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Is it possible to do this integral with polar coord?

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the surface area of that portion of the cylinder x2 + y2 = 16
    that is above the region in the first quadrant bounded on the graphs of x=0, x=2, y=0, y=5

    I know how to solve this via the coordinate system given, but it simplifies to 20∫02 1/(16-x2).5, which means that I have to use the sin-1 formula, and this is not something I'd recognize to do on a test. I am wondering if I can switch it to polar coordinates and solve. I've tried a few times, but not with much luck



    2. The attempt at a solution
    x2 + y2 = r2cos2∅ + r2sin2∅ = 42

    I set up the double integral as:

    ∫∫( 1 + r2).5rdrd∅, but obviously this does not yield the right answer. What else could I try?

    Thanks for any insight you can give
     
  2. jcsd
  3. Apr 18, 2012 #2

    D H

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    Aside: This is a terribly worded problem. No jibe against you. I can see by google search that you have expressed the problem verbatim. What cylinder? This is a plane figure. There's no cylinder here.


    No, it doesn't. Show your work.

    This is something you should be able to recognize on a test. On seeing an integral involving [itex]a^2+x^2[/itex], [itex]\sqrt{a^2+x^2}[/itex], [itex]a^2-x^2[/itex], or [itex]\sqrt{a^2-x^2}[/itex] you should immediately think "trig substitution". You should also know the trig functions for simple angles such as 30° (pi/6), 45° (pi/4), and 60° (pi/3).

    It helps to draw a picture. Draw a picture of this problem and you can cheat. This problem can be viewed as the sum of the area of two figures, a triangle with base 2 and height 2√3, and a circular sector of radius 4 and angle 30°. It's always a good idea in tests (and in real life!) to see if the problem can be solved with the simple techniques you learned years ago.

    On a test the instructor just might want to see the calculus solution. Knowing the high school algebra answer is still worthwhile. It lets you know whether your calculus solution is correct.

    BTW, the polar coordinates integral will give you the high school algebra solution, but via a convoluted path. The circular sector is trivial but the triangle is not.
     
  4. Apr 18, 2012 #3
    z=(16-x2).5

    Our book says that the formula for area is: ∫∫√(1 + fx2 + fy2)dA

    Accordingly:

    A = ∫0205 ( 1 + 0 + x/(16-x2).5 dydx

    =∫0205 ((16-x2 + x2)/((16-x2).5 dydx

    =4∫0205 ((1/(16-x2).5 dydx
    =20∫02 (1/(16-x2).5 dx

    By this line of reasoning, I think it does wind up equaling what I had in the problem statement, but my way of solving it A) might not be correct or B) might be different than the way you would approach the integral

    From here, I know as I said before that the integral of (1/(16-x2).5 dx

    Is equal to 20sin-1x/4 from 0 to 2, which equals 10pi/3, which is the answer


    I had never even considered just drawing the picture and solving it that way. Thank you for telling me, that is a great way to check my work. I will use that on the final.
     
    Last edited: Apr 18, 2012
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