Soluzioni 6-AM4

Laura Di Gregorio 30 ottobre 2003

1. Si parametrizzi la superficie Γ tramite

φ(t, u) = (t, u, 1 − t − u)

da cui segue che dσ = \sqrt{3} dudt. Quindi si ha

∫_{Γ} yz\sqrt{1 − x} dσ = \sqrt{3} ∫_{0}^{1} ∫_{0}^{1−t} u\sqrt{1 − t}(1 − t − u)dudt
= \sqrt{3} ∫_{0}^{1} ∫_{0}^{1−t} u\sqrt{1 − t} dudt − \sqrt{3} ∫_{0}^{1} ∫_{0}^{1−t} ut\sqrt{1 − t} dudt

−\sqrt{3} ∫_{0}^{1} ∫_{0}^{1−t} u^{2}\sqrt{1 − t} dudt

= \sqrt{3} ∫_{0}^{1} \frac{(1 − t)^{\frac{5}{2}}}{2} dt − \sqrt{3} ∫_{0}^{1} ∫_{0}^{1−t} t(1 − t)^{\frac{5}{2}} dt

−\sqrt{3} ∫_{0}^{1} \frac{(1 − t)^{\frac{7}{2}}}{2} dt

= [\frac{\sqrt{3}}{7}(1 − t)^{\frac{7}{2}} + 2\frac{\sqrt{3}}{7}t(1 − t)^{\frac{7}{2}} + \frac{4\sqrt{3}}{63} − \frac{2\sqrt{3}}{9}(1 − t)^{\frac{9}{2}} ]_{0}^{1}

2. Si calcola ds = e^{t}\sqrt{3} dt. L’integrale diventa

∫_{γ} x^{2} ln z ds = \sqrt{3} ∫_{0}^{T} te^{3t} cos^{2} t dt
= \frac{\sqrt{3}}{2} ∫_{0}^{T} te^{3t}(1 + cos 2t) dt
= \frac{\sqrt{3}}{2} ∫_{0}^{T} te^{3t} dt + \frac{\sqrt{3}}{2} ∫_{0}^{T} te^{3t} cos 2t

= \frac{\sqrt{3}}{2}[\frac{t}{3}e^{3t}]_{0}^{T} − ∫_{0}^{T} \frac{e^{3t}}{3} dt + \frac{\sqrt{3}}{2} ∫_{0}^{T} te^{3t} cos 2t dt
= \frac{\sqrt{3}}{2}[\frac{t}{3}e^{3t} − \frac{e^{3t}}{9} dt]_{0}^{T} + \frac{\sqrt{3}}{2} ∫_{0}^{T} te^{3t} cos 2t dt

Si calcoli ora

∫ e^{3t} cos 2t dt = \frac{e^{3t}}{3} cos 2t + 2 ∫ \frac{e^{3t}}{3} sin 2t dt
= \frac{e^{3t}}{3} cos 2t + \frac{2}{3} ∫ e^{3t} sin 2t dt
= \frac{e^{3t}}{3} cos 2t + \frac{2}{9}e^{3t} sin 2t − \frac{4}{9} ∫ e^{3t} cos 2t dt

da cui segue che

∫ e^{3t} cos 2t dt = \frac{9}{39}e^{3t}( cos 2t + \frac{2}{3} sin 2t)

Quindi, integrando per parti, si ha che:

\frac{\sqrt{3}}{2}[\frac{t}{3}e^{3t} − \frac{e^{3t}}{9} dt]_{0}^{T} +\frac{\sqrt{3}}{2} ∫ te^{3t} cos 2t dt =
= \frac{\sqrt{3}}{2}[\frac{t}{3}e^{3t} − \frac{e^{3t}}{9} + \frac{9}{39}te^{3t}( cos 2t + \frac{2}{3} sin 2t)]_{0}^{T}

−\frac{\sqrt{3}}{2} \frac{9}{39} ∫_{0}^{T} e^{3t}( cos 2t + \frac{2}{3} sin 2t) dt

= \frac{\sqrt{3}}{2}[\frac{t}{3}e^{3t} − \frac{e^{3t}}{9} + +\frac{9}{39}te^{3t}( cos 2t + \frac{2}{3} sin 2t)

−\frac{9^{2}}{39^{2}} e^{3t}( cos 2t + \frac{2}{3} sin 2t)]_{0}^{T} + \frac{2}{3} ∫_{0}^{T} e^{3t} sin 2t dt

Ora basta calcolare

∫_{0}^{T} e^{3t} sin 2t dt = [\frac{e^{3t}}{3} sin 2t]_{0}^{T} − \frac{2}{3} ∫_{0}^{T} e^{3t} cos 2t dt
= [\frac{e^{3t}}{3} sin 2t − \frac{2}{9} e^{3t} cos 2t]_{0}^{T} − \frac{4}{9} ∫_{0}^{T} e^{3t} sin 2t dt

da cui segue che

∫_{0}^{T} e^{3t} sin 2t dt = [\frac{9}{39} e^{3t}( sin 2t − \frac{2}{3} cos 2t)]_{0}^{T} .

Dunque

∫_{γ} x^{2} ln z ds = \frac{\sqrt{3}}{2}[\frac{t}{3} e^{3t} − \frac{e^{3t}}{9} + \frac{9}{39}te^{3t}( cos 2t + \frac{2}{3} sin 2t)

−\frac{9^{2}}{39^{2}} e^{3t}( cos 2t + \frac{2}{3} sin 2t) + \frac{2}{3} · \frac{9}{39} e^{3t}( sin 2t − \frac{2}{3} cos 2t)]_{0}^{T}

3. Ponendo e^{t} = x si ha, per x > 0

∫ e^{t} ln \frac{1}{\sqrt{1 − e^{−t}}} dt = −\frac{1}{2} ∫ ln \frac{x + 1}{x} dx
= −\frac{1}{2} ∫ ln(x + 1) dx + \frac{1}{2} ∫ ln x dx
= −\frac{1}{2}x ln(x + 1) + \frac{1}{2} \frac{x}{x + 1} + \frac{1}{2}x ln x − \frac{1}{2}

4. Integrando per parti si ha

∫ \frac{ln^{2} x}{x^{2}} dx = −\frac{1}{x} ln^{2} x + 2 ∫ \frac{1}{x^{2}} ln x dx
= −\frac{1}{x} ln^{2} x − \frac{2}{x} ln x + 2 ∫ \frac{1}{x^{2}} dx
= −\frac{1}{x} ln^{2} x − \frac{2}{x} ln x − \frac{2}{x} dx