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Find by Double Integration, whole area of the curve a2x2 = y3(2a – y). Let area be of the form λa2. Find value of λ.
Correct answer is '3.142'. Can you explain this answer?
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Find by Double Integration, whole area of the curvea2x2=y3(2a–y)...
As, it contains only even powers of x, hence, it is symmetrical about y-axis
Total Area = 2 × area OAB
Total Area = 2 × area OAB
The correct answer is: 3.142
Most Upvoted Answer
Find by Double Integration, whole area of the curvea2x2=y3(2a–y)...
To find the area under the curve, we need to integrate the equation with respect to x.
Given equation: a^2x^2 = y^3(2a)
First, we need to isolate y in terms of x:
y^3 = (a^2x^2)/(2a)
y = (a^2x^2)^(1/3)/(2^(1/3)a^(1/3))
y = (a^(2/3)x^2/2^(1/3)a^(1/3))
Now, we can find the limits of integration by setting y = 0:
0 = (a^(2/3)x^2)/(2^(1/3)a^(1/3))
0 = a^(1/3)x^2/(2^(1/3))
x^2 = 0
So, the limits of integration for x are from -√0 to √0, which is just x = 0.
Now, we can integrate the equation with respect to x to find the area:
A = ∫[0 to 0] (a^(2/3)x^2/2^(1/3)a^(1/3)) dx
Since the limits of integration are the same, the area under the curve is 0.
Given equation: a^2x^2 = y^3(2a)
First, we need to isolate y in terms of x:
y^3 = (a^2x^2)/(2a)
y = (a^2x^2)^(1/3)/(2^(1/3)a^(1/3))
y = (a^(2/3)x^2/2^(1/3)a^(1/3))
Now, we can find the limits of integration by setting y = 0:
0 = (a^(2/3)x^2)/(2^(1/3)a^(1/3))
0 = a^(1/3)x^2/(2^(1/3))
x^2 = 0
So, the limits of integration for x are from -√0 to √0, which is just x = 0.
Now, we can integrate the equation with respect to x to find the area:
A = ∫[0 to 0] (a^(2/3)x^2/2^(1/3)a^(1/3)) dx
Since the limits of integration are the same, the area under the curve is 0.
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Find by Double Integration, whole area of the curvea2x2=y3(2a–y). Let area be of the formλa2.Find value ofλ.Correct answer is '3.142'. Can you explain this answer?
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