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^2è - sin^2è).
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.
a) To find the area of region R using an integral expression with respect to y, we need to determine the limits of integration for y. We can do this by finding the points of intersection between the two curves x = 5/3 y and x = (1+y^2)^(1/2).
Setting the equations equal to each other:
5/3 y = (1+y^2)^(1/2)
Squaring both sides:
(25/9) y^2 = 1 + y^2
Simplifying:
(16/9) y^2 = 1
Dividing both sides by (16/9):
y^2 = 9/16
Taking the square root:
y = ±3/4
The y-values of the points of intersection are y = 3/4 and y = -3/4.
Now, let's evaluate the integral expression for the area of R using these limits of integration:
∫[from -3/4 to 3/4] (x - (5/3 y)) dy
To find x in terms of y, we can substitute the equation x = (1+y^2)^(1/2) into the integral expression:
∫[from -3/4 to 3/4] ((1+y^2)^(1/2) - (5/3 y)) dy
Evaluating this integral will give us the area of region R.
b) To show that x^2 - y^2 = 1 can be written as the polar equation r^2 = 1/(cos^2θ - sin^2θ), we need to convert the Cartesian equation into polar form.
Starting with the Cartesian equation:
x^2 - y^2 = 1
Substituting x = r cosθ and y = r sinθ (using x = r cosθ and y = r sinθ as the definitions of polar coordinates):
(r cosθ)^2 - (r sinθ)^2 = 1
Simplifying:
r^2 cos^2θ - r^2 sin^2θ = 1
Factoring out r^2:
r^2 (cos^2θ - sin^2θ) = 1
Dividing both sides by (cos^2θ - sin^2θ):
r^2 = 1/(cos^2θ - sin^2θ)
This shows that x^2 - y^2 = 1 can be written as r^2 = 1/(cos^2θ - sin^2θ) in polar form.
c) To set up an integral expression with respect to θ that represents the area of region R using the polar equation r^2 = 1/(cos^2θ - sin^2θ):
We need to determine the limits of integration for θ. These limits depend on the shape of region R.
From the given equations, we know that x = (1+y^2)^(1/2), so in polar coordinates, x = r cosθ.
Setting r cosθ = (1+y^2)^(1/2):
r cosθ = (1+(r sinθ)^2)^(1/2)
Squaring both sides:
(r cosθ)^2 = 1 + (r sinθ)^2
Expanding and simplifying:
r^2 cos^2θ = 1 + r^2 sin^2θ
Rearranging:
r^2 (cos^2θ - sin^2θ) = 1
Comparing this equation to the polar equation given in part b), we can see that the polar equation r^2 = 1/(cos^2θ - sin^2θ) represents the curve C.
To find the limits of integration for θ, we need to determine the range of values for which the polar equation r^2 = 1/(cos^2θ - sin^2θ) is valid.
Since r represents the distance from the origin and cannot be negative, we focus on the denominator of the equation: cos^2θ - sin^2θ.
Using trigonometric identities, we know that cos^2θ - sin^2θ = cos(2θ).
Therefore, the equation r^2 = 1/(cos^2θ - sin^2θ) is valid as long as cos(2θ) ≠ 0.
This means that θ cannot equal 45° or 135°, since cos(90°) = 0.
Therefore, the limits of integration for θ are from 0 to 45° and from 45° to 90°.
To set up the integral expression, we integrate with respect to θ:
∫[from 0 to π/4] ∫[from 0 to 1/(cos^2θ - sin^2θ)] r dr dθ
Evaluating this double integral will give us the area of region R.