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Question Number 126632 by BHOOPENDRA last updated on 22/Dec/20

Commented by BHOOPENDRA last updated on 22/Dec/20

$t h a n k u s i r$
	Answered by Ar Brandon last updated on 22/Dec/20
	$In polar coordinates; { ((x=rcosθ),(r≥0)),((y=rsinθ),(θ∈[0,2π])) :} 1≤x^2 +y^2 ≤4 ⇒ 1≤r^2 ≤4 ⇒ 1≤r≤2 x≤0 ⇒ rcosθ≤0 ⇒ cosθ≤0 ⇒ θ∈[(π/2),((3π)/2)] I=∫∫_R (x+y)dA =∫_((3π)/2) ^(π/2) ∫_1 ^2 r(rcosθ+rsinθ)drdθ =∫_((3π)/2) ^(π/2) ∫_1 ^2 r^2 (cosθ+sinθ)drdθ =[(r^3 /3)]_1 ^2 [sinθ−cosθ]_((3π)/2) ^(π/2) =((8/3)−(1/3))(1−−1)=((14)/3)$
	$In polar coordinates;$ ${\begin{cases} x = rcos θ & r ⩾ 0 \\ y = rsin θ & θ \in [0, 2 π] \end{cases}$ $1 ⩽ x^{2} + y^{2} ⩽ 4 \Rightarrow 1 ⩽ r^{2} ⩽ 4 \Rightarrow 1 ⩽ r ⩽ 2$ $x ⩽ 0 \Rightarrow rcos θ ⩽ 0 \Rightarrow \cos θ ⩽ 0 \Rightarrow θ \in [\frac{π}{2}, \frac{3 π}{2}]$ $I = \int \int_{R} (x + y) dA$ $= \int_{\frac{3 π}{2}}^{\frac{π}{2}} \int_{1}^{2} r (rcos θ + rsin θ) drd θ$ $= \int_{\frac{3 π}{2}}^{\frac{π}{2}} \int_{1}^{2} r^{2} (\cos θ + \sin θ) drd θ$ $= {[\frac{r^{3}}{3}]}_{1}^{2} {[\sin θ - \cos θ]}_{\frac{3 π}{2}}^{\frac{π}{2}}$ $= (\frac{8}{3} - \frac{1}{3}) (1 - - 1) = \frac{14}{3}$
	Answered by mathmax by abdo last updated on 22/Dec/20
	$A =∫∫_R (x+y)dxdy with R ={(x,y)/1≤x^2 +y^2 ≤4 and x≤0} we use the diffeomorphism { ((x=rcosθ)),((y=rsinθ)) :} 1≤x^2 +y^2 ≤4 ⇒1≤r^2 ≤4 ⇒1≤r≤2 x≤0 ⇒cosθ ≤0 ⇒θ ∈[(π/2),((3π)/2)] ⇒ A =∫_1 ^2 ∫_(π/2) ^((3π)/2) (rcosθ +rsinθ)rdrdθ =∫_1 ^2 r^2 dr ∫_(π/2) ^((3π)/2) (cosθ +sinθ)dθ =[(r^3 /3)]_1 ^2 ×[sinθ −cosθ]_(π/2) ^((3π)/2) =(1/3)(8−1)(sin(((3π)/2))−cos(((3π)/2))−sin((π/2))+cos((π/2))) =(7/3)(−1−o−1)=−((14)/3)$
	$A = \int \int_{R} (x + y) dxdy with R = {(x, y) / 1 ⩽ x^{2} + y^{2} ⩽ 4 and x ⩽ 0}$ $we use the diffeomorphism {\begin{cases} x = rcos θ \\ y = rsin θ \end{cases}$ $1 ⩽ x^{2} + y^{2} ⩽ 4 \Rightarrow 1 ⩽ r^{2} ⩽ 4 \Rightarrow 1 ⩽ r ⩽ 2$ $x ⩽ 0 \Rightarrow \cos θ ⩽ 0 \Rightarrow θ \in [\frac{π}{2}, \frac{3 π}{2}] \Rightarrow$ $A = \int_{1}^{2} \int_{\frac{π}{2}}^{\frac{3 π}{2}} (rcos θ + rsin θ) rdrd θ = \int_{1}^{2} r^{2} dr \int_{\frac{π}{2}}^{\frac{3 π}{2}} (\cos θ + \sin θ) d θ$ $= {[\frac{r^{3}}{3}]}_{1}^{2} \times {[\sin θ - \cos θ]}_{\frac{π}{2}}^{\frac{3 π}{2}} = \frac{1}{3} (8 - 1) (\sin (\frac{3 π}{2}) - \cos (\frac{3 π}{2}) - \sin (\frac{π}{2}) + \cos (\frac{π}{2}))$ $= \frac{7}{3} (- 1 - o - 1) = - \frac{14}{3}$
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