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Surface Area Confusion

  1. Dec 7, 2006 #1
    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?
  2. jcsd
  3. Dec 8, 2006 #2


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    "The process of Sqrt[1 + dz/dx ^2 + dz/dy ^2]." Is there a verb in that sentence?:rolleyes:

    I see no reason to use Cartesian coordinates at all. Find the area of the surface z= 9- r2 above the cone z= 8r. The paraboloid cuts the cone when 9-r2= 8r or r2+ 8r- 9= (r+9)(r-1)= 0 or r= -9, r= 1. Since r must be positive, the paraboloid and cone intersect on the circle r= 1 and the paraboloid is above the cone for r< 1.

    Use the polar coordinates, r and [itex]\theta[/itex] as parameters, the paraboloid is described by [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex], [itex]z= 9- r^2[/itex]. The "position vector" is [itex]\vec{r}= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ (9- r^2)\vec{k}[/itex]. The derivatives with respect to r and [itex]\theta[/itex] are [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}- 2r\vec{k}[/itex] and [itex]-r sin(\theta)\vec{i}+ r cos(\theta)\vec{j}[/itex] respectively. The "fundamental vector product" is the cross product of those derivatives, [itex]-2r^2 cos(\theta)\vec{i}- 2r^2 sin(\theta)\vec{j}+ r\vec{k}[/itex] and the differential of surface area is its length times [itex]drd\theta[/itex], [itex]\sqrt{4r^2+ r^2}drd\theta = \sqrt{5} rdrd\theta[/itex].

    The surface area, then, is
    [tex]\sqrt{5}\int_{r=0}^1\int_{\theta= 0}^{2\pi}r drd\theta= \pi \sqrt{5}[/tex].
    Last edited: Dec 8, 2006
  4. Dec 8, 2006 #3
    I'm wondering how you got the expression Sqrt[4r^2 + r^2] dr dtheta. I found the magnitude of the cross product to be Sqrt[4r^4 + r^2]. Integrating over r from 0 to 1 and theta from 0 to 2pi gave me (1/6)(-1 +5*Sqrt[5])*pi. The same answer as when I use x=x, y=y, and z=f(x,y)= 9-x^2 -y^2 as my parametrizations and then convert to polar once I have the integral set up.

    My problem is that once I use the parameters x=r Cos[theta], y=r Sin[theta], and z=9-r^2, wouldn't the transformation change it so that once I go to integrate the surface area I'm really integrating over rectangular co-ordinates instead of polar? This would also give me the integral of Sqrt[4r^4 + r^2]*r dr dtheta instead of what my teacher got: Sqrt[4r^4 + r^2] * r^2 dr dtheta. The former being the way I would expect it to be done and how my book(Stewart) does it for a parabaloid once it is in polar co-ordinates.

  5. Dec 9, 2006 #4


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    If you write any surface in terms of two parameters: x= f(u,v), y= g(u,v), z= h(u,v) then you are integrating with respect to the two parameters u and v, not "rectangular coordinates" or "polar coordinates".
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