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Integrating Implicit Functions

  1. Feb 26, 2016 #1
    • Originally posted in a non-homework section, so missing the template
    In one of the homework sheets my teacher gave us, we had to calculate area geometrically (meaning no integration was used). Some parts, she said, we needed to just eyeball which I hate doing. In this case the top left portion of a circle described by the equation...
    https://lh4.googleusercontent.com/RksM0NW9ZgHNhfNFiZk6LZScXMRRTw9bIc76W9q8udnF4Qy48n4HmG5f526lezK8bEMipFLHJvp-hPN97gcSZYtb6ix0O9TQxnrV6IXAG7VGMPdtkASNYGxDtZgOvIsrc0gHHqa4
    was shown and we need to find the area from [6,8]. So I integrated the equation from 6 to 8 with respect to x...
    https://lh5.googleusercontent.com/JVPHXihl_rTOFyRbQMGlATw5i1spj0qU4c5taOghVF_q7f-jyI4VTvtBeK69XOI76hnqw2FhdUt5Vzayfa-cBZ1iCC7Fhw5RKsOFo6OxlPBndtCLk-gvFrNwR4Z3KaVlS00t7Jxk
    https://lh6.googleusercontent.com/QI58405jGPaitde0HrZfC_zJrsB4iEoixqs9AkqV8eyr4dCLYeSd64WccercpncO3bElIcfRd_HJttVC8hnMpmX9aZJJYwhpBXlO1Y5WYI99wubqFDl5bErVTOaxZUs9vSOLLmQo
    I am not sure how to integrate this by hand so I used Mathway.com and it spit out this.
    https://lh4.googleusercontent.com/hVf_R-YZA8zDWUU3Sm-0lQt8SpkIiVlJy-JXYCBMrYwiBlx_jzuUOUwn7dJjxaVRv5JxYQ1aO2TQl6i94iGe_Fri_CgcP-y6kO2xuKJvvgWmQUISoAN7JGEpSVDCJX47VX_2P5s8
    I don't completely understand what this means. I graphed it to find where the zeros are and got
    https://lh6.googleusercontent.com/El282NKbIJn3H7_uPivwQaP_VqX0WPFllVwqKbn5R7FO3u4_-yI_TgeoJLoiwx7nFBHFo901-kxUEkAOQufVZE8Ku4q078oa29X3lmDPZjXndfpNEz97JuWPQPa-AFhbL9pFfvb3
    At first it looked to me as though the negative number is the area under the x-axis from 6 to 8 and the positive number is the area above the x-axis from 6 to 8. However, when I do eyeball it, it looks like those values should be doubled which is what causes my confusion. If anyone can explain any flaws in my math or what this means, I would appreciate it. Thanks.
     
  2. jcsd
  3. Feb 26, 2016 #2

    Mark44

    Staff: Mentor

    I'm not sure what your teacher had in mind. If it was finding the area of the upper left quarter circle, that's easy enough to do. If it's estimating the area of the upper left quarter, between x = 6 and x = 8, you can approximate this area by calculating the area of a triangle. The triangle is inside the region we're interested in, so its area will be a little less than the area we're interested in.

    I don't know why you say you hate doing this -- being able to use geometry to estimate areas is a very important skill. Some problems can be solved very easily by invoking geometry, while solving them analytically can be very difficult or worse.
    This isn't right. The integrand should be a function of x only. You can get this by solving the original equation for y. Because the equation represents a circle, for most x values there will be two y values. Since you're dealing with the upper left quadrant of the circle, the y values will be positive.
     
  4. Feb 26, 2016 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The integration you wrote is not the area; it is something, but I don't know what.

    For a correct approach, see, eg.,
    http://tutorial.math.lamar.edu/Classes/CalcI/AreaBetweenCurves.aspx
    In your case the two curves are either (a) the lower and upper semicircles, or else (b) the upper semicircle (on top) and the x-axis (on the bottom), depending on exactly how you interpret the question.

    Anyway, were you not supposed to avoid integration?
     
  5. Feb 26, 2016 #4
    The area we trying to find is not the top left quadrant. I apologize if my phrasing was misleading. I said top left portion which was terribly vague. It is the area of the circle from 6 to 8 divided by two.

    Its not so much that I hate using geometry. If it is possible to find exact answers rather than estimations, I personally prefer to do that. If using geometry is necessary or more efficient, I will gladly use it. I did solve the problem using geometry but I challenged myself to find the exact answer using integration.
     
  6. Feb 26, 2016 #5

    Mark44

    Staff: Mentor

    Estimates have the advantage of being quick to perform. Getting a quick, approximate answer is especially helpful to show that an exact, but totally wrong, answer (like your attempt) has been calculated.
    Again, solve for y as a function of x from your equation for the circle. There willl be a square root involved, but since you're dealing with the upper half of the circle, you need only the pos.. square root. I understand that you're finding the area of only a portion of the upper left quadrant. To do the integration you will probably need to use a trig substitution or an entry from a table of integration formulas.
     
  7. Feb 26, 2016 #6

    Mark44

    Staff: Mentor

    Doing the problem in two ways, with geometry (finding the areas of a sector of a circle and a triangle) and with calculus, is a good idea. If both techniques produce the same answer, that's a good sign!
     
  8. Feb 26, 2016 #7

    Ray Vickson

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    Science Advisor
    Homework Helper

    If you have already been given (or have already developed for yourself) a formula for the area of a circular sector, you can use that to get an exact answer. This could be done without using integration, but of course, integration has been used to get the formula for a sectoral area.
     
  9. Feb 26, 2016 #8
    Thanks for the help guys :)
     
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