- #1
Sculptured
- 23
- 0
Just had my test on Vector Fields and there was one question which really confused me. It asked to find the surface area of the parabaloid z = 9-x^2 -y^2 which is above the cone z = 8Sqrt[x^2 + y^2]. My memory told me to use the differential in rectangular coordinates and then convert to cylindrical. The process of Sqrt[1 + dz/dx ^2 + dz/dy ^2]. This leads to an integral of r Sqrt[1 + 4r^2] in cylindrical coordinates. After doing so and getting an answer my teacher said he messed up when creating the problem and that the integral turned out to be r^2 * Sqrt[1 + 4r^2]. This seemed non-intuitive and took me a good amount of time before I went through the process of parametrizing the variables to: x = tcos(theta), y = tsin(theta), and z = 9-t^2. (t being the same as r). Finding the partial derivatives and solving for the magnitude of the cross product led me to t Sqrt[1 + 4t^2] for my area integral. Making my area differential t dt dtheta gave me that extra t to make the t squared; however, this seems confusing because each method created different answers. The first thing coming to mind would be that because my parameters are already in polar coordinates the extra t in the differential shouldn't be there and I should be just able to integrate with dt dtheta. Which way is correct and why?