To set up the iterated double integral in polar coordinates that gives the volume of the solid enclosed by the hyperboloid $z=\sqrt{1+\frac{x^2}{2}+\frac{y^2}{2}}$ and under the plane $z=5$, we need to express the equations in terms of polar coordinates $(r,\theta )$.
In polar coordinates, the equations for the hyperboloid and the plane can be written as follows:
Hyperboloid: $z=\sqrt{1+\frac{r^2{\cos }^2(\theta )}{2}+\frac{r^2{\sin }^2(\theta )}{2}}$
Plane: $z=5$
To find the limits of integration, we need to determine the range of $r$ and $\theta$ values that correspond to the region enclosed by the hyperboloid and under the plane.
First, let's consider the equation of the hyperboloid:
$z=\sqrt{1+\frac{r^2{\cos }^2(\theta )}{2}+\frac{r^2{\sin }^2(\theta )}{2}}$
To simplify this expression, we can notice that $\frac{r^2{\cos }^2(\theta )}{2}+\frac{r^2{\sin }^2(\theta )}{2}=\frac{r^2}{2}({\cos }^2(\theta )+{\sin }^2(\theta ))=\frac{r^2}{2}$.
Substituting this back into the equation, we have:
$z=\sqrt{1+\frac{r^2}{2}}$
Now, we can determine the range of $r$ and $\theta$ values. Since we are interested in the region enclosed by the hyperboloid and under the plane, the volume extends to the values of $r$ where the hyperboloid intersects with the plane $z=5$.
Setting $z=5$ in the equation of the hyperboloid, we have:
$5=\sqrt{1+\frac{r^2}{2}}$
Squaring both sides of the equation, we get:
$25=1+\frac{r^2}{2}$
Simplifying, we obtain:
$\frac{r^2}{2}=24$
$r^2=48$
$r=\sqrt{48}=4\sqrt{3}$
So, the range of $r$ values is from $0$ to $4\sqrt{3}$.
Next, let's consider the range of $\theta$ values. Since we want the region enclosed by the hyperboloid, we need to determine the full range of $\theta$ that covers a complete revolution in the $xy$-plane. This range is typically from $0$ to $2\pi$.
Now, we can set up the iterated double integral in polar coordinates to calculate the volume:
$V={\iint }_{D}f(r,\theta )rdrd\theta$
where $f(r,\theta )$ represents the height function given by $f(r,\theta )=\sqrt{1+\frac{r^2}{2}}$, and $D$ represents the region in the $r\theta$-plane that corresponds to the volume enclosed by the hyperboloid.
The limits of integration for $r$ are from $0$ to $4\sqrt{3}$, and for $\theta$, the limits are from $0$ to $2\pi$. Therefore, the iterated double integral becomes:
$V={\int }_0^{2\pi }{\int }_0^{4\sqrt{3}}\sqrt{1+\frac{r^2}{2}}rdrd\theta$
This is the setup of the iterated double integral in polar coordinates that gives the volume of the solid enclosed by the hyperboloid $z=\sqrt{1+\frac{x^2}{2}+\frac{y^2}{2}}$ and under the plane $z=5$.