Find the area of the region that lies inside both curves.
r = (3)^(1/2) cos (theta),
r = sin (theta)
To find the area of the region that lies inside both curves, we need to determine the points of intersection between the curves.
Let's find the points of intersection by setting the two equations equal to each other:
(3)^(1/2) cos(theta) = sin(theta)
To simplify this equation, we can square both sides and use the trigonometric identity cos^2(theta) + sin^2(theta) = 1:
(3/2) cos^2(theta) = 1 - cos^2(theta)
Multiplying through by 2:
3 cos^2(theta) = 2 - 2 cos^2(theta)
Combining like terms:
5 cos^2(theta) = 2
Dividing by 5:
cos^2(theta) = 2/5
Taking the square root of both sides:
cos(theta) = sqrt(2/5)
Now, let's find the values of theta where cosine is equal to sqrt(2/5). We need to consider the range of theta where both curves overlap.
Using the inverse cosine function, we can find one solution in the range [0, π/2]:
theta = acos(sqrt(2/5))
Now, let's find the area of the region by integrating the inner curve (r = (3)^(1/2) cos(theta)) from theta = 0 to theta = acos(sqrt(2/5)), and then integrating the outer curve (r = sin(theta)) from theta = acos(sqrt(2/5)) to theta = π/2.
The formula for finding the area enclosed by a polar curve is:
Area = ∫[inner curve to outer curve] (1/2) r^2 dθ
Substituting the values of the curves, the integral becomes:
Area = 1/2 ∫[0 to acos(sqrt(2/5))] [(3)^(1/2) cos(theta)]^2 dθ + 1/2 ∫[acos(sqrt(2/5)) to π/2] [sin(theta)]^2 dθ
We can simplify these integrals and evaluate them to find the area of the region.