Can someone help me start is problem off. Find the volume bounded by the paraboloid $x^2 + y^2 + z =7$ and the plane $z=1$.



First, notice that the two surfaces intersect in the circle $$x^2+y^2=6,\quad z=1$$

Now, the region is symmetric respect to the planes $xz$ and $yz$, then it will be sufficient to determine the volume laying in the first octant: \begin{align} \frac{1}{4}V&=\int_0^{\sqrt 6}\int_0^{\sqrt{6-x^2}}\int_1^{7-x^2-y^2}\,dz\,dy\,dx \end{align} Where $V$ denotes the volume of the region.

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  • $\begingroup$ First thank you for responding! Second i think you are missing a "dx" and i am just confused how to get the bounds for these types of problems and where did you get the 1/4 from? $\endgroup$ – Beshoy Awad Nov 16 '15 at 0:06
  • 1
    $\begingroup$ The $\frac{1}{4}$ factor is due to the symmetry. $\endgroup$ – Ángel Mario Gallegos Nov 16 '15 at 0:11

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