Question

1. let $u$ be the solid enclosed by the paraboloids $z = x^{2}+y^{2}$ and $z = 8-(x^{2}+y^{2})$. (note: the paraboloids intersect where $z = 4$.) write $iiint_{u}f(x,y,z) dv$ as an iterated integral in the order $dz dy dx$.

Explanation:

Step1: Find the range of \(z\)

The lower - bound of \(z\) is given by \(z = x^{2}+y^{2}\) and the upper - bound is \(z = 8-(x^{2}+y^{2})\).

Step2: Find the intersection of the paraboloids in the \(xy\) - plane

Set \(x^{2}+y^{2}=8-(x^{2}+y^{2})\), then \(2(x^{2}+y^{2}) = 8\), or \(x^{2}+y^{2}=4\). To find the range of \(y\) in terms of \(x\), we solve \(y^{2}=4 - x^{2}\), so \(y=-\sqrt{4 - x^{2}}\) to \(y = \sqrt{4 - x^{2}}\).

Step3: Find the range of \(x\)

The circle \(x^{2}+y^{2}=4\) in the \(xy\) - plane has \(x\) ranging from \(x=-2\) to \(x = 2\).

Step4: Write the triple - integral

The triple - integral \(\iiint_{U}f(x,y,z)dV\) in the order \(dzdydx\) is \(\int_{-2}^{2}\int_{-\sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}}\int_{x^{2}+y^{2}}^{8-(x^{2}+y^{2})}f(x,y,z)dzdydx\).

Answer:

\(\int_{-2}^{2}\int_{-\sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}}\int_{x^{2}+y^{2}}^{8-(x^{2}+y^{2})}f(x,y,z)dzdydx\)