Answer
$\frac{64}{3}$
Work Step by Step
The paraboloid $z=2+x^2+(y-2)^2$ is above the plane $z=1$.
Then, the volume of the solid enclosed by the paraboloid and the planes $z=1$, $x=1$, $x=-1$, $y=0$, and $y=4$ is represented by the integral $V=\int_{-1}^1\int_0^42+x^2+(y-2)^2-1dydx$.
Find the volume by evaluating this integral:
$V=\int_{-1}^1\int_0^41+x^2+(y-2)^2dydx=\int_{-1}^1y+x^2y+\frac{(y-2)^3}{3}]_0^4dx$
$=\int_{-1}^1(4+4x^2+\frac{16}{3})dx=\int_{-1}^14x^2+\frac{28}{3}dx=\frac{4x^3+28x}{3}]_{-1}^1=\frac{32}{3}-(-\frac{32}{3})=\frac{64}{3}$
