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 $C_1 : x^2 + y^2 = 64$. There is a circular pond with a fountain in the middle of the roundabout whose equation is given by $C_2 : x^2 + y^2 = 4$.Based on the given information, answer the following questions :
(i) Represent the given equations $C_1$ and $C_2$ with the help of a diagram. [1 mark]
(ii) Express y as a function of $x$, ($y = f(x)$), for both $C_1$ an $C_2$. [1 mark]
(iii) (a) Using integration find the area of region covered by the roundabout. [2 marks]
OR
(iii) (b) Using integration, find the area of region covered by circular pond. [2 marks]
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Detailed Solution
Step 1: Diagram Representation
The equations $x^2 + y^2 = 64$ and $x^2 + y^2 = 4$ represent two concentric circles centered at the origin $(0,0)$ with radii $r_1 = 8$ and $r_2 = 2$ respectively. The diagram consists of a larger circle of radius 8 and a smaller circle of radius 2 inside it, both sharing the same center.
Step 2: Expressing y as a function of x
For $C_1: x^2 + y^2 = 64$, we have $y^2 = 64 - x^2$, so $y = \pm\sqrt{64 - x^2}$. For $C_2: x^2 + y^2 = 4$, we have $y^2 = 4 - x^2$, so $y = \pm\sqrt{4 - x^2}$.
Step 3: Area Calculation (Integration)
The area of a circle $x^2 + y^2 = r^2$ using integration is given by $4 \int_{0}^{r} \sqrt{r^2 - x^2} dx$.
For $C_1$ (Roundabout): $Area = 4 \int_{0}^{8} \sqrt{8^2 - x^2} dx = 4 \left[ \frac{x}{2}\sqrt{64-x^2} + \frac{64}{2}\sin^{-1}(\frac{x}{8}) \right]_0^8 = 4(0 + 32(\frac{\pi}{2})) = 64\pi$ square units.
For $C_2$ (Pond): $Area = 4 \int_{0}^{2} \sqrt{2^2 - x^2} dx = 4 \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) \right]_0^2 = 4(0 + 2(\frac{\pi}{2})) = 4\pi$ square units.
Final Answer: Area of Roundabout = 64π sq. units; Area of Pond = 4π sq. units
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