91Ó°ÊÓ

First, let's sketch the region of integration. The region is bound by the cone \(z=r\) and the sphere \(\rho=2\). Since the range for \(z\) is non-negative values only, we know that our region is in the upper half-space. Also, note that when the cone intersects the sphere, they form a circle in the \(xy\)-plane. Sketching the region, we get the following solid: [insert 3D plot of the region above the cone \(z=r\) and below the sphere \(\rho=2\) for \(z\geq0\)] Now, let's find the limits of integration for each order.

To express the region in the order \(dz\,dr\,d\theta\), first identify the limits of integration: - For \(z\), we know the range is between the cone \(z=r\) and the sphere \(\rho=2\). So, \(z\) starts at the cone (\(z=r\)) and goes up to the sphere. In cylindrical coordinates, the sphere is given by \(z^2 + r^2 = 4\). Solving for \(z\), we get \(z=\sqrt{4-r^2}\). Thus, the bounds for \(z\) are \(r\leq z\leq \sqrt{4-r^2}\). - For \(r\), it ranges from \(r=0\) on the \(z\)-axis up to the circle formed by the intersection of the cone and sphere. To find this point of intersection, set \(z=r\) and \(z=\sqrt{4-r^2}\) and solve for \(r\). We get \(r^2=4-r^2\), which gives \(r=\sqrt{2}\), so the bounds for \(r\) are \(0\leq r\leq \sqrt{2}\). - Lastly, for \(\theta\), it covers the entire circular cross-section, so the bounds are \(0\leq \theta \leq 2\pi\). So, the iterated integral in the order \(dz\,dr\,d\theta\) is: \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} f(r,\theta,z)\,dz\,dr\,d\theta\)

Now, we'll find the limits of integration for the order \(dr\,dz\,d\theta\): - For \(r\), the range is from the cone \(z=r\) to the outer radius of the region based on the \(z\) value. To find the outer radius, we use the equation of the sphere, \(z^2 + r^2 = 4\). Solving for \(r\), we get \(r=\sqrt{4-z^2}\). Thus, the bounds for \(r\) are \(z\leq r\leq \sqrt{4-z^2}\). - For \(z\), it ranges from the vertex of the cone (\(z=0\)) up to the height where the cone and sphere intersect. From the previous step, we calculated this as \(z=\sqrt{2}\). Thus, the bounds for \(z\) are \(0\leq z\leq \sqrt{2}\). - For \(\theta\), it covers the entire circular cross-section as before, so the bounds are \(0\leq \theta \leq 2\pi\). So, the iterated integral in the order \(dr\,dz\,d\theta\) is: \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_z^{\sqrt{4-z^2}} f(r,\theta,z)\,dr\,dz\,d\theta\)

Lastly, we'll find the limits of integration for the order \(d\theta\,dz\,dr\): - For \(\theta\), it covers the entire circular cross-section, so the bounds are \(0\leq \theta \leq 2\pi\). - For \(z\), given \(r\), the range is from the cone (\(z=r\)) to the sphere, which was found previously as \(z=\sqrt{4-r^2}\), so the bounds for \(z\) are \(r\leq z\leq \sqrt{4-r^2}\). - For \(r\), it ranges from \(0\) at the \(z\)-axis to the maximum radius value, which we computed as \(r=\sqrt{2}\) in the previous steps. Thus, the bounds for \(r\) are \(0\leq r\leq \sqrt{2}\). So, the iterated integral in the order \(d\theta\,dz\,dr\) is: \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} f(r,\theta,z)\,d\theta\,dz\,dr\)