Answer:
[tex]4\pi- \frac{8}{3}[/tex]
Step-by-step explanation:
The solid that you described is shown in the image of below. To compute the volume we use triple integrals. Observe that
[tex]0\leq x\leq 2[/tex]
[tex]0\leq y \leq \sqrt{4-x^2}[/tex]
[tex]0\leq z \leq 4-x[/tex]
So, the volumen of the solid is given by
[tex]\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{4-x}1\, dzdydx=\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}4-x\, dydx=\int_{0}^{2}(4-x)\sqrt{4-x^2}\,dx[/tex]
This last integral equals [tex]4\pi -\frac{8}{3}[/tex]
