Prove-that-0-2-1-y-1-2-0-xy-2-dxdy-4-5-after-changing-the-integral-to-polar-form-

Question Number 72466 by Learner-123 last updated on 29/Oct/19

P r o v e t h a t \int_{0}^{2} \int_{- \sqrt{1 - {(y - 1)}^{2}}}^{0} {x y}^{2} d x d y = - \frac{4}{5}

a f t e r changing the integral to polar form .

Commented by Abdo msup. last updated on 30/Oct/19

I = \int_{0}^{2} (\int_{- \sqrt{1 - {(y - 1)}^{2}}}^{0} x d x) y^{2} d y b u t

\int_{- \sqrt{1 - {(y - 1)}^{2}}}^{0} x d x = {[\frac{x^{2}}{2}]}_{- \sqrt{1 - {(y - 1)}^{2}}}^{0} = - \frac{1 - {(y - 1)}^{2}}{2}

= - \frac{1 - y^{2} + 2 y - 1}{2} = \frac{y^{2} - 2 y}{2} \Rightarrow

I = \frac{1}{2} \int_{0}^{2} (y^{2} - 2 y) y^{2} d y = \frac{1}{2} \int_{0}^{2} (y^{4} - 2 y^{3}) d y

= \frac{1}{2} {[\frac{y^{5}}{5} - \frac{1}{2} y^{4}]}_{0}^{2} = \frac{1}{2} {\frac{32}{5} - 8} = \frac{1}{2} \times \frac{- 8}{5} = - \frac{4}{5}

w i t h o u t u s e o f p o l a r c o o r d i n a t e s .

Commented by Learner-123 last updated on 30/Oct/19

t h a n k s s i r .

Answered by mind is power last updated on 29/Oct/19

x = rcos (θ)

y = rsin (θ)

dxdy = rdrd θ

{0 ⩾ x ⩾ - \sqrt{1 - (y - 1})}^{2}

0 ⩽ x^{2} ⩽ 1 - {(y - 1)}^{2} = 2 y - y^{2} \Rightarrow 0 ⩽ x^{2} + y^{2} ⩽ 2 y

0 ⩽ r^{2} ⩽ 2 rsin (θ) \Rightarrow 0 ⩽ r ⩽ 2 \sin (θ)

0 ⩽ y ⩽ 2

\Rightarrow 0 ⩽ rsin (θ) ⩽ 2 \Rightarrow \sin (θ) ⩾ 0 \Rightarrow θ \in [0, π]

but x ⩽ 0 \Rightarrow \cos (θ) ⩽ 0 \Rightarrow θ \in [\frac{π}{2}, 2]

our integral \Leftrightarrow \int_{\frac{π}{2}}^{π} \int_{0}^{2 \sin (θ)} . rcos (θ) . r^{2} \sin^{2} (θ) . rdrd θ

= \int_{\frac{π}{2}}^{π} \int_{0}^{2 \sin (θ)} . r^{4} \cos (θ) \sin^{2} (θ) drd θ

Commented by Learner-123 last updated on 29/Oct/19

O n e m o r e Q u e s . ::

\int_{- a}^{a} \int_{- \sqrt{a^{2} - x^{2}}}^{\sqrt{a^{2} - x^{2}}} d y d x .

Commented by Learner-123 last updated on 29/Oct/19

t h a n k y o u s i r!

Commented by mind is power last updated on 29/Oct/19

y^{'} re Welcom

Commented by mind is power last updated on 29/Oct/19

x = rcos (b)

y = rsin (b)

we have - a < x < a

- \sqrt{a^{2} - x^{2}} < y < \sqrt{a^{2} - x^{2}} \Rightarrow 0 < x^{2} + y^{2} ⩽ a^{2}

\Rightarrow 0 < R < a

b \in [0, 2 π]

\int_{0}^{2 π} \int_{0}^{a} rdrdb

= 2 π . (\frac{a^{2}}{2}) = π a^{2}

Commented by Learner-123 last updated on 29/Oct/19

t h a n k s a g a i n, s i r!

Commented by mind is power last updated on 29/Oct/19

most Welcom