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The area bounded by two curves y2=4ax and x2=4ay is
- a)(32/3)a2 sq.unit
- b)(16/3)sq.unit
- c)(32/3)sq.unit
- d)(16/3)a2 sq.unit
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2sq.unitb)...
Explanation:
The area bounded by two curves can be found by integrating the difference of the two curves with respect to the variable of integration. In this case, we have two curves:
1. y^2 = 4ax
2. x^2 = 4ay
Finding the points of intersection:
To find the points of intersection of the two curves, we can substitute one equation into the other to eliminate one variable. By substituting x^2 = 4ay into y^2 = 4ax, we get:
(4ay)^2 = 4ax
16a^2y^2 = 4ax
4ay = x
y = x/4a
Substitute this back into x^2 = 4ay:
x^2 = 4a(x/4a)
x^2 = x
x = 0 or x = 1
Therefore, the points of intersection are (0, 0) and (1, 1/4a).
Calculating the area:
The area bounded by the two curves can be found by integrating the difference of the two curves with respect to y from 0 to 1/4a (the limits of integration determined by the points of intersection):
Area = ∫[0,1/4a] (4ax - x^2) dy
Area = ∫[0,1/4a] (4ay^2 - y^2) dy
Area = 32a/3
Therefore, the correct answer is option (A) (32/3)a^2 sq. unit.
The area bounded by two curves can be found by integrating the difference of the two curves with respect to the variable of integration. In this case, we have two curves:
1. y^2 = 4ax
2. x^2 = 4ay
Finding the points of intersection:
To find the points of intersection of the two curves, we can substitute one equation into the other to eliminate one variable. By substituting x^2 = 4ay into y^2 = 4ax, we get:
(4ay)^2 = 4ax
16a^2y^2 = 4ax
4ay = x
y = x/4a
Substitute this back into x^2 = 4ay:
x^2 = 4a(x/4a)
x^2 = x
x = 0 or x = 1
Therefore, the points of intersection are (0, 0) and (1, 1/4a).
Calculating the area:
The area bounded by the two curves can be found by integrating the difference of the two curves with respect to y from 0 to 1/4a (the limits of integration determined by the points of intersection):
Area = ∫[0,1/4a] (4ax - x^2) dy
Area = ∫[0,1/4a] (4ay^2 - y^2) dy
Area = 32a/3
Therefore, the correct answer is option (A) (32/3)a^2 sq. unit.
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The area bounded by two curves y2=4ax and x2=4ay isa)(32/3)a2sq.unitb)(16/3)sq.unitc)(32/3)sq.unitd)(16/3)a2sq.unitCorrect answer is option 'A'. Can you explain this answer?
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