find-4-x-2-y-2-dxdy-with-x-y-R-2-x-2-y-2-2x-

Question Number 27500 by abdo imad last updated on 07/Jan/18

f i n d \int \int_{Δ} \sqrt{4 - x^{2} - y^{2}} d x d y w i t h

Δ = {(x, y) \in R^{2} / x^{2} + y^{2} ⩽ 2 x}

Commented by abdo imad last updated on 10/Jan/18

l e t u s e t h e c h a n g e m e n t x = r c o s θ a n d y = r s i n θ

x^{2} + y^{2} ⩽ 2 x \Leftrightarrow r^{2} ⩽ 2 r c o s θ \Leftrightarrow 0 < r ⩽ 2 c o s θ

I = \int \int_{-_{} \frac{π}{2} < θ < \frac{π}{2} a n d 0 < r ⩽ 2 c o s θ} \sqrt{4 - r^{2}} r d r d θ

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\int_{0}^{2 c o s θ} r \sqrt{4 - r^{2}} d r) d θ b u t

\int_{0}^{2 c o s θ} r \sqrt{4 - r^{2}} d r = {[- \frac{1}{3} {(4 - r^{2})}^{\frac{3}{2}}]}_{0}^{2 c o s θ}

= - \frac{1}{3} ({(4 - 4 {c o s}^{2} θ)}^{\frac{3}{2}} - 4^{\frac{3}{2}}

= - \frac{1}{3} (4^{\frac{3}{2}} {({s i n}^{2})}^{\frac{3}{2}} - 8) = - \frac{1}{3} (8 {s i n}^{3} θ - 8)

I = - \frac{8}{3} \int_{- \frac{π}{2}}^{\frac{π}{2}} ({s i n}^{3} θ - 1) d θ = \frac{8 π}{3} - \frac{8}{3} \int_{- \frac{π}{2}}^{\frac{π}{2}} {s i n}^{3} θ d θ

w e f i n d t h e v a l u e o f I b y l i n e a r i s a t i o n o f {s i n}^{3} θ \dots .

Commented by abdo imad last updated on 10/Jan/18

t h e f o n c t i o n i s i m p a r s o \int_{- \frac{π}{2}}^{\frac{π}{2}} {s i n}^{3} d x = 0 a n d I = \frac{8 π}{3} .