find-the-value-of-D-x-2-y-dxdy-on-the-domain-D-x-y-R-2-x-2-y-2-2x-0-and-y-0-

Question Number 26398 by abdo imad last updated on 25/Dec/17

f i n d t h e v a l u e o f \int \int_{D} x^{2} y d x d y o n t h e d o m a i n

D = {(x . y) \in R^{2} / x^{2} + y^{2} - 2 x ⩽ 0 a n d y ⩾ 0}

Commented by abdo imad last updated on 25/Dec/17

l e t p u t f (x ., y) = x^{2} y a n d ω = {(r, θ) / 0 ⩽ r ⩽ 1 a n d 0 ⩽ θ ⩽ π t h e

a p l i c a t i o n φ w - - > D / (r, θ) - > φ (r, θ) = (x, y) a n d

x = 1 + r c o s θ, y = r s i n θ i s a C^{1} d f f e o m o r p h s m e \Rightarrow

\int \int_{D} f (x, y) d x d y == \int \int_{w} f o φ (r, θ) / j_{φ} / d r d θ

M_{j (φ)} = (_{s n θ r c o s θ}^{c o s θ - r s i n θ}) a n d d e t (M j (φ)) = r \Rightarrow

\int \int_{D} f (x, y) d x d y = \int \int_{0 ⩽ r ⩽ 1_{0 ⩽ θ ⩽ π}} {(1 + r c o s θ)}^{2} r s i n θ r d r d θ

= \int \int_{w} r^{2} (1 + 2 r c o s θ + r^{2} {c o s}^{2} θ) s i n θ d r d θ

= \int_{0}^{1} r^{2} d r . \int_{0}^{π} s i n θ d θ + 2 \int_{0}^{1} r^{3} d r \int_{0}^{π} c o s θ s i n θ d θ + \int_{0}^{1} r^{4} d r . \int_{0}^{π} {c o s}^{2} θ s i n θ d θ

= \frac{2}{3} + \frac{1}{4} . 0 + \frac{1}{5} . (- \frac{1}{3}) (- 2) = \frac{2}{3} + \frac{2}{15} = \frac{4}{5}

Commented by ajfour last updated on 25/Dec/17

Answered by kaivan.ahmadi last updated on 25/Dec/17

Commented by prakash jain last updated on 25/Dec/17

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Answered by ajfour last updated on 25/Dec/17

x = r \cos θ, y = r \sin θ, r_{m a x} = 2 \cos θ_{m a x}

d x d y = (r d r) d θ

\int \int_{D} (x^{2} y) (d x d y)

= \int_{0}^{π / 2} \int_{0}^{2 \cos θ_{m a x}} (r^{3} \cos^{2} θ \sin θ) (r d r d θ)

= \int_{0}^{π / 2} [\int_{0}^{2 \cos θ_{m a x}} (r^{4} d r)] \cos^{2} θ \sin θ d θ

= \int_{0}^{π / 2} \frac{{(2 \cos θ)}^{5} \cos^{2} θ \sin θ}{5} d θ

= \frac{32}{5} \times \frac{{(\cos θ)}^{8}}{8} ∣_{π / 2}^{0} = \frac{4}{5} .

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