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Roundabouts are often made on busy roads to ease the traffic and avoid red lights. One such round-about is made such that equation representing its boundary is given by C1 ;  x2 + y2 = 64. - Mathematics

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Question

Roundabouts are often made on busy roads to ease the traffic and avoid red lights.

One such round-about is made such that equation representing its boundary is given by C1 ;  x2 + y2 = 64.

There is a circular pond with a fountain in the middle of the roundabout whose equation is given by C2 ;  x2 + y2 = 64.

Based on the given information, answer the following questions:

  1. Represent the given equations C1 and C2 with the help of a diagram.
  2. Express y as a function of x, (y = f(x)), for both C1 and C2.
    1. Using integration find the area of region covered by the roundabout.
      OR
    2. Using integration, find the area of region covered by circular pond.
Case Study
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Solution

(i)

The equations represent two concentric circles centered at the origin (0, 0):

  • C1 (Roundabout): x2 + y2 = 82 (Radius = 8 units)
  • C2 (Circular Pond): x2 + y2 = 22 (Radius = 2 units)

Image of a graph lightbox

(ii)

To express y = f(x), we isolate y in both equations:

  • For C1: y = `+-sqrt(64 - x^2)`
  • For C2: y = `+-sqrt(4 - x^2)`

(iii) (a)

To find the area of the entire roundabout (C1), we calculate the area in the first quadrant and multiply by 4:

Area = `4 int_0^8 sqrt(8^2 - x^2)  dx`

Using the identity `int sqrt(a^2 - x^2)  dx = x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1 (x/a)`:

Area = `4[x/2 sqrt(64 - x^2) + 64/2 sin^-1 (x/8)]_0^8`

Area = `4[(0 + 32 * π/2) - (0 + 0)]`

Area = 4(16π)

∴ Area = 64π sq. units

OR

(iii) (b)

Similarly, for the pond (C2) with radius r = 2:

Area = `4 int_0^2 sqrt(2^2 - x^2)  dx`

Area = `4[ x/2 sqrt(4 - x^2) + 4/2 sin^-1 (x/2)]_0^2`

Area = `4[(0 + 2 * π/2) - (0 + 0)]`

Area = 4(π)

∴ Area = 4π sq. units

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2025-2026 (March) 65/1/1
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