How do I integrate surface area using rectangular or polar coordinates?

In summary: Since I used r and \theta as parameters, I was integrating with respect to those. I did not "convert to polar coordinates" after setting up the integral- I set up the integral with respect to r and \theta as parameters. The surface area element is the square root of the square of the length of the cross product of the two "fundamental vector products". There are two possible surfaces, the one above the cone and the one below the cone. The one above the cone has parameter values r between 0 and 1 and \theta between 0 and 2\pi. The one below the cone has parameter values r between 1 and 2 and \theta between 0 and 2
  • #1
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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?
 
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  • #2
"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].
 
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  • #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.

Thanks.
 
  • #4
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".
 

What is surface area confusion?

Surface area confusion refers to a common misconception among students about the concept of surface area in mathematics and science. It often involves confusion between surface area and volume, leading to incorrect calculations and misunderstandings.

How is surface area different from volume?

Surface area is the measurement of the total area of all the exposed surfaces of a three-dimensional object, while volume is the measurement of the space occupied by the object. Surface area is a two-dimensional measurement, while volume is a three-dimensional measurement.

Why is understanding surface area important?

Understanding surface area is important in many fields, including engineering, architecture, and chemistry. It is crucial for accurately calculating things like material quantities, heat transfer, and chemical reactions. It also helps with visualizing and understanding the properties of different shapes and objects.

What are some common mistakes people make when dealing with surface area?

One common mistake is confusing surface area with volume, as mentioned before. Another mistake is forgetting to include all the necessary surfaces when calculating surface area, such as the base or curved surfaces. Using the wrong formula for a specific shape can also lead to errors.

How can someone improve their understanding of surface area?

To improve understanding of surface area, one can practice calculating surface area with different shapes and objects. It is also helpful to visualize the shapes and surfaces in three dimensions. Seeking help from a teacher or tutor can also be beneficial in clearing up any confusion or misunderstandings.

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