Using integration, find the area of the region enclosed between the two circles:
Given equations of the circles are:
Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C(2, 0) and radius 2.
Solving (1) and (2), we have:
This gives
Thus, the points of intersection of the given circles are
Required area
= Area of the region OACA'O
= 2[area of the region ODCAO]
=2[area of the region ODAO + area of the region DCAD]
Find the area bounded by the circle x2 + y2 = 16 and the line √3y=x in the first quadrant, using integration.
Using integration, find the area of region bounded by the triangle whose vertices are (–2, 1), (0, 4) and (2, 3).
Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 =32
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Prove that the curves y²= 4x and x²= 4y divide the area of the square bonded by x = 0, x = 4, y = 4, and y = 0 into three equal parts.
Using integration find the area of the triangular region whose sides have equations y=2x+1, y=3x+1 and x=4.
Using the method of method of integration, find the area of the region bounded by the following lines:
3x – y – 3 = 0,
2x + y – 12 = 0,
x – 2y – 1 = 0