Question Number 61645 by maxmathsup by imad last updated on 05/Jun/19
$${calculate}\:\int\int_{{D}} \int\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz} \\ $$$${with}\:{D}\:=\left\{\left({x},{y},{z}\right)\:/\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:,\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\:\:,\:\mathrm{2}\leqslant{z}\leqslant\mathrm{3}\:\right\} \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/19
![let I =∫∫∫_D (√(x^2 +y^2 +z^2 ))dxdydz ⇒ I =∫_2 ^3 A(z)dz with A(z) =∫∫_w (√(x^2 +y^2 +z^2 ))dxdy let consider the diffeomorphism x =r cosθ and y =rsinθ we have 0≤x≤1 and 1≤y≤2 ⇒ 1 ≤x^2 +y^2 ≤5 ⇒1≤r^2 ≤5 ⇒1≤r≤(√(5 )) ⇒ A(z) =∫∫_(1≤r≤(√5) and 0≤θ≤(π/2)) (√(r^2 +z^2 ))rdrdθ =(π/2) ∫_1 ^(√5) r(r^2 +z^2 )^(1/2) dr =(π/2)[ (1/3)(r^2 +z^2 )^(3/2) ]_1 ^(√5) =(π/6){(z^2 +5)^(3/2) −(z^2 +1)^(3/2) } ⇒ I = (π/6)∫_2 ^3 (z^2 +5)^(3/2) dz −(π/6) ∫_2 ^3 (z^2 +1)^(3/2) dz ∫_2 ^3 (1+z^2 )^(3/2) dz =_(z =sh(t)) ∫_(argsh(2)) ^(argsh(3)) (ch^2 t)^(3/2) ch(t)dt =∫_(ln(2+(√5))) ^(ln(3+(√(10)))) ch^4 t dt =∫_(ln(2+(√5))) ^(ln(3+10)) (((1+ch(2t))/2))^2 dt =(1/4) ∫_(ln(2+(√5))) ^(ln(3+(√(10)))) (1+2ch(2t)+ch^2 (2t))dt =(1/4){ln(3+(√(10)))−ln(2+(√5))) +(1/4)[sh(2t)]_(ln(2+(√5))) ^(ln(3+(√(10)))) + (1/8) ∫_(ln(2+(√5))) ^(ln(3+(√(10)))) (1+ch(4t))dt = .....rest to acheive the calculus for the integral ∫_2 ^3 (z^2 +5)^(3/2) dz we do the chang.z =(√5)sh(t)....](Q61680.png)
$${let}\:{I}\:=\int\int\int_{{D}} \sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{dxdydz}\:\Rightarrow\:{I}\:=\int_{\mathrm{2}} ^{\mathrm{3}} \:{A}\left({z}\right){dz}\:{with} \\ $$$${A}\left({z}\right)\:=\int\int_{{w}} \sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{dxdy}\:\:\:{let}\:{consider}\:{the}\:{diffeomorphism} \\ $$$${x}\:={r}\:{cos}\theta\:\:{and}\:{y}\:={rsin}\theta\:\:\:\:\:{we}\:{have}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:{and}\:\mathrm{1}\leqslant{y}\leqslant\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{1}\:\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{5}\:\Rightarrow\mathrm{1}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{5}\:\Rightarrow\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{5}\:}\:\Rightarrow \\ $$$${A}\left({z}\right)\:=\int\int_{\mathrm{1}\leqslant{r}\leqslant\sqrt{\mathrm{5}}\:\:{and}\:\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\:\sqrt{{r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }{rdrd}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:{r}\left({r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dr}\:=\frac{\pi}{\mathrm{2}}\left[\:\:\frac{\mathrm{1}}{\mathrm{3}}\left({r}^{\mathrm{2}} \:+{z}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} \:=\frac{\pi}{\mathrm{6}}\left\{\left({z}^{\mathrm{2}} \:+\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}\:\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mathrm{6}}\int_{\mathrm{2}} ^{\mathrm{3}} \:\:\left({z}^{\mathrm{2}} \:+\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dz}\:−\frac{\pi}{\mathrm{6}}\:\int_{\mathrm{2}} ^{\mathrm{3}} \:\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dz} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{3}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dz}\:=_{{z}\:={sh}\left({t}\right)} \:\:\:\:\int_{{argsh}\left(\mathrm{2}\right)} ^{{argsh}\left(\mathrm{3}\right)} \:\:\:\left({ch}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{ch}\left({t}\right){dt}\:=\int_{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} ^{{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right)} \:{ch}^{\mathrm{4}} {t}\:{dt} \\ $$$$=\int_{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} ^{{ln}\left(\mathrm{3}+\mathrm{10}\right)} \:\:\left(\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} ^{{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right)} \left(\mathrm{1}+\mathrm{2}{ch}\left(\mathrm{2}{t}\right)+{ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right)−{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} ^{{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right)} \:+\:\frac{\mathrm{1}}{\mathrm{8}}\:\int_{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} ^{{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right)} \left(\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)\right){dt} \\ $$$$=\:\:.....{rest}\:{to}\:{acheive}\:{the}\:{calculus}\:\:{for}\:{the}\:{integral}\:\int_{\mathrm{2}} ^{\mathrm{3}} \left({z}^{\mathrm{2}} \:+\mathrm{5}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dz} \\ $$$${we}\:{do}\:{the}\:{chang}.{z}\:=\sqrt{\mathrm{5}}{sh}\left({t}\right).... \\ $$