Posted by PLEASE HELP! on Sunday, March 2, 2008 at 7:48pm.
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 xaxis.
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.

MATHHELP!  Damon, Sunday, March 2, 2008 at 8:30pm
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^21) 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
or
[sqrt(1+y^2) (5/3)y ] dy from 0 to y = 3/4

MATHHELP!  Damon, Sunday, March 2, 2008 at 8:32pm
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

MATHHELP!  Damon, Sunday, March 2, 2008 at 8:40pm
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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