Let E be the region bounded above by the paraboloid $z=4- x^2 - y^2$ in the first octant and $f(x,y,z)=5\sqrt{x^2+y^2}$ be the density of the solid E.

a) Set up an integral for the volume E.

b) Find the mass of the solid E.

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    $\begingroup$ What are your thoughts on this problem? What have you tried so far? $\endgroup$ – Alex Wertheim Aug 17 '13 at 2:02

a) The volume is given by

$$V = \iiint_E dV = \int_0^2 dr \, r \, \int_0^{4-r^2} dz \, \int_0^{\pi/2} d\theta$$

b) The mass of this object is given by the volume integral over the density, or

$$M = 5 \iiint_E \rho\, dV = 5 \int_0^2 dr \, r^2 \, \int_0^{4-r^2} dz \, \int_0^{\pi/2} d\theta$$


$$M = \frac{5 \pi}{2} \int_0^2 dr \, r^2 (4-r^2) $$

Can you take it from here? I get $M = 32 \pi/3$.

  • $\begingroup$ wait howd you know the r goes from 2 to 0 $\endgroup$ – user90705 Aug 17 '13 at 2:33
  • $\begingroup$ Because in the plane $z=0$ where the cross-section is biggest, $x^2+y^2=r^2=4$. $\endgroup$ – Ron Gordon Aug 17 '13 at 2:36

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