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Question Number 125981 by 0731619177 last updated on 16/Dec/20

	Answered by snipers237 last updated on 16/Dec/20
	$That is still equal to A=∫∫∫_D (√(x^2 +y^2 +z^2 )) dxdydz with D={(x,y,z)/ ∣y∣≤3 ,∣x∣≤(√(9−y^2 )) ,∣z∣≤(√(9−x^2 −y^2 )) } and it′s easy to show that D(O,3)⊆D⊆D(O,3) Then D={x^2 +y^2 +z^2 ≤3^2 } using the spheric coordonnate x=rsinθcosϕ ; y=rsinθsinϕ; z=rcosθ D≊{(r,θ,ϕ)/ 0≤r≤3 ,0≤θ≤π ,0≤ϕ≤2π} D(x,y,z)=Jac(r,θ,ϕ)D(r,θ,ϕ) ⇒dxdydz=r^2 sinθdrdθdϕ A=∫_0 ^3 ∫_0 ^π ∫_0 ^(2π) (√r^2 ) (r^2 sinθ)drdθdϕ =(∫_0 ^(2π) dϕ)(∫_0 ^3 r^3 dr)(∫_0 ^π sinθdθ) LFYTC$
	$T h a t i s s t i l l e q u a l t o$ $A = \int \int \int_{D} \sqrt{x^{2} + y^{2} + z^{2}} d x d y d z$ $w i t h D = {(x, y, z) / ∣ y ∣⩽ 3, ∣ x ∣⩽ \sqrt{9 - y^{2}}, ∣ z ∣⩽ \sqrt{9 - x^{2} - y^{2}}}$ $a n d {i t}^{'} s e a s y t o s h o w t h a t D (O, 3) \subseteq D \subseteq D (O, 3)$ $T h e n D = {x^{2} + y^{2} + z^{2} ⩽ 3^{2}}$ $u s i n g t h e s p h e r i c c o o r d o n n a t e x = r s i n θ c o s φ; y = r s i n θ s i n φ; z = r c o s θ$ $D ≊ {(r, θ, φ) / 0 ⩽ r ⩽ 3, 0 ⩽ θ ⩽ π, 0 ⩽ φ ⩽ 2 π}$ $D (x, y, z) = J a c (r, θ, φ) D (r, θ, φ) \Rightarrow d x d y d z = r^{2} s i n θ d r d θ d φ$ $A = \int_{0}^{3} \int_{0}^{π} \int_{0}^{2 π} \sqrt{r^{2}} (r^{2} s i n θ) d r d θ d φ$ $= (\int_{0}^{2 π} d φ) (\int_{0}^{3} r^{3} d r) (\int_{0}^{π} s i n θ d θ)$ $L F Y T C$
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