Question

Sketch the graph defined by \[ \left\{(x,y): \frac{x^2}{25}+\frac{y^2}{25}=1\right\}. \] Find the area of the region of the minor segment cut off by the line \[ x=\frac{5}{2}, \] using integration. 

Hint:

Always simplify the equation first. While the initial form looked like an ellipse, the equal denominators indicate a circle, which is much simpler to calculate.

Answer 1

Concept: 
• The equation \( \frac{x^2}{25} + \frac{y^2}{25} = 1 \) simplifies to \( x^2 + y^2 = 25 \), which is a circle with center \( (0,0) \) and radius \( 5 \). 
• The area of a segment is found by integrating the function \( y = f(x) \) between specified limits. 
• Use the standard integral formula: \( \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) + C \). 

Step 1: Sketch and identify the region
The graph is a circle with radius 5. The line \( x = 5/2 \) is a vertical line.
The region is the area between the line \( x = 5/2 \) and the right-hand part of the circle (where \( x = 5 \)).
Due to symmetry about the x-axis, the total area is twice the area in the first quadrant. 

Step 2: Set up the integral
Area \( A = 2 \int_{5/2}^{5} y \, dx = 2 \int_{5/2}^{5} \sqrt{25 - x^2} \, dx \).
Apply the integration formula:
\[ A = 2 \left[ \frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{5/2}^{5} \]
\[ A = \left[ x\sqrt{25 - x^2} + 25\sin^{-1}\left(\frac{x}{5}\right) \right]_{5/2}^{5} \] 

Step 3: Evaluate the definite integral
Upper limit (\( x = 5 \)): \( 5\sqrt{0} + 25\sin^{-1}(1) = 25(\frac{\pi}{2}) = \frac{25\pi}{2} \).
Lower limit (\( x = 5/2 \)): \( \frac{5}{2}\sqrt{25 - \frac{25}{4}} + 25\sin^{-1}(\frac{1}{2}) = \frac{5}{2}\left(\frac{5\sqrt{3}}{2}\right) + 25(\frac{\pi}{6}) = \frac{25\sqrt{3}}{4} + \frac{25\pi}{6} \).
Total Area = \( \frac{25\pi}{2} - \left( \frac{25\sqrt{3}}{4} + \frac{25\pi}{6} \right) = \frac{75\pi - 25\pi}{6} - \frac{25\sqrt{3}}{4} \)
Area = \( \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} \) sq. units.