 Hello and welcome to the screencast on section 11.6, surfaces defined parametrically and surface area. We've already seen the parametric equations of lines and other vector-valued functions. In this screencast, we'll look at how to parameterize surfaces. As a quick review, any single variable function y equal to g of x can be represented parametrically x component t and y component g of t, where t is our parameter. More importantly, curves that are not functions may be represented parametrically, such as the unit circle, which is parameterized by x component given by cosine of t and y component given by sine of t, where t is some parameter that runs between 0 and 2 pi. Switching to two variable functions, any surface z equal to f of x, y may also be expressed parametrically. Using two parameters, we choose to use s and t as our parameters here. Then, the x component can be given by the variable s. The y component is given by the variable t and the z component is given by f of s, t. But again, even in the two variable case, what is more important are the surfaces that cannot be generated by a two variable function, but can be represented parametrically. For example, consider a torus, also known as a donut, where the radius from the center of the hole to the center of the tube is 1 and the radius of the tube is also 1. Note that the torus cannot be expressed as a single function of two variables. However, the torus can be represented parametrically by the following equations. You'll investigate the parametric equations of other such surfaces in section 11.6. We'll now turn our attention to determining the surface area of parametrically defined surfaces. So, consider a surface r defined by the following parametric equations. As r is a function of two variables, we can consider its partial derivatives, which are displayed here. Recall that a differentiable function is locally linear. That is, if we zoom in on the surface around the point, the surface looks like it's a tangent plane. We now exploit this idea in order to determine the surface area generated by a parameterization. The basic idea we use is a familiar one. We will subdivide the surface into small pieces by subdividing the intervals of s and t. We next estimate the entire surface area by adding up the areas of the smaller approximations. Ultimately, we use an integral to sum these approximations and determine the exact surface area. We'll proceed with the basic idea here. See your textbook for the full details. When we start by subdividing the intervals of s and t, the area of the sub-rectangles that remain can be approximated using the area of this parallelogram, whose lengths of sides are given in terms of the partial derivatives that we have here. Recalling facts about the cross product, we can find the area of this parallelogram using the magnitude of the cross product of these two vectors. We sum up the areas of the small parallelograms using a double Riemann sum so to estimate the total surface area generated by r. Taking a limit, increasing the number of parallelograms without bound, our approximation approaches the exact value of the surface area given by the following double integral. We end this screencast with a formal statement of the surface area here.