Question

1. in this problem, well look at the iterated integral $int_{0}^{1}int_{0}^{z}int_{y^{2}}^{1}f(x,y,z)dx dy dz$. (a) rewrite the iterated integral in the order $dx dz dy$. (b) rewrite the iterated integral in the order $dz dy dx$. 5. let $mathcal{u}$ be the solid contained in $x^{2}+y^{2}-z^{2}=16$ and lying between the planes $z = - 3$ and $z = 3$. sketch $mathcal{u}$ and write an iterated integral which expresses its volume. in which orders of integration can you write just a single iterated integral (as opposed to a sum of iterated integrals)?

Explanation:

Step1: Analyze the limits of integration for the original integral

The original integral is $\int_{0}^{1}\int_{0}^{z}\int_{y^{2}}^{1}f(x,y,z)dxdy dz$. The limits are: $0\leq z\leq1$, $0\leq y\leq z$, $y^{2}\leq x\leq1$.

Step2: Rewrite for $dxdzdy$ order

First, find the range of $y$. From the original limits, $y$ ranges from $0$ to $1$. For a fixed $y$, $x$ ranges from $y^{2}$ to $1$, and for a fixed $y$ and $x$, $z$ ranges from $y$ to $1$. So the integral in $dxdzdy$ order is $\int_{0}^{1}\int_{y}^{1}\int_{y^{2}}^{1}f(x,y,z)dxdzdy$.

Step3: Rewrite for $dzdydx$ order

Find the range of $x$. From the original limits, $x$ ranges from $0$ to $1$. For a fixed $x$, we need to find the range of $y$ and $z$. Since $y^{2}\leq x$, then $0\leq y\leq\sqrt{x}$. And for a fixed $x$ and $y$, $z$ ranges from $y$ to $1$. So the integral in $dzdydx$ order is $\int_{0}^{1}\int_{0}^{\sqrt{x}}\int_{y}^{1}f(x,y,z)dzdydx$.

Answer:

(a) $\int_{0}^{1}\int_{y}^{1}\int_{y^{2}}^{1}f(x,y,z)dxdzdy$
(b) $\int_{0}^{1}\int_{0}^{\sqrt{x}}\int_{y}^{1}f(x,y,z)dzdydx$