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