Answer
Refer below for the answer.
Work Step by Step
The six different iterated triple integrals for volume $V$ are defined as:
1. $V= \int_{0}^{3} \int_{0}^{\frac{z}{3}} \int_{0}^{2-2x-\frac{2z}{3}} \ dy \ dx \ dz$
2. $V= \int_{0}^{2} \int_{0}^{3-\frac{3y}{2}} \int_{0}^{1-\frac{y}{2}-\frac{z}{3}} \ dx \ dz \ dy$
3. $V=\int_{0}^{3} \int_{0}^{2-\frac{2z}{3}} \int_{0}^{1-\frac{y}{2}-\frac{z}{3}} \ dx \ dy \ dz$
4. $V= \int_{0}^{2} \int_{0}^{1-\frac{y}{2}} \int_{0}^{2-2x-\frac{2z}{3}} \ dz \ dx \ dz$
5. $V= \int_{0}^{1} \int_{0}^{3-3x} \int_{0}^{2-2x-\frac{2z}{3}} \ dy \ dz \ dx$
6. $V= \int_{0}^{1} \int_{0}^{2-2x} \int_{0}^{3-3x-\frac{3y}{2}} \ dz \ dy \ dx$
Now, $$V=\int_{0}^{1} \int_{0}^{3-3x} \int_{0}^{(2-2x-\frac{2z}{3})} \ dy \ dz \ dx =\int_{0}^{1} \int_{0}^{3-3x} [y]_{0}^{{(2-2x-\frac{2z}{3})}} \ dz \ dx =\int_{0}^{1} (2z-2xz-\dfrac{2}{3} \times \dfrac{z^2}{2}]_0^{3-3x} \ dx =\int_0^1 [2(3-3x)-2x(3-3x) -\dfrac{(3-3x)^2}{3}) \ dx =\int_0^1 [6x^2-12x+6-3(1-2x+x^2)] \ dx \\=\int_0^1 (3x^2-6x+3) dx =[x^3]_0^1-3[x^2]_0^1+3[x]_0^1 =1$$
