A} {(x + y)}^{2} dxdy

A = {(x; y); \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1}

E n c h a n g e a n t l e s c o o r d o n n e e s d e c a r t e s i e n n e \overset{`}{a} p o l a i r e

o n a : {\begin{cases} x = a r c o s θ \\ y = b r s i n θ \end{cases} 1 > r > 0 θ \in [0; 2 π]

e t d x d y =∣ | \begin{matrix} a c o s θ - a r s i n θ \\ b s i n θ b r c o s θ \end{matrix} | ∣ d r d θ

d x d y = a b r d r d θ

A = {(r; θ) 0 < r < 1; 0 ⩽ θ ⩽ 2 π} M o i v r e

\int \int_{A} {a b r}^{3} {(a c o s θ + b s i n θ)}^{2} d r d θ = k

k = \int_{0}^{2 π} {[\frac{a b}{4} r^{4} {(a c o s θ + b s i n θ)}^{2}]}_{0}^{1} d θ

= \frac{a b}{4} \int_{0}^{2 π} (a^{2} c o s \overset{2}{θ} + 2 a b c o s θ s i n θ + b^{2} s i \overset{2}{n θ}) d θ

\frac{a b}{4} {[a^{2} (\frac{1}{2} θ + \frac{1}{4} s i n 2 θ) - \frac{a b c o s 2 θ}{2} + b^{2} (\frac{1}{2} θ - \frac{1}{4} s i n 2 θ)]}_{0}^{2 π}

= \frac{a b}{4} (π (a^{2} + b)) ★ C e d r i c j u n i o r

= \frac{π}{4} a b (a^{2} + b^{2})