Class 12 Applications Of Integrals · Case Study4 Markscase_based_area
Class 12 Maths · Applications Of Integrals · Case Study #6
An external elevator has an upper-semicircular floor. Its curved boundary is the upper half of x²+y²=4 and its straight boundary is y=0. Use the complete floor width when taking horizontal strips.
Part (1) 1 Marks
Find both intersections of the curved and straight boundaries.
Answer: (−2,0) and (2,0)
- Set y=0 in x²+y²=4, giving x²=4.
- Thus the intersections are (−2,0) and (2,0).
Part (2) 1 Marks
Find the length of a vertical strip at coordinate x.
Answer: √(4−x²), −2≤x≤2
- The upper boundary is y=√(4−x²) and the lower boundary is y=0.
- The vertical length is √(4−x²), for −2≤x≤2.
Part (3) 2 Marks
Find the differential area of (a) a full vertical strip and (b) a full horizontal strip of the floor.
Answer: (a) dA=√(4−x²)dx; (b) dA=2√(4−y²)dy.
- For a vertical strip, height is √(4−x²) and thickness is dx.
- For a horizontal strip at 0≤y≤2, endpoints are x=±√(4−y²), so its full width is 2√(4−y²).
- Therefore dA=√(4−x²)dx vertically, or dA=2√(4−y²)dy horizontally.
Part (4) 2 Marks
Alternatively, find the entire floor area using integration.
Answer: 2π square units
- A=∫₋₂²√(4−x²)dx. Use the primitive F=x√(4−x²)/2+2arcsin(x/2).
- At x=2, F=π; at x=−2, F=−π. Subtract to obtain 2π.
- The floor area is 2π square units.