Answer
Volume $=3 \pi $
1. $V= \int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dx \ dz \ dy$
2. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt {4-x^2}} \ dx \ dy \ dz$
3. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt {4-x^2}} \ dz \ dy \ dx$
4. $V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dz \ dx \ dy$
5. $V= \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{3} \ dy \ dx \ dz$
6. $V= \int_{0}^{1} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{3} \ dy \ dz \ dx$
Work Step by Step
The six different iterated triple integrals for volume $V$ can be written in different forms as:
1. $V= \int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dx \ dz \ dy$
2. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt {4-x^2}} \ dx \ dy \ dz$
3. $V= \int_{0}^{2} \int_{0}^{3} \int_{0}^{\sqrt {4-x^2}} \ dz \ dy \ dx$
4. $V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \ dz \ dx \ dy$
5. $V= \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{3} \ dy \ dx \ dz$
6. $V= \int_{0}^{1} \int_{0}^{\sqrt {4-x^2}} \int_{0}^{3} \ dy \ dz \ dx$
Now, $V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{\sqrt {4-x^2}} dz dx dy$
or, $=\int_{0}^{3} \int_{0}^{2} [z]_{0}^{\sqrt {4-x^2}} dx dy$
or, $=\int_{0}^{3} \int_0^2 \sqrt {4-x^2} dx\ dy$
Suppose that $x =2 \sin a \\ 2 \cos a da = dx$
$V=\int_0^{3} \int_0^{\pi/2} (2 \cos a) \sqrt {4-(2 \sin a)^2} da \ dy$
or, $=\int_0^{3} \int_0^{\pi/2} (2 \cos a) \times \sqrt {4-4 \sin^2 (a)} da \ dy$
or, $=4 \times \int_0^3 [\dfrac{a}{2}+\dfrac{\sin (2a)}{4}]_0^{\pi/2} \ dy$
or, $=4 \times \int_0^{3} \dfrac{1}{4} \times \pi \ dy \\=4 \int_0^3 [0+\dfrac{\pi/2}{2}] \ dy$
or, $=4 \times \int_0^{3} \dfrac{\pi}{4} \ dy $
or, $=\pi (3-0)$
or, $=3\pi$
