Question

Find a vector-valued function whose graph is the indicated surface. The part of the paraboloid \(z=x^{2}+y^{2}\) that lies inside the cylinder \(x^{2}+y^{2}=9\)

Short answer

The vector-valued function whose graph is the part of the paraboloid \(z = x^{2} + y^{2}\) that lies inside the cylinder \(x^{2} + y^{2}=9\) is \(\vec{r}(r, \theta) = r \cos(\theta)\vec{i} + r \sin(\theta)\vec{j} + r^{2}\vec{k}\) where \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2 \pi\).

Step-by-step solution

Understanding the task

The task is to find a vector-valued function for a paraboloid \(z = x^{2} + y^{2}\) but only for the part that lies inside the cylinder \(x^{2} + y^{2} = 9\). Here, the limits of the surfaces are determined by the equation of the cylinder.

Setup the vector-valued function

The vector-valued function for the paraboloid may be expressed in the form \(\vec{r}(x, y) = x\vec{i} + y\vec{j} + (x^{2} + y^{2})\vec{k}\). However, the part that lies inside the cylinder means that the values of x and y should satisfy the equation \(x^{2} + y^{2} \leq 9\). That's a circle equation in the xy plane, so it's easier to use polar coordinates here. Therefore, let \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) so the vector-valued function becomes \(\vec{r}(r, \theta) = r \cos(\theta)\vec{i} + r \sin(\theta)\vec{j} + r^{2}\vec{k}\), where \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2 \pi\).

Converting vector-valued function to polar form

Having substituted the polar coordinates, the vector-valued function now gets the following form: \(\vec{r}(r, \theta) = r \cos(\theta)\vec{i} + r \sin(\theta)\vec{j} + r^{2}\vec{k}\). This vector valued function gives the same paraboloid, but only for the part that lies inside the cylinder \(x^{2} + y^{2}=9\).