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Calculus still confused

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Find the length in the first quadrant of the circle described by the polar equation
r=(2 sin theta)+(4 cos theta)
A. (sqrt2)(pi)
B. (sqrt5)(pi)
C. (2)(pi)
D. (5)(pi)

  • Calculus still confused -

    I am confused also. That does not look to me like the polar equation for a circle.

  • Calculus still confused -

    It is the equation of a circle.

    r = 2sinθ + 4cosθ
    r*r = 2rsinθ + 4rcosθ

    x^2 + y^2 = 2y + 4x

    (x^2 - 4x + 4) + (y^2 - 2y + 1) = 4+1
    (x-2)^2 + (y-1)^2 = 5

    However, we need to find the area using polar coordinates:

    As I noted in my earlier posting, the first quadrant means 0 <= θ <= pi/2

    So, we integrate

    A = 1/2 Int(r^2 dθ)[0,pi/2]
    = 1/2 Int(4sin^2θ + 16sinθ cosθ + 16 cos^2θ dθ)[0,pi/2]

    Recalling some trig identities:

    sin^2θ + cos^2θ = 1
    sin 2θ = 2sinθ cosθ
    cos^2θ = (1 + cos2θ)/2

    A = 1/2 Int(4 + 8sin2θ + 6(1 + cos2θ) dθ)[0,pi/2]
    = 1/2 Int(10 + 8sin2θ + 6cos2θ)[0,pi/2]

    = 1/2(10θ - 4cos2θ + 3sin2θ)[0,pi/2]
    = (5θ - 2cos2θ + 3/2 sin2θ)[0,pi/2]
    = (5*pi/2 - 2(-1) + 0) - (0 - 2(1) + 0)
    = 5pi/2 + 4
    = 11.854

    I don't get any of the given choices. And, my answer agrees with wolfram dot com:

    integrate .5*((2 sin theta)+(4 cos theta) )^2 dtheta, theta=0..pi/2

    Take off the limits of integration to see that their indefinite integral also agrees with mine.

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