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Math Help - Calc 1 area of plane curve

  1. #1
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    Joined
    Oct 2008
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    Calc 1 area of plane curve

    Consider the parametric equation:

    x = 3(cos(theta) + (theta)sin(theta))
    y = 3(sin(theta) - (theta)cos(theta))

    what is the length of the curve for theta = 0 to theta = 3/2(pi)?

    I put the derivative of x and y in a length formula to get: I'll put "t" for theta

    int_0^3/2pi sqrt[(3(tcost))^2 + (3(-tsint))^2]

    then i distributed the squares to get:

    int_0^3/2pi sqrt[9t^2(cos^2t+sin^2t)]

    then i took out the 9t^2 to get:

    3 int_0^3/2pi sqrt[(cos^2t + sin^2t)]

    and integrated...

    then i took out the 9t^2 to get:

    3 int_0^3/2pi sqrt[(sin^2t - cos^2t)]

    I put in t = 3/2pi to get: 29.09768846
    my online submission was declared incorrect. Can anybody help me see what I did wrong?

    Thanks!
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  2. #2
    MHF Contributor
    Joined
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    France
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    Everything is correct except what I put in red color

    I put the derivative of x and y in a length formula to get: I'll put "t" for theta

    int_0^3/2pi sqrt[(3(tcost))^2 + (3(+tsint))^2]

    then i distributed the squares to get:

    int_0^3/2pi sqrt[9t^2(cos^2t+sin^2t)]

    then i took out the 9t^2 to get:

    3 int_0^3/2pi t sqrt[(cos^2t + sin^2t)]

    and integrated...

    then i took out the 9t^2 to get:

    3 int_0^3/2pi t sqrt[(sin^2t + cos^2t)]

    I put in t = 3/2pi to get: 29.09768846

    Remember that sin^2t + cos^2t = 1
    Therefore you get
    3 int_0^3/2pi t dt = 3/2 (3pi/2)² = 27pi²/8 = 33.31
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  3. #3
    Member
    Joined
    Oct 2008
    Posts
    76
    Thank you!

    One quick question. If the integral of sinx is -cosx, why is the integral of sin^2x cos^2x and not -cos^2x?
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