Class 12 Applications Of Integrals · Case Study4 Markscase_based_area
Class 12 Maths · Applications Of Integrals · Case Study #7
A simplified honeybee-nest profile is bounded by the parabola x²=4y and the horizontal branch line y=4. Use these mathematical boundaries for the calculations.
Part (1) 1 Marks
Find both intersections of the parabola and line.
Answer: (−4,4) and (4,4)
- Substitute y=4 into x²=4y to obtain x²=16.
- The intersections are (−4,4) and (4,4).
Part (2) 1 Marks
Find the differential area of a full vertical strip.
Answer: dA=(4−x²/4)dx, −4≤x≤4
- The top boundary is y=4 and the bottom is y=x²/4.
- Thus dA=(4−x²/4)dx for −4≤x≤4.
Part (3) 2 Marks
Find (a) the full horizontal strip length and (b) the vertical strip length.
Answer: (a) 4√y for 0≤y≤4; (b) 4−x²/4 for −4≤x≤4.
- At fixed y, x=±2√y, so the full horizontal length is 2√y−(−2√y)=4√y.
- At fixed x, vertical length is 4−x²/4. The intervals are 0≤y≤4 and −4≤x≤4, respectively.
- Therefore the lengths are (a) 4√y and (b) 4−x²/4.
Part (4) 2 Marks
Alternatively, calculate the entire bounded area.
Answer: 64/3 square units
- Use A=∫₋₄⁴(4−x²/4)dx, or equivalently ∫₀⁴4√y dy.
- The vertical primitive is 4x−x³/12; the endpoint difference is 2(16−16/3)=64/3.
- The area is 64/3 square units.