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The area of the planar region bounded by the curves x = 6y2 – 2 and x = 2y2 is
- a)4√2/3
- b)2√2/3
- c)√2/3
- d)√2
Correct answer is option 'A'. Can you explain this answer?
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The area of the planar region bounded by the curves x = 6y2 – 2 ...
And x = 2 + 6y2 is given by:
First, we need to find the intersection points of the two curves:
6y2 = 2 + 6y2
0 = 2
This is a contradiction, so the two curves do not intersect.
Next, we need to find the bounds of the region. To do this, we can find the y-values where each curve is equal to zero:
x = 6y2
y = 0
x = 2 + 6y2
y = ±sqrt((x-2)/6)
Since the curves do not intersect, the region is split into two parts. We can integrate each separately:
For the part of the region where y is between 0 and sqrt((x-2)/6), we integrate x = 6y2 from y = 0 to y = sqrt((x-2)/6):
∫0^(sqrt((x-2)/6)) 6y2 dy = 2(x-2)
For the part of the region where y is between -sqrt((x-2)/6) and 0, we integrate x = 2 + 6y2 from y = -sqrt((x-2)/6) to y = 0:
∫(-sqrt((x-2)/6))^0 (2 + 6y2) dy = 2/3(x-2)
Putting these together, the area of the region is given by:
A = ∫2^(14/3) 2(x-2) dx + ∫14 2/3(x-2) dx
A = 56/3 + 28/3
A = 28
Therefore, the area of the planar region bounded by the curves x = 6y2 and x = 2 + 6y2 is 28 square units.
First, we need to find the intersection points of the two curves:
6y2 = 2 + 6y2
0 = 2
This is a contradiction, so the two curves do not intersect.
Next, we need to find the bounds of the region. To do this, we can find the y-values where each curve is equal to zero:
x = 6y2
y = 0
x = 2 + 6y2
y = ±sqrt((x-2)/6)
Since the curves do not intersect, the region is split into two parts. We can integrate each separately:
For the part of the region where y is between 0 and sqrt((x-2)/6), we integrate x = 6y2 from y = 0 to y = sqrt((x-2)/6):
∫0^(sqrt((x-2)/6)) 6y2 dy = 2(x-2)
For the part of the region where y is between -sqrt((x-2)/6) and 0, we integrate x = 2 + 6y2 from y = -sqrt((x-2)/6) to y = 0:
∫(-sqrt((x-2)/6))^0 (2 + 6y2) dy = 2/3(x-2)
Putting these together, the area of the region is given by:
A = ∫2^(14/3) 2(x-2) dx + ∫14 2/3(x-2) dx
A = 56/3 + 28/3
A = 28
Therefore, the area of the planar region bounded by the curves x = 6y2 and x = 2 + 6y2 is 28 square units.
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The area of the planar region bounded by the curves x = 6y2 – 2 ...
Hence correct option is A.
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The area of the planar region bounded by the curves x = 6y2 – 2 and x = 2y2 isa)4√2/3b)2√2/3c)√2/3d)√2Correct answer is option 'A'. Can you explain this answer?
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