Roundabouts are often made on busy roads to ease the traffic and avoid red lights.
One such round-about is made such that the 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
Question: 1
Represent the given equations \( C_1 \) and \( C_2 \) with the help of a diagram.
Show Hint
For any circle \( x^2 + y^2 = a^2 \), the x-intercepts are \( (\pm a, 0) \) and the y-intercepts are \( (0, \pm a) \).
Concentric circles never intersect because their radii are different while their center is the same.
Concept: • A circle with its center at the origin \( (0, 0) \) and radius \( r \) is represented by the equation \( x^2 + y^2 = r^2 \).
• Concentric circles are circles that share the same center but have different radii.
Step 1: Identify the center and radius for the boundary circle \( C_1 \)
The equation for \( C_1 \) is:
\[ x^2 + y^2 = 64 \]
Comparing this with the standard form \( x^2 + y^2 = r^2 \):
\[ r_1^2 = 64 \implies r_1 = 8 \text{ units} \]
The center is at the origin \( (0, 0) \).
Step 2: Identify the center and radius for the pond circle \( C_2 \)
The equation for \( C_2 \) is:
\[ x^2 + y^2 = 4 \]
Comparing this with the standard form:
\[ r_2^2 = 4 \implies r_2 = 2 \text{ units} \]
The center is also at the origin \( (0, 0) \).
Step 3: Draw the diagram showing concentric circles
Since both circles have the same center, they are concentric.
The pond \( C_2 \) is located inside the roundabout boundary \( C_1 \).
\includegraphics[width=0.5\linewidth]{Q36_Sol.png}
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Question: 2
Express \( y \) as a function of \( x \), (\( y = f(x) \)), for both \( C_1 \) and \( C_2 \).
Show Hint
Mathematically, a full circle is a relation and fails the vertical line test. We use the positive root to represent the upper boundary as a function.
The domain for \( C_1 \) is \( [-8, 8] \) and for \( C_2 \) is \( [-2, 2] \).
Concept: • To express \( y \) as a function of \( x \), we isolate the \( y \) variable on one side of the equation.
• A circle \( x^2 + y^2 = r^2 \) can be split into two functions: \( y = \sqrt{r^2 - x^2} \) (upper semicircle) and \( y = -\sqrt{r^2 - x^2} \) (lower semicircle).
Step 1: Derive the function for the boundary circle \( C_1 \)
The equation for \( C_1 \) is:
\[ x^2 + y^2 = 64 \]
Subtract \( x^2 \) from both sides:
\[ y^2 = 64 - x^2 \]
Taking the square root:
\[ y = \pm \sqrt{64 - x^2} \]
Considering the upper part of the circle to represent it as a function:
\[ f_1(x) = \sqrt{64 - x^2} \]
Step 2: Derive the function for the pond circle \( C_2 \)
The equation for \( C_2 \) is:
\[ x^2 + y^2 = 4 \]
Subtract \( x^2 \) from both sides:
\[ y^2 = 4 - x^2 \]
Taking the square root:
\[ y = \pm \sqrt{4 - x^2} \]
Considering the upper part:
\[ f_2(x) = \sqrt{4 - x^2} \]
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Question: 3
Using integration, find the area of the region covered by the roundabout.
Show Hint
Always verify integration results for circles using the geometric formula \( \pi r^2 \).
The factor 4 accounts for symmetry across the x and y axes.
• The total area of a circle \( x^2 + y^2 = r^2 \) is four times the area in the first quadrant.
• Area in the first quadrant = \( \int_{0}^{r} y \, dx \).
• Standard Integral: \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \).
Step 1: Set up the integral for the total area of circle \( C_1 \)
The radius is \( r = 8 \). The total area \( A_1 \) is given by:
\[ A_1 = 4 \int_{0}^{8} \sqrt{64 - x^2} \, dx \]
Step 2: Apply the integration formula
Using the standard formula with \( a = 8 \):
\[ A_1 = 4 \left[ \frac{x}{2}\sqrt{64 - x^2} + \frac{64}{2}\sin^{-1}\left(\frac{x}{8}\right) \right]_{0}^{8} \]
Step 3: Evaluate the definite integral with limits
Substituting the upper limit \( x = 8 \):
\[ \left( \frac{8}{2}\sqrt{64 - 64} + 32\sin^{-1}(1) \right) = 0 + 32 \cdot \frac{\pi}{2} = 16\pi \]
Substituting the lower limit \( x = 0 \):
\[ \left( \frac{0}{2}\sqrt{64 - 0} + 32\sin^{-1}(0) \right) = 0 + 0 = 0 \]
Step 4: Final multiplication
\[ A_1 = 4 \times (16\pi - 0) \]
\[ A_1 = 64\pi \text{ sq. units} \]
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Question: 4
Using integration, find the area of the region covered by the circular pond.
Show Hint
Integration is the standard tool for finding areas of regions with curved boundaries.
For a circle of radius \( r \), the integration result \( \int_{0}^{r} \sqrt{r^2 - x^2} \, dx \) always yields \( \frac{\pi r^2}{4} \).
Concept: • The pond is bounded by \( C_2 : x^2 + y^2 = 4 \).
• It is a circle with center at origin and radius \( r = 2 \).
• Area = \( 4 \times \int_{0}^{r} \sqrt{r^2 - x^2} \, dx \).
Step 1: Set up the integral for circle \( C_2 \)
With radius \( r = 2 \):
\[ A_2 = 4 \int_{0}^{2} \sqrt{4 - x^2} \, dx \]
Step 2: Apply the integration formula
Using the standard formula with \( a = 2 \):
\[ A_2 = 4 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2} \]
Step 3: Evaluate the definite integral
Substituting the upper limit \( x = 2 \):
\[ \left( \frac{2}{2}\sqrt{4 - 4} + 2\sin^{-1}(1) \right) = 0 + 2 \cdot \frac{\pi}{2} = \pi \]
Substituting the lower limit \( x = 0 \):
\[ \left( 0 + 2\sin^{-1}(0) \right) = 0 \]
Step 4: Final multiplication
\[ A_2 = 4 \times (\pi - 0) \]
\[ A_2 = 4\pi \text{ sq. units} \]
