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Question Number 27598 by abdo imad last updated on 10/Jan/18

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

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} .