[tex]\int \int \int f(x,y,z) = \dfrac{10240}{21}[/tex]
Step-by-step explanation:
Given :
[tex]f(x,y,z) = 20xz[/tex]
the region in the first octant [tex](x,y,z\geq 0)[/tex] above the parabolic cylinder [tex]z = y^2[/tex] and below the paraboloid [tex]z = 8-2x^2-y^2[/tex].
Calculation :
[tex]y^2 = 8-2x^2-y^2[/tex]
[tex]2y^2= 8-2x^2[/tex]
[tex]y = \sqrt{4-x^2}[/tex]
[tex]\bigtriangleup = \left \{ (x,y,z)| 0\leq x\leq 2, 0\leq y\leq \sqrt{4-x^2} ,y^2 \leq z \leq 8-2x^2-y^2\right \}[/tex]
[tex]\int \int \int f(x,y,z) = \int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{y^2}^{8-2x^2-y^2}20xz\; dzdydx[/tex]
[tex]\int \int \int f(x,y,z) = \int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\dfrac{20}{2}x \left(z^2\right)_{y^2}^{8-2x^2-y^2} \; dydx[/tex]
[tex]\int \int \int f(x,y,z) = \int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}10x((8-2x^2-y^2)^2-(y^4)) \; dydx[/tex]
[tex]\int \int \int f(x,y,z) = \int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}(640x-320x^3-160xy^2+40x^5+40x^3y^2)\; dydx[/tex]
[tex]\int \int \int f(x,y,z) = \int_{0}^{2}\left(640xy-320x^3y-\dfrac{160}{3}xy^3+40x^5y+\dfrac{40}{3}x^3y^3 \right)_{0}^{\sqrt{4-x^2} } \; dx[/tex]
[tex]\int \int \int f(x,y,z) = \dfrac{10240}{21}[/tex]
For more information, refer to the link given below
https://brainly.com/question/19484372?referrer=searchResults
