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Use a triple integral to find the volume of the region in the first octant bounded by the coordinate plane y=1-x, and the surface z=cos(pix/2), x=0,1

Respuesta :

LammettHash
The volume is given by

[tex]\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=\cos(\pi x/2)}\mathrm dz\,\mathrm dy\,\mathrm dx=\frac4{\pi^2}[/tex]

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