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
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.
3 answers
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
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
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)
area = integral (dT/(cos^2 T - sin^2 T))
lower limit is T = 0
upper limit is where tan T = (3/4) / (5/4)
0 answers