The area of the smaller region enclosed by the curves y2 = 8x + 4 and
x2+y2+4√3x-4=0
is equal to
The area of the smaller region enclosed by the curves y2 = 8x + 4 and
x2+y2+4√3x-4=0
is equal to
- \(\frac{1}{3}(2-12√3+8π)\)
- \(\frac{1}{3}(2-12√3+6π)\)
- \(\frac{1}{3}(4-12√3+8π)\)
- \(\frac{1}{3}(4-12√3+6π)\)
The Correct Option is C
Solution and Explanation
The correct answer is (C):
\(cosθ =\frac{ 2√3}{4}\)
= √3/4
⇒ θ = 30°
Area of the required region
=\(\frac{ 2}{3}(4×\frac{1}{2})+42×\frac{π}{6}-\frac{1}{2}×4×2√3\)
= \(\frac{4}{3}+\frac{8π}{3}-4√3\)
= \(\frac{1}{3}{4-12√3+8π}\)
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Concepts Used:
Area between Two Curves
Integral calculus is the method that can be used to calculate the area between two curves that fall in between two intersecting curves. Similarly, we can use integration to find the area under two curves where we know the equation of two curves and their intersection points. In the given image, we have two functions f(x) and g(x) where we need to find the area between these two curves given in the shaded portion.
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,
