{ (( If, D : x^2 +y^( 2) + z^( 2) ≤1)),(( ⇒∫∫_D^ ∫(( x^2 + 2y^( 2) )/(x^2 + 4y^2 +z^2 )) dxdydz=?)) :}


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Question Number 210787 by mnjuly1970 last updated on 19/Aug/24
${ (( If, D : x^2 +y^( 2) + z^( 2) ≤1)),(( ⇒∫∫_D^ ∫(( x^2 + 2y^( 2) )/(x^2 + 4y^2 +z^2 )) dxdydz=?)) :}$
${\begin{cases} I f, D : x^{2} + y^{2} + z^{2} ⩽ 1 \\ \Rightarrow \int \int_{\overset{}{D}} \int \frac{x^{2} + 2 y^{2}}{x^{2} + 4 y^{2} + z^{2}} d x d y d z = ? \end{cases}$
	Answered by BHOOPENDRA last updated on 19/Aug/24

	$\frac{2 π}{3}$
	Commented by mnjuly1970 last updated on 19/Aug/24

	$p l e a s e w i t h e x p l a n a t i o n \underset{⏟}{λ}$
	Answered by Berbere last updated on 19/Aug/24

	$(x, y, z) \to (z, y, x)$ $d o n t C h a n g e D b y s y m e t r i e$ $= V = \int \int \int_{D} \frac{x^{2} + 2 y^{2}}{x^{2} + 4 y^{2} + z^{2}} d x d y d z = \int \int \int_{D} \frac{z^{2} + 2 y^{2} + x^{2}}{x^{2} + 4 y^{2} + z^{2}} d x d y d z$ $\Rightarrow 2 V = \int \int \int_{D} \frac{x^{2} + 4 y^{2} + z^{2}}{x^{2} + 4 y^{2} + z^{2}} d x d y d z = 2 V = \int \int \int_{D} d x d y d z$ $= 2 V = V o l u m (D) = \frac{4}{3} π \in S p h e r o f r a d i u s 1$ $V = \frac{2 π}{3}$
	Commented by mnjuly1970 last updated on 19/Aug/24

	$⋚$
	Answered by BHOOPENDRA last updated on 19/Aug/24

	Commented by BHOOPENDRA last updated on 21/Aug/24

	$x^{2} + y^{2} + z^{2} = 1$ $f (x, y) = \frac{x^{2} + 2 y^{2}}{x^{2} + 4 y^{2} + z^{2}}$
	Answered by BHOOPENDRA last updated on 19/Aug/24

	Answered by BHOOPENDRA last updated on 19/Aug/24
	$Given that the region D is a sphere x^2 +y^2 +z^2 ≤1 V=∫∫∫ ((x^2 +2y^2 )/(x^2 +4y^2 +z^2 )) dxdydz if we apply the transformation (x,y,z)→(z,y,x) integral becomes V=∫∫∫_D ((z^2 +2y^2 )/(z^2 +4y^2 +x^2 )) dxdydz By symmetry swapping x and z does not change the value of the integral because integrand and the limits of integration remain the same. This leads to 2V=∫∫∫_D (((x^2 +2y^2 )/(x^2 +4y^2 +z^2 ))+((z^2 +2y^2 )/(z^2 +4y^2 +x^2 )))dxdydz Now ∫∫∫_D 1dxdydz 2V=volume of D {Vol(D)} =(4/3)π r^3 =(4/3)π1^3 =(4/3)π V=(2/3)π$
	$G i v e n t h a t t h e r e g i o n D i s$ $a s p h e r e x^{2} + y^{2} + z^{2} ⩽ 1$ $V = \int \int \int \frac{x^{2} + 2 y^{2}}{x^{2} + 4 y^{2} + z^{2}} d x d y d z$ $i f w e a p p l y t h e t r a n s f o r m a t i o n$ $(x, y, z) \to (z, y, x) i n t e g r a l b e c o m e s$ $V = \int \int \int_{D} \frac{z^{2} + 2 y^{2}}{z^{2} + 4 y^{2} + x^{2}} d x d y d z$ $B y s y m m e t r y s w a p p i n g x a n d z d o e s$ $n o t c h a n g e t h e v a l u e o f t h e i n t e g r a l$ $b e c a u s e i n t e g r a n d a n d t h e l i m i t s$ $o f i n t e g r a t i o n r e m a i n t h e s a m e .$ $T h i s l e a d s t o$ $2 V = \int \int \int_{D} (\frac{x^{2} + 2 y^{2}}{x^{2} + 4 y^{2} + z^{2}} + \frac{z^{2} + 2 y^{2}}{z^{2} + 4 y^{2} + x^{2}}) d x d y d z$ $N o w$ $\int \int \int_{D} 1 d x d y d z$ $2 V = v o l u m e o f D {V o l (D)}$ $= \frac{4}{3} π r^{3} = \frac{4}{3} π 1^{3} = \frac{4}{3} π$ $V = \frac{2}{3} π$
	Commented by BHOOPENDRA last updated on 19/Aug/24

	Commented by BHOOPENDRA last updated on 21/Aug/24

	$p l o t$ $w h e n$ $x^{2} + y^{2} + z^{2} = 1$ $f (z, y) = \frac{z^{2} + 2 y^{2}}{z^{2} + 4 y^{2} + z^{2}}$
	Commented by mnjuly1970 last updated on 19/Aug/24


	Answered by mr W last updated on 19/Aug/24

	$I = \int \int \int \frac{x^{2} + 4 y^{2} + z^{2} - 2 y^{2} - z^{2}}{x^{2} + 4 y^{2} + z^{2}} d x d y d z$ $= \int \int \int (1 - \frac{2 y^{2} + z^{2}}{x^{2} + 4 y^{2} + z^{2}}) d x d y d z$ $= V - I_{1}$ $w i t h I_{1} = \int \int \int (\frac{2 y^{2} + z^{2}}{x^{2} + 4 y^{2} + z^{2}}) d x d y d z$ $I_{1} = \int \int \int (\frac{2 y^{2} + x^{2}}{z^{2} + 4 y^{2} + x^{2}}) d z d y d x$ $I_{1} + I_{1} = \int \int \int (\frac{2 y^{2} + z^{2}}{x^{2} + 4 y^{2} + z^{2}} + \frac{2 y^{2} + x^{2}}{x^{2} + 4 y^{2} + z^{2}}) d x d y d z$ $= \int \int \int (\frac{x^{2} + 4 y^{2} + z^{2}}{x^{2} + 4 y^{2} + z^{2}}) d x d y d z$ $= \int \int \int d x d y d z$ $= V$ $\Rightarrow I_{1} = \frac{V}{2}$ $\Rightarrow I = V - \frac{V}{2} = \frac{V}{2} = \frac{1}{2} \times \frac{4 π \times 1^{3}}{3} = \frac{2 π}{3}$
	Commented by mnjuly1970 last updated on 20/Aug/24

	$g r a t e f u l s i r W$
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