Answer
1. $V= \int_{0}^{3} \int_{0}^{1} \int_{0}^{2} \ dy \ dx \ dz$
2. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{1} \ dx \ dz \ dy$
3. $V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{1} \ dx \ dy \ dz$
4. $V= \int_{0}^{1} \int_{0}^{2} \int_{0}^{3} \ dz \ dy \ dx$
5. $V= \int_{0}^{2} \int_{0}^{1} \int_{0}^{3} \ dz \ dx \ dy$
6. $V= \int_{0}^{3} \int_{0}^{1} \int_{0}^{2} \ dy \ dx \ dz$
$Volume =6$
Work Step by Step
The six different iterated triple integrals for volume $V$ are defined as below:
1. $V= \int_{0}^{3} \int_{0}^{1} \int_{0}^{2} \ dy \ dx \ dz$
2. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{1} \ dx \ dz \ dy$
3. $V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{1} \ dx \ dy \ dz$
4. $V= \int_{0}^{1} \int_{0}^{2} \int_{0}^{3} \ dz \ dy \ dx$
5. $V= \int_{0}^{2} \int_{0}^{1} \int_{0}^{3} \ dz \ dx \ dy$
6. $V= \int_{0}^{3} \int_{0}^{1} \int_{0}^{2} \ dy \ dx \ dz$
We will solve one triple integral among the six different iterated triple integrals for the volume $V$.
$V=\int_{0}^{2} \int_{0}^{1} \int_{0}^{3} \ dz \ dx \ dy =\int_{0}^{2} \int_{0}^{1} [z]_{0}^{3} \ dx \ dy $
or, $=\int_{0}^{2} \int_{0}^{1} (3-0) \ dx \ dy$
or, $=3 \times \int_0^2 [z]_0^1 \ dy $
or, $=3 \times [y]_0^2$
So, $Volume=3(2-0)=6$
