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The region R, is bounded by the graphs of x = 5/3 y and the curve C given by x = (1+y^2)^(1/2), and the x-axis.

a) Set up and evaluate an integral expression with respect to y that gives the area of R.

b) Curve C is part of the curve x^2 - y^2 = 1, Show that x^2 - y^2 = 1 can be written as the polar equation r^2 = 1/(cos^2theta - sin^2theta).

c) Use the polar equation given in part b) to set up an integral expression with respect to theta that represents the area of R.

  • MATH-HELP! -

    x = (1+y^2)^(1/2)
    x^2 = 1 + y^2
    x^2 - y^2 = 1
    so this is a hyperbola centered at the origin and opening right and left.
    since it was defined as you gave it, we are only doing where x is positive, in the first quadrant and since we are looking at the line, the hyperbola and the x axis, only in the first quadrant.
    we are looking at below the line, above the x axis, and left of the first quadrant branch of the hyperbola.
    now y = 3/5 x is our line
    y = sqrt(x^2-1) is our hyperbola and crosses the axis at x = 1
    where does the line hit the hyperbola?
    5/3 y = sqrt(1+y^2)
    25/9 y^2 = 1+y^2
    16/9 y^2 = 1
    y^2 = 9/16
    since we are totally in the first quadrant, y = 3/4
    so we are looking at
    (x hyperbola - x line) dy from y = 0 to y = 3/4
    [sqrt(1+y^2) -(5/3)y ] dy from 0 to y = 3/4

  • MATH-HELP! -

    x^2 - y^2 = 1
    x = r cos T
    y = r sin T
    r^2 cos^2 T - r^2 sin^2 T = 1
    r^2 (cos^2 T - sin^2 T) =1

  • MATH-HELP! -

    area = (1/2) r^2 dT in polar coordinates
    area = integral (dT/(cos^2 T - sin^2 T))
    lower limit is T = 0
    upper limit is where tan T = (3/4) / (5/4)

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