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Area Under a Parametric Curve (did I do it correctly?)

  1. May 28, 2009 #1
    Find the area under the parametric curve (clearly a circle):
    domain of t: [0,2pi]


    integral from alpha to beta (lower to upper) of g(t)h'(t)dt

    integral from 0 to pi/2 of 10cos(t)^2
    10(x/2+sin2x/4) evaluated from 0 to pi/2

    solution: 2.5pi

    however, i'm not sure if that value gets multiplied by either 2 or 4 or is fine as is
  2. jcsd
  3. May 28, 2009 #2


    Staff: Mentor

    Your parametric equations don't describe a circle: they describe an ellipse.

    What's your reasoning for integrating g(t)h'(t)? That would be y*dx/dt, and I don't see how that ties into your ellipse.
  4. May 28, 2009 #3
    sorry i meant an ellipse

    apparently that's the formula

    integral of alpha to beta of g(t)h'(t)dt
  5. May 28, 2009 #4
    The problem says 0 to 2*pi, which is all the way around the ellipse. You did 0 to pi/2 which is only a quarter of the way around the ellipse. Since the other three quarters look the same as the one quarter that you did, what would be the answer for the entire ellipse?
  6. May 28, 2009 #5
    By the way you can look up the formula for the area of an ellipse and check your answer. They should be the same.
  7. May 28, 2009 #6
  8. May 28, 2009 #7
    I was almost tempted to make the mistake of interpreting t as the polar angle and using the area formula for polar coordinates. :blushing:
  9. May 28, 2009 #8
    Thanks what i'm confused with

    i assumed you multiply that solution by 4

    however, i looked up the formula for the area of an oval

    It's the length of the major axis/2 x length of the minor axis/2 x pi

    therefore the answer would be 2.5pi

    that is equal to when i evaluated it from 0 to pi/2.
  10. May 28, 2009 #9
    The length of the major axis of this particular ellipse is 10 and it's minor axis length is 4.
  11. May 28, 2009 #10

    Well that explains all of my confusion


    i accidentally only looked at half of the minor axis and divided by 2
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