Question

33 - 36 express the integral $iiint_e f(x,y,z)dv$ as an iterated integral in six different ways, where $e$ is the solid bounded by the given surfaces. 33. $y = 4 - x^{2}-4z^{2},y = 0$

Explanation:

Step1: Rewrite the surface equation

The surface $y = 4 - x^{2}-4z^{2}$ is an elliptic - paraboloid opening along the positive $y$ - axis with vertex at $(0,4,0)$. When $y = 0$, we have $x^{2}+4z^{2}=4$, which is an ellipse in the $xz$ - plane.

Step2: Find the limits for different order of integrations

Order $dxdydz$:

First, solve the ellipse $x^{2}+4z^{2}=4$ for $x$: $x=\pm\sqrt{4 - 4z^{2}}$.
The limits for $y$ go from $y = 0$ to $y=4 - x^{2}-4z^{2}$, and for $z$ go from $z=-1$ to $z = 1$.
\(\iiint_E f(x,y,z)dV=\int_{-1}^{1}\int_{0}^{4 - 4z^{2}}\int_{-\sqrt{4 - 4z^{2}-y}}^{\sqrt{4 - 4z^{2}-y}}f(x,y,z)dxdydz\)

Order $dxdzdy$:

For a fixed $y$, the cross - section in the $xz$ - plane is the ellipse $x^{2}+4z^{2}=4 - y$. Solving for $x$ gives $x=\pm\sqrt{4 - y-4z^{2}}$ and for $z$ gives $z=\pm\frac{1}{2}\sqrt{4 - y}$.
The limits for $y$ go from $y = 0$ to $y = 4$.
\(\iiint_E f(x,y,z)dV=\int_{0}^{4}\int_{-\frac{1}{2}\sqrt{4 - y}}^{\frac{1}{2}\sqrt{4 - y}}\int_{-\sqrt{4 - y-4z^{2}}}^{\sqrt{4 - y-4z^{2}}}f(x,y,z)dxdzdy\)

Order $dydxdz$:

The limits for $x$ go from $x=-2$ to $x = 2$, for $z$ go from $z=-\frac{1}{2}\sqrt{4 - x^{2}}$ to $z=\frac{1}{2}\sqrt{4 - x^{2}}$, and for $y$ go from $y = 0$ to $y=4 - x^{2}-4z^{2}$.
\(\iiint_E f(x,y,z)dV=\int_{-2}^{2}\int_{-\frac{1}{2}\sqrt{4 - x^{2}}}^{\frac{1}{2}\sqrt{4 - x^{2}}}\int_{0}^{4 - x^{2}-4z^{2}}f(x,y,z)dydxdz\)

Order $dydzdx$:

The limits for $x$ go from $x=-2$ to $x = 2$, for $y$ go from $y = 0$ to $y=4 - x^{2}$, and for $z$ go from $z=-\frac{1}{2}\sqrt{4 - x^{2}-y}$ to $z=\frac{1}{2}\sqrt{4 - x^{2}-y}$.
\(\iiint_E f(x,y,z)dV=\int_{-2}^{2}\int_{0}^{4 - x^{2}}\int_{-\frac{1}{2}\sqrt{4 - x^{2}-y}}^{\frac{1}{2}\sqrt{4 - x^{2}-y}}f(x,y,z)dydzdx\)

Order $dzdxdy$:

The limits for $y$ go from $y = 0$ to $y = 4$, for $x$ go from $x=-\sqrt{4 - y}$ to $x=\sqrt{4 - y}$, and for $z$ go from $z=-\frac{1}{2}\sqrt{4 - y - x^{2}}$ to $z=\frac{1}{2}\sqrt{4 - y - x^{2}}$.
\(\iiint_E f(x,y,z)dV=\int_{0}^{4}\int_{-\sqrt{4 - y}}^{\sqrt{4 - y}}\int_{-\frac{1}{2}\sqrt{4 - y - x^{2}}}^{\frac{1}{2}\sqrt{4 - y - x^{2}}}f(x,y,z)dzdxdy\)

Order $dzdydx$:

The limits for $x$ go from $x=-2$ to $x = 2$, for $y$ go from $y = 0$ to $y=4 - x^{2}$, and for $z$ go from $z = 0$ to $z=\frac{1}{2}\sqrt{4 - x^{2}-y}$.
\(\iiint_E f(x,y,z)dV=\int_{-2}^{2}\int_{0}^{4 - x^{2}}\int_{0}^{\frac{1}{2}\sqrt{4 - x^{2}-y}}f(x,y,z)dzdydx\)

Answer:

The six iterated integrals are:

1. \(\int_{-1}^{1}\int_{0}^{4 - 4z^{2}}\int_{-\sqrt{4 - 4z^{2}-y}}^{\sqrt{4 - 4z^{2}-y}}f(x,y,z)dxdydz\)
2. \(\int_{0}^{4}\int_{-\frac{1}{2}\sqrt{4 - y}}^{\frac{1}{2}\sqrt{4 - y}}\int_{-\sqrt{4 - y-4z^{2}}}^{\sqrt{4 - y-4z^{2}}}f(x,y,z)dxdzdy\)
3. \(\int_{-2}^{2}\int_{-\frac{1}{2}\sqrt{4 - x^{2}}}^{\frac{1}{2}\sqrt{4 - x^{2}}}\int_{0}^{4 - x^{2}-4z^{2}}f(x,y,z)dydxdz\)
4. \(\int_{-2}^{2}\int_{0}^{4 - x^{2}}\int_{-\frac{1}{2}\sqrt{4 - x^{2}-y}}^{\frac{1}{2}\sqrt{4 - x^{2}-y}}f(x,y,z)dydzdx\)
5. \(\int_{0}^{4}\int_{-\sqrt{4 - y}}^{\sqrt{4 - y}}\int_{-\frac{1}{2}\sqrt{4 - y - x^{2}}}^{\frac{1}{2}\sqrt{4 - y - x^{2}}}f(x,y,z)dzdxdy\)
6. \(\int_{-2}^{2}\int_{0}^{4 - x^{2}}\int_{0}^{\frac{1}{2}\sqrt{4 - x^{2}-y}}f(x,y,z)dzdydx\)