Find the area of the region inside the circle x^2+y^2=2 AND above the line y=1.
the radius of the circle is √2
Join radii to the points where the line y=1 cuts the circle.
from the origin to the line along the y-axis is 1, the radius is √2, so the other side of the triangle must be 1.
so you have a 45, 45 90 triangle, so is the other one
Both of them make a 90º at the centre.
the area of those two triangles is 1/2(2)(1) = 1
The area of the sector with the 90 angle is 1/4(the circle) = 1/4(2pi) = pi/2
So the area of the part above the line is
pi/2 - 1
Ohh!! That makes perfect sense. Thanks so much for helping Reiny :-) I appreciate it!
To find the area of the region inside the circle x^2 + y^2 = 2 and above the line y = 1, we need to determine the points of intersection between the circle and the line.
First, let's find the x-coordinate of the points of intersection. To do so, substitute y = 1 into the equation of the circle:
x^2 + (1)^2 = 2
x^2 + 1 = 2
x^2 = 1
x = ± 1
So, the points of intersection are (-1, 1) and (1, 1).
Next, we need to determine the limits of integration. The region inside the circle and above the line can be split into two equal halves. We will calculate the area for one half and then double it to get the total area.
Let's calculate the area for the half where x is between -1 and 1. To do so, we integrate the difference between the equation of the circle and the line:
A = 2 * ∫[from -1 to 1] [(2 - x^2) - 1] dx
Simplifying this:
A = 2 * ∫[from -1 to 1] (1 - x^2) dx
Using the power rule, we integrate term by term:
A = 2 * [x - (1/3)x^3] evaluated from -1 to 1
Substituting our limits of integration:
A = 2 * [(1 - (1/3)(1)^3) - ((-1) - (1/3)(-1)^3)]
A = 2 * (1 - (1/3) - (-1 + (1/3)))
A = 2 * (1 - 1/3 + 1 - 1/3)
A = 2 * (2 - 2/3)
A = 2 * (4/3)
A = 8/3
Therefore, the area of the region inside the circle x^2 + y^2 = 2 and above the line y = 1 is 8/3 square units.
To find the area of the region inside the circle x^2+y^2=2 and above the line y=1, we need to determine the points of intersection between the circle and the line.
First, let's represent the equation of the circle and the line in a standard form:
Circle: x^2 + y^2 = 2
Line: y = 1
Next, we substitute y=1 into the equation of the circle:
x^2 + 1^2 = 2
Simplifying the equation, we get:
x^2 + 1 = 2
x^2 = 1
x = ±√1
x = ±1
So, the x-coordinate of the points of intersection is ±1.
Now, to find the y-coordinate of the points of intersection, we substitute x=1 into the equation of the line:
y = 1
Therefore, the point of intersection (1, 1) lies on the circle and above the line.
Similarly, substituting x=-1 into the equation of the line, we get:
y = 1
So, the point of intersection (-1, 1) also lies on the circle and above the line.
Now, we have two points of intersection: (1, 1) and (-1, 1). These are the points where the circle intersects with the line.
To find the area of the region inside the circle and above the line, we can calculate the area of the sector formed by these two points, subtract the area of the triangle formed by these two points and the center of the circle, and add the area of the segment formed by the circle and the triangle.
Let's start by calculating the area of the sector. The formula for the area of a sector is given by:
Area of sector = (θ/360) * πr^2
Since the angle of the sector in this case is 90 degrees (a quarter of a full circle), the area of the sector can be calculated as:
Area of sector = (90/360) * π * 2^2
Simplifying the equation, we get:
Area of sector = (1/4) * π * 4
Area of sector = π
Next, we need to calculate the area of the triangle formed by the two points of intersection and the center of the circle (0, 0). Since the base of the triangle is the diameter of the circle, which has a length of 2, and the height is the difference between the y-coordinate of the points of intersection and the y-coordinate of the center, which is 0 - 1 = -1, the area of the triangle can be calculated as:
Area of triangle = (1/2) * base * height
Area of triangle = (1/2) * 2 * (-1)
Area of triangle = -1
Finally, we can calculate the area of the segment by subtracting the area of the triangle from the area of the sector:
Area of segment = Area of sector - Area of triangle
Area of segment = π - (-1)
Area of segment = π + 1
Therefore, the area of the region inside the circle x^2 + y^2 = 2 and above the line y = 1 is given by the area of the segment, which is π + 1 square units.