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Thread: Location of r = 4sin(theta)

  1. #1
    Jan 2009

    Location of r = 4sin(theta)

    EDIT: Let me rephrase the topic, PLOT of r = 4sin(theta)

    Hi, I'm not sure if this is the right section, but here's my problem.

    I'm having trouble understanding why the graph (in polar coordinates) of r = 4sin(theta) is just a circle split by the y-axis ontop of the x-axis with a diameter of 4.

    My intuition tells me that the circle should be reflected on the x-axis, leaving a figure-8 shape, but every program I've graphed it in to verify tells me that it's simply the upper circle.

    Why would the domain of theta be restricted to 0-pi.

    The context of the question is a problem (which I've solved):
    Find the surface area of r*cos(3*theta) above the domain r = rsin(theta).
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009
    I'll use t for theta

    consider r = 4sin(t)

    r^2 = 4rsin(t)

    x^2 + y^2 = 4y

    x^2 + y^2 - 4y = 0

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

    which is the eqn of a circle centered at (0,2) with r = 2

    As to your second question as t varies from 0 to pi/2 r varies from 0 to 4

    completing half the circle . as t varies from pi/2 to pi r varies from 4 back to 0 completing the circle.
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