find ∫∫∫_D (x^2 +y^2 )dxdxy with D={x,y,z)∈R^3 /x^2 +y^2 +z^2 ≤1 and z≥0 }


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Question Number 27598 by abdo imad last updated on 10/Jan/18
$find ∫∫∫_D (x^2 +y^2 )dxdxy with D={x,y,z)∈R^3 /x^2 +y^2 +z^2 ≤1 and z≥0 }$
$f i n d \int \int \int_{D} (x^{2} + y^{2}) d x d x y w i t h$ $D = {x, y, z) \in R^{3} / x^{2} + y^{2} + z^{2} ⩽ 1 a n d z ⩾ 0}$
Commented by abdo imad last updated on 21/Jan/18
$we have z^2 ≤ 1−(x^2 +y^2 ) ≤1 and z≥0 ⇒ 0≤z≤1 so I= ∫∫∫ (x^2 +y^2 +z^2 )dxdydz = ∫_0 ^1 (∫∫_W (x^2 +y^2 )dxdy)dz with W ={ (x,y)∈R^2 / x^2 +y^2 ≤1−z^2 and z≥0} let define a diffeomorphisme on W we use the polar coordinate x=r cosθ and y =rsinθ we must have 0<r ≤(√(1−z^2 )) and 0≤θ≤2π ∫∫_W ( x^2 +y^2 )dxdy= ∫∫_(0<r≤(√(1−z^2 )) and 0≤θ≤2π) r^2 rdrdθ = 2π ∫_0 ^(√(1−z^2 )) r^3 dr = (π/2) [ r^4 ]_0 ^(√(1−z^2 )) = (π/2)(1−z^2 )^2 = (π/2)( z^4 −2z^2 +1) and I= (π/2) ∫_0 ^1 (z^4 −2z^2 +1)dz =(π/2) [ (z^5 /5) −(2/3) z^3 +z]_0 ^1 = (π/2) ( (1/5) −(2/3) +1) = (π/2)( (1/5) +(1/3)) = ((8π)/(30)) = ((4π)/(15)) .$
$w e h a v e z^{2} ⩽ 1 - (x^{2} + y^{2}) ⩽ 1 a n d z ⩾ 0 \Rightarrow 0 ⩽ z ⩽ 1 s o$ $I = \int \int \int (x^{2} + y^{2} + z^{2}) d x d y d z = \int_{0}^{1} (\int \int_{W} (x^{2} + y^{2}) d x d y) d z$ $w i t h W = {(x, y) \in R^{2} / x^{2} + y^{2} ⩽ 1 - z^{2} a n d z ⩾ 0} l e t d e f i n e$ $a d i f f e o m o r p h i s m e o n W w e u s e t h e p o l a r c o o r d i n a t e$ $x = r c o s θ a n d y = r s i n θ w e m u s t h a v e 0 < r ⩽ \sqrt{1 - z^{2}}$ $a n d 0 ⩽ θ ⩽ 2 π$ $\int \int_{W} (x^{2} + y^{2}) d x d y = \int \int_{0 < r ⩽ \sqrt{1 - z^{2}} a n d 0 ⩽ θ ⩽ 2 π} r^{2} r d r d θ$ $= 2 π \int_{0}^{\sqrt{1 - z^{2}}} r^{3} d r = \frac{π}{2} {[r^{4}]}_{0}^{\sqrt{1 - z^{2}}} = \frac{π}{2} {(1 - z^{2})}^{2}$ $= \frac{π}{2} (z^{4} - 2 z^{2} + 1) a n d$ $I = \frac{π}{2} \int_{0}^{1} (z^{4} - 2 z^{2} + 1) d z = \frac{π}{2} {[\frac{z^{5}}{5} - \frac{2}{3} z^{3} + z]}_{0}^{1}$ $= \frac{π}{2} (\frac{1}{5} - \frac{2}{3} + 1) = \frac{π}{2} (\frac{1}{5} + \frac{1}{3}) = \frac{8 π}{30} = \frac{4 π}{15} .$
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