Question

1. in this problem, well look at the iterated integral $int_{0}^{1}int_{0}^{x}int_{y^{2}}^{1}f(x,y,z)dzdydx$. (a) rewrite the iterated integral in the order $dxdzdy$. (b) rewrite the iterated integral in the order $dzdydx$. 5. let $u$ be the solid contained in $x^{2}+y^{2}-z^{2}=16$ and lying between the planes $z = - 3$ and $z = 3$. sketch $u$ 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 limits of original integral

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

Step2: Rewrite for $dzdxdy$

First, find the region in the $xy -$ plane. The intersection of $y = 0,y=x$ and $x = 0,x = 1$ gives the region in the $xy -$ plane. For a fixed $y\in[0,1]$, the lower - bound of $x$ is $x = y$ and the upper - bound is $x = 1$. The lower - bound of $z$ is $z=y^{2}$ and the upper - bound is $z = 1$. So the integral in the order $dzdxdy$ is $\int_{0}^{1}\int_{y}^{1}\int_{y^{2}}^{1}f(x,y,z)dzdxdy$.

Step3: Rewrite for $dydzdx$

From the original limits, for a fixed $x\in[0,1]$, the region in the $yz -$ plane is bounded by $y = 0,y=\sqrt{z}$ (from $z = y^{2}$) and $0\leq z\leq1$. So the integral in the order $dydzdx$ is $\int_{0}^{1}\int_{0}^{1}\int_{0}^{\min(\sqrt{z},x)}f(x,y,z)dydzdx$.

Answer:

(a) $\int_{0}^{1}\int_{y}^{1}\int_{y^{2}}^{1}f(x,y,z)dzdxdy$
(b) $\int_{0}^{1}\int_{0}^{1}\int_{0}^{\min(\sqrt{z},x)}f(x,y,z)dydzdx$