The area of the region bounded by y2 = 8x and y2 = 16(3 – x) is equal to
\(\frac{32}{3}\)
\(\frac{40}{3}\)
- 16
- 19
The Correct Option is C
Solution and Explanation
The correct answer is (C) : 16
c1 : y² = 8x
c2 : y² = 16(3 - x)
Solving c1 and c2
48 – 16x = 8x
x = 2
∴ y = ± 4
∴ Area of shaded region
= \(2\) \(\int_{0}^{4} \left\{ \left( 48 - \frac{y^2}{16} \right) - \left( \frac{y^2}{8} \right) \right\} \,dy\)
= \(\frac{1}{8}\) \([48y - y^3]_{0}^{4} = 16\)
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Concepts Used:
Area between Two Curves
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Area Between Two Curves With Respect to Y is
If f(y) and g(y) are continuous on [c, d] and g(y) < f(y) for all y in [c, d], then,
