Evaluate-1-2-2-4-x-2-4-x-2-3-x-dydx-after-changing-the-integral-to-polar-form-2-0-4-0-4-x-0-4-y-2-4-dzdydx-

Question Number 73428 by Learner-123 last updated on 12/Nov/19

E v a l u a t e :

1) \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} (3 - x) d y d x .

(a f t e r c h a n g i n g t h e i n t e g r a l t o p o l a r f o r m) .

2) \int_{0}^{4} \int_{0}^{4 - x} \int_{0}^{4 - \frac{y^{2}}{4}} d z d y d x .

Commented by mathmax by abdo last updated on 12/Nov/19

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

= 2 \int_{- 2}^{2} (3 - x) \sqrt{4 - x^{2}} d x =_{x = 2 s i n θ} 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} (3 - 2 s i n θ) 2 c o s θ (2 c o s θ) d θ

= 8 \int_{- \frac{π}{2}}^{\frac{π}{2}} (3 - 2 s i n θ) {c o s}^{2} θ d θ = 24 \int_{- \frac{π}{2}}^{\frac{π}{2}} {c o s}^{2} θ d θ - 16 \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n θ {c o s}^{2} θ d θ

w e h a v e \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n θ {c o s}^{2} θ d θ = {[- \frac{1}{3} {c o s}^{3} θ]}_{- \frac{π}{2}}^{\frac{π}{2}} = 0

\int_{- \frac{π}{2}}^{\frac{π}{2}} {c o s}^{2} θ d θ = 2 \int_{0}^{\frac{π}{2}} \frac{1 + c o s (2 θ)}{2} d θ = \frac{π}{2} + \int_{0}^{\frac{π}{2}} c o s (2 θ) d θ

= \frac{π}{2} + {[\frac{1}{2} s i n (2 θ)]}_{0}^{\frac{π}{2}} = \frac{π}{2} + 0 = \frac{π}{2} \Rightarrow I = 24 \times \frac{π}{2} \Rightarrow I = 12 π

Commented by Learner-123 last updated on 12/Nov/19

t h n k s s i r .

Answered by Henri Boucatchou last updated on 12/Nov/19

1) I = \int_{- 2}^{2} 2 (3 - x) \sqrt{4 - x^{2}} d x

x = 2 c o s θ, - 1 ⩽ c o s θ ⩽ 1, d x = - 2 s i n θ d θ

I = 4 \int_{- π}^{0} (3 - 2 c o s θ) ∣ s i n θ ∣ (- 2 s i n θ) d θ

= 8 [\int_{- π}^{0} (3 - 2 c o s θ) {s i n}^{2} θ d θ]

= 8 [3 \int_{- π}^{0} {s i n}^{2} θ d θ - 2 \int_{- π}^{0} c o s θ {s i n}^{2} θ d θ]

= 8 [\frac{3}{2} \int_{- π}^{0} (1 - c o s 2 θ) d θ - 2.0]

= 8 . \frac{3}{2} {[θ - \frac{1}{2} s i n 2 θ]}_{- π}^{0}

= 12 π

- - - - - - - - - - - - - - -

\int_{0}^{4} \int_{0}^{4 - x} \int_{0}^{4 - y^{2} / 4} d z d y d x =

\int_{0}^{4} \int_{0}^{4 - x} (4 - \frac{y^{2}}{4}) d y d x =

\int_{0}^{4} {[4 y - \frac{y^{3}}{12}]}_{0}^{4 - x} d x =

\int_{0}^{4} (4 (4 - x) - \frac{1}{12} {(4 - x)}^{3}) d x =

{4 (x - \frac{1}{2} x^{2}) + \frac{1}{12.4} {(4 - x)}^{4}]}_{0}^{4} =

4 (4 - \frac{1}{2} 4^{2}) - \frac{1}{12} =

- 16 - \frac{1}{12} =

- \frac{193}{12}

P L E A S E C H E C K I F {T H E R E}^{'} S N O E R R O R

Commented by Learner-123 last updated on 12/Nov/19

{m a}^{'} a m, t h e r e i s s l i g h t m i s t a k e i n 4 t h l a s t

l i n e .