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 $dz dy dx$.

1. let $mathcal{u}$ be the solid contained in $x^{2}+y^{2}-z^{2}=16$ and lying between the planes $z = - 3$ and $z$. sketch $mathcal{u}$ and write an iterated integral which expresses its volume. in which orders of integration you write just a single iterated integral (as opposed to a sum of iterated integrals)?

Explanation:

Step1: Analyze original limits

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

Step2: Rewrite for order $dx dz dy$

First, find the relationship between variables for the new - order.
From $0\leq y\leq z$ and $0\leq z\leq1$, we get $y\leq z\leq1$ and $0\leq y\leq1$. Also, $y^{2}\leq x\leq1$. So the integral is $\int_{0}^{1}\int_{y}^{1}\int_{y^{2}}^{1}f(x,y,z)dxdzdy$.

Step3: Rewrite for order $dz dy dx$

We need to express $z$ and $y$ in terms of $x$.
From $y^{2}\leq x\leq1$, we have $0\leq y\leq\sqrt{x}$. And from $y\leq z\leq1$, the integral becomes $\int_{0}^{1}\int_{0}^{\sqrt{x}}\int_{y}^{1}f(x,y,z)dzdydx$.

Answer:

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