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 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! -
**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^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

or

[sqrt(1+y^2) -(5/3)y ] dy from 0 to y = 3/4

- MATH-HELP! -
**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

- MATH-HELP! -
**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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