3. let $\\mathcal{u}$ be the solid enclosed by the paraboloids $z = x^{2}+y^{2}$ and $z = 8-(x^{2}+y^{2})$…

3. let $\\mathcal{u}$ 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 $\\iiint_{\\mathcal{u}}f(x,y,z)dv$ as an iterated integral in the order $dz\\ dy\\ dx$.

3. let $\\mathcal{u}$ 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 $\\iiint_{\\mathcal{u}}f(x,y,z)dv$ as an iterated integral in the order $dz\\ dy\\ dx$.

Answer

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})). So, (x^{2}+y^{2}\leq z\leq8-(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).

Step3: Find the range of (y) for a given (x)

For a fixed (x), from (x^{2}+y^{2}=4), we have (y=-\sqrt{4 - x^{2}}) to (y = \sqrt{4 - x^{2}}), and (x) ranges from (-2) to (2).

Step4: Write the triple - integral

The triple - integral (\iiint_{U}f(x,y,z)dV=\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)