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Question Number 170663 by kndramaths last updated on 28/May/22

Commented by kaivan.ahmadi last updated on 28/May/22

$a .$ $0 ⩽ x ⩽ a, 0 ⩽ y ⩽ \sqrt{a^{2} - x^{2}}$ $\int \frac{d x}{a^{2} + x^{2}} d x = \frac{1}{a} a r c t g (\frac{x}{a})$ $\Rightarrow a = \int \frac{y}{a} a r c t g (\frac{x}{a}) ∣_{0}^{a} d y = \int \frac{y}{π} a r c t g 1 d y =$ $\int \frac{π}{4} \times \frac{y}{π} d y = \frac{y^{2}}{8} ∣_{0}^{\sqrt{a^{2} - x^{2}}} = \frac{a^{2} - x^{2}}{8}$
	Answered by Mathspace last updated on 29/May/22

	$A = \int \int_{D} \frac{y}{x^{2} + a^{2}} d x d y$ $w e d o t h e c h a n g e m e n t x = a r c o s θ$ $e t y = a r s i n θ \Rightarrow$ $x^{2} + y^{2} ⩽ a^{2} \Rightarrow a^{2} r^{2} ⩽ a^{2} \Rightarrow 0 ⩽ r ⩽ 1$ $x ⩾ 0 a n d y ⩾ 0 \Rightarrow 0 ⩽ θ ⩽ \frac{π}{2}$ $\Rightarrow A = \int \int_{o ⩽ r ⩽ 1 a n d o ⩽ θ ⩽ \frac{π}{2}} \frac{a r s i n θ}{a^{2} r^{2} {c o s}^{2} θ + a^{2}} a^{2} r d r d θ$ $= a \int \int_{o ⩽ r ⩽ 1 a n d 0 ⩽ θ ⩽ \frac{π}{2}} \frac{r^{2} s i n θ}{1 + r^{2} {c o s}^{2} θ} d r d θ$ $= \int_{0}^{1} (\int_{0}^{\frac{π}{2}} \frac{r^{2} s i n θ}{1 + r^{2} {c o s}^{2} θ} d θ) d r$ $b u t \int_{0}^{\frac{π}{2}} \frac{r^{2} s i n θ}{1 + r^{2} {c o s}^{2} θ} =_{r c o s θ = u}$ $= \int_{r}^{0} \frac{- r d u}{1 + u^{2}} = r \int_{0}^{r} \frac{d u}{1 + u^{2}}$ $= r a r c t a n r \Rightarrow$ $A = \int_{0}^{1} r a r c t a n (r) d r$ $= {[\frac{r^{2}}{2} a r c t a n r]}_{0}^{1} - \int_{0}^{1} \frac{r^{2}}{2} \frac{d r}{1 + r^{2}}$ $= \frac{π}{8} - \frac{1}{2} \int_{0}^{1} \frac{1 + r^{2} - 1}{1 + r^{2}} d r$ $= \frac{π}{8} - \frac{1}{2} + \frac{1}{2} \int_{0}^{1} \frac{d r}{1 + r^{2}}$ $= \frac{π}{8} - \frac{1}{2} + \frac{π}{8} = \frac{π}{4} - \frac{1}{2}$
	Commented by Mathspace last updated on 29/May/22

	$A = (\frac{π}{4} - \frac{1}{2}) a$
	Answered by Mathspace last updated on 29/May/22

	$C = \int \int \int_{V} x y d x d y d z$ $w e d o t h e c h a n g e m e n t x = r c o s θ$ $y = r s i n θ \Rightarrow C = \int_{0}^{1} (\int \int_{w} x y d x d y) d z$ $x^{2} + y^{2} ⩽ z^{2} \Rightarrow o ⩽ r ⩽ z \Rightarrow$ $\int \int_{w} x y d x d y = \int \int_{o ⩽ r ⩽ z a n d o ⩽ θ ⩽ 2 π} (r c o s θ) (r s i n θ) r d r d θ$ $= \int_{0}^{z} r^{3} d r \int_{0}^{2 π} c o s θ s i n θ$ $= \frac{z^{4}}{8} \int_{0}^{2 π} s i n (2 θ) d θ$ $= \frac{z^{4}}{8} {[- \frac{1}{2} c o s (2 θ)]}_{0}^{2 π} = 0 \Rightarrow C = 0$
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