R ={(x,y): (x−2)^2 +y^2 ≤4} ∫∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 127160 by kaivan.ahmadi last updated on 27/Dec/20
$R ={(x,y): (x−2)^2 +y^2 ≤4} ∫∫_R (x^2 +y^2 )^2 dydx=?$
$R = {(x, y) : {(x - 2)}^{2} + y^{2} ⩽ 4}$ $\int \int_{R} {(x^{2} + y^{2})}^{2} d y d x = ?$
	Answered by mathmax by abdo last updated on 27/Dec/20
	$we use the diffeomorphism { ((x−2=rcosθ)),((y=rsinθ)) :} (x−2)^2 +y^2 ≤4 ⇒r^2 ≤4 ⇒0≤r≤2 ∫∫_R (x^2 +y^2 )^2 dxdy =∫_0 ^2 ∫_(−π) ^π ((rcosθ +2)^2 +r^2 sin^2 θ)^2 rdrdθ = ∫_0 ^2 ∫_(−π) ^π {r^2 cos^2 θ +4rcosθ +4+r^2 sin^2 θ)^2 drdθ = ∫_0 ^2 ∫_(−π) ^π {r^2 +4rcosθ +4)^2 drdθ =∫_0 ^2 ∫_(−π) ^π {(r^2 +4)^2 +8rcosθ(r^2 +4)+16r^2 cos^2 θ}drdθ =2π ∫_0 ^2 (r^2 +4)^(2 ) dr +8 ∫_0 ^2 (r^3 +4r)dr∫_(−π) ^π cosθ dθ(→0)+16∫_0 ^2 r^2 dr∫_(−π) ^π cos^2 θ dθ ∫_0 ^2 (r^2 +4)^2 dr =∫_0 ^2 (r^4 +8r^2 +16)dr =[(r^5 /5)+(8/3)r^3 +16r]_0 ^2 =.... ∫_0 ^2 r^2 dr =[(r^3 /3)]_0 ^2 =(8/3) ∫_(−π) ^π cos^2 θ dθ =2∫_0 ^π ((1+cos(2θ))/2)dθ =π rest to collect the values...$
	$we use the diffeomorphism {\begin{cases} x - 2 = rcos θ \\ y = rsin θ \end{cases}$ ${(x - 2)}^{2} + y^{2} ⩽ 4 \Rightarrow r^{2} ⩽ 4 \Rightarrow 0 ⩽ r ⩽ 2$ $\int \int_{R} {(x^{2} + y^{2})}^{2} dxdy = \int_{0}^{2} \int_{- π}^{π} {({(rcos θ + 2)}^{2} + r^{2} \sin^{2} θ)}^{2} rdrd θ$ $= \int_{0}^{2} \int_{- π}^{π} {{r^{2} \cos^{2} θ + 4 rcos θ + 4 + r^{2} \sin^{2} θ)}^{2} drd θ$ $= \int_{0}^{2} \int_{- π}^{π} {{r^{2} + 4 rcos θ + 4)}^{2} drd θ$ $= \int_{0}^{2} \int_{- π}^{π} {{(r^{2} + 4)}^{2} + 8 rcos θ (r^{2} + 4) + {16 r}^{2} \cos^{2} θ} drd θ$ $= 2 π \int_{0}^{2} {(r^{2} + 4)}^{2} dr + 8 \int_{0}^{2} (r^{3} + 4 r) dr \int_{- π}^{π} \cos θ d θ (\to 0) + 16 \int_{0}^{2} r^{2} dr \int_{- π}^{π} \cos^{2} θ d θ$ $\int_{0}^{2} {(r^{2} + 4)}^{2} dr = \int_{0}^{2} (r^{4} + {8 r}^{2} + 16) dr = {[\frac{r^{5}}{5} + \frac{8}{3} r^{3} + 16 r]}_{0}^{2} = . . . .$ $\int_{0}^{2} r^{2} dr = {[\frac{r^{3}}{3}]}_{0}^{2} = \frac{8}{3}$ $\int_{- π}^{π} \cos^{2} θ d θ = 2 \int_{0}^{π} \frac{1 + \cos (2 θ)}{2} d θ = π rest to collect the values . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum