Answer
$3\pi$
Work Step by Step
The six different iterated triple integrals for the volume $V$ are defined as:
1. $V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dz \ dx \ dy$
2. $V= \int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dx \ dz \ dy$
3. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt {4-x^2}} \ dx \ dy \ dz$
4. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt {4-x^2}} \ dz \ dy \ dx$
5. $V= \int_{0}^{1} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{3} \ dy \ dz \ dx$
6. $V= \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{3} \ dy \ dx \ dz$
We will solve one triple integral among the six different iterated triple integrals for the volume $V$.
$V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dz \ dx \ dy \\=\int_{0}^{3} \int_{0}^{2} [z]_{0}^{\sqrt {4-x^2}} \ dx \ dy \\=\int_{0}^{3} \int_0^2 \sqrt {4-x^2} dxdy$
Let us consider that $x =2 \sin a $ and $dx=2 \cos a da$
$V=\int_0^{3} \int_0^{\pi/2} (2 \cos a) \sqrt {4-(2 \sin a)^2} da dy \\=\int_0^{3} \int_0^{\pi/2} (2 \cos a) \sqrt {4-4 \sin^2 (a)} da dy \\=4 \int_0^3 [\dfrac{a}{2}+\dfrac{\sin (2a)}{4}]_0^{\pi/2} \ dy\\=4 \int_0^{3} \dfrac{\pi}{4} \ dy \\=4 \int_0^3 [0+\dfrac{\pi/2}{2}] \ dy \\=4 \int_0^{3} \dfrac{\pi}{4} \ dy \\=[\pi y ]_0^3 \\=\pi (3-0) \\=3\pi$
