Which has more area, the region in the first quadrant enclosed by the line x+y=1 and the circle x^2+y^2=1, or the region in the first quadrant enclosed by the line x+y=1 and the curve sqrt(x)+sqrt(y)=1? Justify your answer.
Nice Question!!!
The first one we can do without Calculus, it is simply the segment between the chord from (0,1) to (1,0), that is
pi/4 - 1/2 = (pi - 2)/4
the second equation is
√x + √y = 1
or
√y = 1-√x
y = (1-√x)^2 = 1 - 2√x + x for 0 <= x <= 1
the integral of 1 - 2√x + x is x - (4/3)x^(3/2) + (1/2)x^2
and the area enclosed by √x + √y = 1 , the x-axis, and the y-axis is (from 0 to 1)
(1 - 4/3 + 1/2) - 0
= 1/6
So the area between the curve √x + √y = 1 and the line x+y=1 is
1/2 - 1/6 = 1/3
then (pi-2)/4 = .2854
1/3 = .33333
So who is bigger?
To compare the areas of the two regions, it is necessary to find the points where each curve intersects the line x+y=1, and then determine the area of the enclosed region for each curve.
1. Line x+y=1 and Circle x^2+y^2=1:
To find the points of intersection, we can set up a system of equations:
x + y = 1 (Equation of the line)
x^2 + y^2 = 1 (Equation of the circle)
From the equation of the line, we can express x in terms of y as x = 1-y, and substitute it into the equation of the circle:
(1-y)^2 + y^2 = 1
Simplifying the equation gives us:
1 + y^2 - 2y + y^2 = 1
2y^2 - 2y = 0
Factoring out 2y:
2y(y - 1) = 0
So, y = 0 or y = 1.
Substituting these values back into the equation of the line, we can find the corresponding x-values:
- When y = 0: x = 1 - y = 1 - 0 = 1
- When y = 1: x = 1 - y = 1 - 1 = 0
Therefore, the points of intersection are (1, 0) and (0, 1).
Now, we need to calculate the areas of the enclosed regions for both the circle and the line:
- For the circle:
The enclosed area is the difference between the area of the circle sector and the area of the triangle created by the line and the x-axis.
To find the area of the sector, we can calculate the angle formed by the x-axis and the line connecting the origin to the point of intersection (1, 0):
θ = arctan(1/0) = π/4
The area of the sector is (1/2) * r^2 * θ, where r is the radius of the circle (r = 1):
Area of sector = (1/2) * 1^2 * (π/4) = π/8
To find the area of the triangle, we use the formula:
Area of triangle = (1/2) * base * height
The base of the triangle is the x-coordinate of the point of intersection (1), and the height is the y-coordinate (0):
Area of triangle = (1/2) * 1 * 0 = 0
Therefore, the area enclosed by the circle and the line in the first quadrant is:π/8.
- For the curve sqrt(x) + sqrt(y) = 1:
Similar to the previous calculation, we need to find the points of intersection between the curve and the line x + y = 1.
Substituting x = 1 - y into the curve equation:
sqrt(1 - y) + sqrt(y) = 1
Squaring both sides to eliminate the square roots:
1 - y + 2*sqrt((1-y)*y) + y = 1
2*sqrt((1-y)*y) = 0
sqrt((1-y)*y) = 0
(1-y)*y = 0
y*(1-y) = 0
This equation gives us two solutions: y = 0 and y = 1.
Substituting these values back into the equation x = 1 - y, we obtain the corresponding x-values:
- When y = 0: x = 1 - y = 1 - 0 = 1
- When y = 1: x = 1 - y = 1 - 1 = 0
So, the points of intersection with the curve are (1, 0) and (0, 1).
Now, let's calculate the area of the enclosed region for the curve:
Since the curve is not a closed shape, finding the exact enclosed region is not straightforward. However, the x and y coordinates of the points of intersection give us an idea of the shape.
The curve sqrt(x) + sqrt(y) = 1 consists of two straight lines: the x-axis and the y-axis. Therefore, the enclosed region is the triangle formed by the line x+y=1 in the first quadrant, and the x and y axes.
The area of the triangle can be calculated using the formula:
Area = (1/2) * base * height
The base of the triangle is the x-axis, which has a length equal to the x-coordinate of the point of intersection (1), and the height is the y-axis, which has a length equal to the y-coordinate (1):
Area = (1/2) * 1 * 1 = 1/2
Hence, the area enclosed by the curve and the line in the first quadrant is: 1/2.
Comparing the two areas:
The area enclosed by the circle and the line: π/8
The area enclosed by the curve and the line: 1/2
Therefore, the region in the first quadrant enclosed by the line x + y = 1 and the circle x^2 + y^2 = 1 has a smaller area than the region enclosed by the line x + y = 1 and the curve sqrt(x) + sqrt(y) = 1.