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The area bounded by the parabola x = 4 − y2 and the y-axis, in square units is
- a)3/32
- b)32/3
- c)33/2
- d)16/3
Correct answer is option 'B'. Can you explain this answer?
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The area bounded by the parabola x = 4 − y2 and the y-axis, in s...
Calculating the Area Bounded by the Parabola
To find the area bounded by the parabola x = 4 - y^2 and the y-axis, we need to determine the points of intersection between the parabola and the y-axis.
Finding Points of Intersection
Setting x = 0 in the equation of the parabola gives 0 = 4 - y^2. Solving for y, we get y = ±2. Therefore, the points of intersection are (0, 2) and (0, -2).
Setting Up the Integral
To find the area, we need to integrate the absolute value of the function x = 4 - y^2 with respect to y, from y = -2 to y = 2.
∫[from -2 to 2] |4 - y^2| dy
Integration
Breaking the integral into two parts, we have:
∫[from -2 to 0] (y^2 - 4) dy + ∫[from 0 to 2] (4 - y^2) dy
Integrating each part separately, we get:
[1/3 y^3 - 4y] (from -2 to 0) + [4y - 1/3 y^3] (from 0 to 2)
Plugging in the limits and simplifying, we get:
(8/3 + 8) + (8 - 8/3) = 32/3
Therefore, the area bounded by the parabola x = 4 - y^2 and the y-axis is 32/3 square units. Hence, option B is the correct answer.
To find the area bounded by the parabola x = 4 - y^2 and the y-axis, we need to determine the points of intersection between the parabola and the y-axis.
Finding Points of Intersection
Setting x = 0 in the equation of the parabola gives 0 = 4 - y^2. Solving for y, we get y = ±2. Therefore, the points of intersection are (0, 2) and (0, -2).
Setting Up the Integral
To find the area, we need to integrate the absolute value of the function x = 4 - y^2 with respect to y, from y = -2 to y = 2.
∫[from -2 to 2] |4 - y^2| dy
Integration
Breaking the integral into two parts, we have:
∫[from -2 to 0] (y^2 - 4) dy + ∫[from 0 to 2] (4 - y^2) dy
Integrating each part separately, we get:
[1/3 y^3 - 4y] (from -2 to 0) + [4y - 1/3 y^3] (from 0 to 2)
Plugging in the limits and simplifying, we get:
(8/3 + 8) + (8 - 8/3) = 32/3
Therefore, the area bounded by the parabola x = 4 - y^2 and the y-axis is 32/3 square units. Hence, option B is the correct answer.
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The area bounded by the parabola x = 4 − y2 and the y-axis, in square units isa)3/32b)32/3c)33/2d)16/3Correct answer is option 'B'. Can you explain this answer?
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