Question

Set up a triple integral for the volume of the solid. The solid bounded by the paraboloid \(z=9-x^{2}-y^{2}\) and the plane \(z=0\)

Short answer

The volume of the solid bounded by the paraboloid \(z=9-x^{2}-y^{2}\) and the plane \(z=0\) can be computed by the triple integral \(V=\int_{0}^{2\pi} \int_{0}^{3} \int_{0}^{9-r^{2}} rdrd\theta dz \)

Step-by-step solution

Understanding the solid

The first step is to understand the shapes of the paraboloid \(z=9-x^{2}-y^{2}\) and the plane \(z=0\). Here, \(z=9-x^{2}-y^{2}\) represents a paraboloid that opens downward, with its vertex at the point (0,0,9), and \(z=0\) is a flat plane on the xy-plane.

Setting up the boundaries

To integrate the solid bounded by these two surfaces, cylindrical coordinates are more efficient due to the symmetry of the paraboloid. In cylindrical coordinates \((r, \theta, z)\), \(x^{2}+y^{2} = r^{2}\). So the equation of the paraboloid becomes \(z=9-r^{2}\). The xy projection of the solid is a circle, thus \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\). For z, the solid lies between the plane \(z=0\) and the paraboloid \(z=9-r^{2}\). Therefore, \(0 \leq z \leq 9-r^{2}\).

Setting up the triple integral

Now it's time to set up the triple integral for the volume. The volume element in cylindrical coordinates is \(dV=rdrd\theta dz\). Thus, the volume of the solid is given by the triple integral \(V=\int\int\int dV=\int_{0}^{2\pi} \int_{0}^{3} \int_{0}^{9-r^{2}} rdrd\theta dz \)