Ensino Superior ⇒ Integral dupla

[Natan]

Calcule:

[tex3]\int_{0}^{1}\int_{0}^{1} \frac{xydydx}{\sqrt{x^2+y^2+1}}[/tex3]

[Rafa2604]

Seja [tex3]\int_{0}^{1}\int_{0}^{1} \;\frac{xy}{\sqrt{x^{2}+y^{2}+1}}\;dydx[/tex3].

Utilizaremos as substituições [tex3]u = x^{2}+y^{2}+1, du = 2ydy[/tex3], e quando chegarmos na integração do dx,
[tex3]v = x^{2}+2, dv = 2xdx\; e \;w = x^{2}+1, dw = 2xdx[/tex3]

Portanto, temos que:
[tex3]\int_{0}^{1}\int_{0}^{1} \;\frac{xy}{\sqrt{x^{2}+y^{2}+1}}\;dydx = \int_{0}^{1}\int_{0}^{1} \;\frac{xy}{\sqrt{u}}\;\frac{du}{2y}dx = \int_{0}^{1}\int_{0}^{1} \;\frac{x}{2u^{1/2}}\;dudx = \int_{0}^{1}\;\frac{x}{2}\;\int_{0}^{1} \;u^{-1/2}\;dudx = \int_{0}^{1}\;\frac{x}{2}[2u^{1/2}]_{0}^{1} \;dx = \\\\ \; = \; \int_{0}^{1}\;x[\sqrt{x^{2}+y^{2}+1}]_{0}^{1} \;dx = \int_{0}^{1}\;x[\sqrt{x^{2}+2} - \sqrt{x^{2}+1} ]\; dx = \int_{0}^{1}\;x\sqrt{x^{2}+2}dx - \int_{0}^{1}\;x\sqrt{x^{2}+1}dx = \int_{0}^{1}x\sqrt{v}\frac{dv}{2x} - \int_{0}^{1}x\sqrt{w}\frac{dw}{2x} = \\\\ \; = \; \int_{0}^{1}\sqrt{v}\frac{dv}{2} - \int_{0}^{1}\sqrt{w}\frac{dw}{2} = \left[\frac{v^{3/2}}{3} - \frac{w^{3/2}}{3} \right]_{0}^{1} = \left[\frac{(x^{2}+2)\sqrt{x^{2}+2}}{3} - \frac{(x^{2}+1)\sqrt{x^{2}+1}}{3} \right]_{0}^{1} = \left[\sqrt{3}-\frac{2\sqrt{2}}{3} - \frac{2\sqrt{2}}{3}+\frac{1}{3} \right] = \frac{4}{3}(\sqrt{3}-\sqrt{2})[/tex3]