z=(\(x^{2}\)+\(y^{2}\))^(\(\frac{1}{2}\)) とした時の、0≦x≦a,0≦y≦b の範囲での
∬zdxdy が分かれば十分、と考えて計算する。
※他の範囲の場合は、領域の切り貼りで対応できるため。
　例えば、0≦x1≦x≦x2,0≦y1≦y≦y2 であれば、
　　0≦x≦x2,0≦y≦y2 の部分から、
　　0≦x≦x1,0≦y≦y2、0≦x≦x2,0≦y≦y1 の部分を引き、
　　0≦x≦x1,0≦y≦y1 の部分を足すことで対処できる。

まずは、極座標に変換し、3つの領域で分けて考える。
∬[0≦x≦a,0≦y≦b] (\(x^{2}\)+\(y^{2}\))^(\(\frac{1}{2}\)) dxdy
= ∬[0≦\(x^{2}\)+\(y^{2}\)≦\(a^{2}\)+\(b^{2}\)] (\(x^{2}\)+\(y^{2}\))^(\(\frac{1}{2}\)) dxdy
　- ∬[\(a^{2}\)≦\(x^{2}\)+\(y^{2}\)≦\(a^{2}\)+\(b^{2}\),x≧a] (\(x^{2}\)+\(y^{2}\))^(\(\frac{1}{2}\)) dxdy
　- ∬[\(b^{2}\)≦\(x^{2}\)+\(y^{2}\)≦\(a^{2}\)+\(b^{2}\),y≧b] (\(x^{2}\)+\(y^{2}\))^(\(\frac{1}{2}\)) dxdy
= ∬[0≦r≦(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\)),0≦θ≦π/2] \(r^{2}\) drdθ
　- ∬[a≦r≦(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\)),0≦θ≦arccos(\(\frac{a}{r}\))] \(r^{2}\) drdθ
　- ∬[b≦r≦(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\)),arcsin(\(\frac{b}{r}\))≦θ≦π/2] \(r^{2}\) drdθ
= ∫[0,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] π/2・\(r^{2}\) dr
　- ∫[a,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] \(r^{2}\)・arccos(\(\frac{a}{r}\)) dr
　- ∫[b,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] \(r^{2}\)・(π/2 - arcsin(\(\frac{b}{r}\))) dr
= π/6・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\))
　- ∫[a,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] \(r^{2}\)・arccos(\(\frac{a}{r}\)) dr
　- ∫[b,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] \(r^{2}\)・arccos(\(\frac{b}{r}\)) dr　…(1)

ここで、∫\(r^{2}\)・arccos(\(\frac{p}{r}\)) dr を求める。
t = \(\frac{p}{r}\) として置換、
r = \(\frac{p}{t}\), dt = -p/\(r^{2}\)・dr

∫\(r^{2}\)・arccos(\(\frac{p}{r}\)) dr
= ∫\(r^{2}\)・arccos(\(\frac{p}{r}\))・(-\(r^{2}\)/p) dt
= -\(\frac{1}{p}\)・∫\(r^{4}\)・arccos(\(\frac{p}{r}\)) dt
= -\(\frac{1}{p}\)・∫(\(\frac{p}{t}\)\()^{4}\)・arccos(t) dt
= -\(p^{3}\)・∫t^(-4)・arccos(t) dt …(2)

次に、∫t^(-4)・arccos(t) dt を求める。
部分積分により、
∫t^(-4)・arccos(t) dt
= -\(\frac{1}{3}\)・t^(-3)・arccos(t) - \(\frac{1}{3}\)・∫t^(-3)・(1-\(t^{2}\))^(-\(\frac{1}{2}\)) dt
= -\(\frac{1}{3}\)・t^(-3)・arccos(t)
　+ \(\frac{1}{6}\)・t^(-2)・(1-\(t^{2}\))^(-\(\frac{1}{2}\))
　- \(\frac{1}{6}\)・∫\(\frac{1}{t}\)・(1-\(t^{2}\))^(-\(\frac{3}{2}\)) dt …(3)

更に、∫\(\frac{1}{t}\)・(1-\(t^{2}\))^(-\(\frac{3}{2}\)) dt を求める。

その前に、
∫\(\frac{1}{c}\)osφ dφ = \(\frac{1}{2}\)・log( (1+sinφ)/(1-sinφ) )
なぜなら
∫\(\frac{1}{c}\)osφ dφ
= ∫\(\frac{1}{s}\)in(π/2+φ) dφ
= ∫\(\frac{1}{2}\)sin(π/4+φ/2)cos(π/4+φ/2) dφ
= ∫( (sin(π/4+φ/2)\()^{2}\)+(cos(π/4+φ/2)\()^{2}\) )/2sin(π/4+φ/2)cos(π/4+φ/2) dφ
= ∫( sin(π/4+φ/2)/2cos(π/4+φ/2) + cos(π/4+φ/2)/2sin(π/4+φ/2) )dφ
= -log|cos(π/4+φ/2)| + log|sin(π/4+φ/2)| )
= log| sin(π/4+φ/2)/cos(π/4+φ/2) | +C
= \(\frac{1}{2}\)・log( 2(sin(π/4+φ/2)\()^{2}\)/2(cos(π/4+φ/2)\()^{2}\) ) +C
= \(\frac{1}{2}\)・log( (1-cos(π/2+φ))/(1+cos(π/2+φ)) ) +C
= \(\frac{1}{2}\)・log( (1+sinφ)/(1-sinφ) )

t = cosφ として置換
dt = -sinφ dφ, sinφ=(1-\(t^{2}\))^(\(\frac{1}{2}\))

∫\(\frac{1}{t}\)・(1-\(t^{2}\))^(-\(\frac{3}{2}\)) dt
= -∫1/(cosφ(sinφ\()^{2}\))dφ
= -∫((cosφ\()^{2}\)+(sinφ\()^{2}\))/(cosφ(sinφ\()^{2}\))dφ
= -∫( cosφ(sinφ)^(-2) + \(\frac{1}{c}\)osφ )dφ
= 1/(sinφ) - \(\frac{1}{2}\)・log( (1+sinφ)/(1-sinφ) ) +C
= (1-\(t^{2}\))^(-\(\frac{1}{2}\)) - \(\frac{1}{2}\)・log( (1+(1-\(t^{2}\))^(\(\frac{1}{2}\)))/(1-(1-\(t^{2}\))^(\(\frac{1}{2}\))) ) +C　…(4)

(3),(4) より
∫t^(-4)・arccos(t) dt
= -\(\frac{1}{3}\)・t^(-3)・arccos(t)
　+ \(\frac{1}{6}\)・t^(-2)・(1-\(t^{2}\))^(-\(\frac{1}{2}\))
　- \(\frac{1}{6}\)・∫\(\frac{1}{t}\)・(1-\(t^{2}\))^(-\(\frac{3}{2}\)) dt
= -\(\frac{1}{3}\)・t^(-3)・arccos(t)
　+ \(\frac{1}{6}\)・t^(-2)・(1-\(t^{2}\))^(-\(\frac{1}{2}\))
　- \(\frac{1}{6}\)・(1-\(t^{2}\))^(-\(\frac{1}{2}\))
　+ \(\frac{1}{12}\)・log( (1+(1-\(t^{2}\))^(\(\frac{1}{2}\)))/(1-(1-\(t^{2}\)))^(\(\frac{1}{2}\)) ) +C
= -\(\frac{1}{3}\)・t^(-3)・arccos(t)
　+ \(\frac{1}{6}\)・t^(-2)・(1-\(t^{2}\))^(\(\frac{1}{2}\))
　+ \(\frac{1}{12}\)・log( (1+(1-\(t^{2}\))^(\(\frac{1}{2}\)))/(1-(1-\(t^{2}\)))^(\(\frac{1}{2}\)) ) +C　…(5)

(2),(5)より、
∫\(r^{2}\)・arccos(\(\frac{p}{r}\)) dr
= -\(p^{3}\)・∫t^(-4)・arccos(t) dt
= \(p^{3}\)/3・t^(-3)・arccos(t)
　- \(p^{3}\)/6・t^(-2)・(1-\(t^{2}\))^(\(\frac{1}{2}\))
　- \(p^{3}\)/12・log( (1+(1-\(t^{2}\))^(\(\frac{1}{2}\)))/(1-(1-\(t^{2}\))^(\(\frac{1}{2}\))) ) +C
= \(r^{3}\)/3・arccos(\(\frac{p}{r}\))
　- p\(\frac{r}{6}\)・(\(r^{2}\)-\(p^{2}\))^(\(\frac{1}{2}\))
　- \(p^{3}\)/12・log( (r+(\(r^{2}\)-\(p^{2}\))^(\(\frac{1}{2}\)))/(r-(\(r^{2}\)-\(p^{2}\))^(\(\frac{1}{2}\))) ) +C　…(6)

(1),(6)より、
∬[0≦x≦a,0≦y≦b] (\(x^{2}\)+\(y^{2}\))^(\(\frac{1}{2}\)) dxdy
= π/6・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\))
　- ∫[a,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] \(r^{2}\)・arccos(\(\frac{a}{r}\)) dr
　- ∫[b,(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))] \(r^{2}\)・arccos(\(\frac{b}{r}\)) dr
= π/6・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\))
　- \(\frac{1}{3}\)・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\))・arccos(a(\(a^{2}\)+\(b^{2}\))^(-\(\frac{1}{2}\)))
　+ \(\frac{1}{6}\)・ab(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))
　+ \(a^{3}\)/12・log( ((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+b)/((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))-b) )
　- \(\frac{1}{3}\)・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\)) /3・arccos(b(\(a^{2}\)+\(b^{2}\))^(-\(\frac{1}{2}\)))
　+ \(\frac{1}{6}\)・ab(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))
　+ \(b^{3}\)/12・log( ((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+a)/((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))-a) )
= π/6・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\))
　- \(\frac{1}{3}\)・(\(a^{2}\)+\(b^{2}\))^(\(\frac{3}{2}\))・( arccos(a(\(a^{2}\)+\(b^{2}\))^(-\(\frac{1}{2}\)))
　+ arccos(b(\(a^{2}\)+\(b^{2}\))^(-\(\frac{1}{2}\))) )+ \(\frac{1}{3}\)・ab(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))
　+ \(a^{3}\)/12・log( ((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+b\()^{2}\)/((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+b)((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))-b) )
　+ \(b^{3}\)/12・log( ((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+a\()^{2}\)/((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+a)((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))-a) )
= \(\frac{1}{3}\)・ab(\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))
　+ \(a^{3}\)/6・log( ((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+b)/a )
　+ \(b^{3}\)/6・log( ((\(a^{2}\)+\(b^{2}\))^(\(\frac{1}{2}\))+a)/b )　　…(答え)