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The volume of the real formed by revolving part of the parabola y2 = 4ax cut off by the latus rectum about the tangent at the vertex is given by:
- a)1/5πa3
- b)2/5πa3
- c)4/5πa3
- d)2/3πa3
Correct answer is option 'C'. Can you explain this answer?
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The volume of the real formed by revolving part of the parabola y2 = 4...
V = 4/5πa3
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The volume of the real formed by revolving part of the parabola y2 = 4...
The equation of the parabola is y^2 = 4ax, where a is a constant.
The latus rectum of the parabola is a line segment perpendicular to the axis of symmetry and passing through the focus of the parabola.
The focus of the parabola is (a/4, 0), so the equation of the latus rectum is x = a/4.
Let's find the equation of the tangent at the vertex of the parabola. The vertex of the parabola is (0, 0), so the equation of the tangent at the vertex is y = 0.
To find the volume of the solid formed by revolving the part of the parabola cut off by the latus rectum about the tangent at the vertex, we can use the method of cylindrical shells.
Consider a small element of width Δx on the parabola, located at a distance x from the vertex. The height of this element is given by y = √(4ax).
When this element is revolved about the tangent at the vertex, it sweeps out a cylindrical shell with radius x and height √(4ax).
The volume of this cylindrical shell is given by V = 2πx * √(4ax) * Δx.
To find the total volume, we integrate this expression over the interval [0, a/4]:
V = ∫[0, a/4] 2πx * √(4ax) dx.
To evaluate this integral, we can use the substitution u = 4ax, du = 4a dx:
V = ∫[0, a] π/2 * √u du.
Using the power rule, we can evaluate this integral:
V = π/2 * (2/3) * u^(3/2) |[0, a]
= π/3 * (a)^(3/2).
Therefore, the volume of the solid formed by revolving part of the parabola y^2 = 4ax cut off by the latus rectum about the tangent at the vertex is given by V = π/3 * (a)^(3/2).
So the answer is not 1/5.
The latus rectum of the parabola is a line segment perpendicular to the axis of symmetry and passing through the focus of the parabola.
The focus of the parabola is (a/4, 0), so the equation of the latus rectum is x = a/4.
Let's find the equation of the tangent at the vertex of the parabola. The vertex of the parabola is (0, 0), so the equation of the tangent at the vertex is y = 0.
To find the volume of the solid formed by revolving the part of the parabola cut off by the latus rectum about the tangent at the vertex, we can use the method of cylindrical shells.
Consider a small element of width Δx on the parabola, located at a distance x from the vertex. The height of this element is given by y = √(4ax).
When this element is revolved about the tangent at the vertex, it sweeps out a cylindrical shell with radius x and height √(4ax).
The volume of this cylindrical shell is given by V = 2πx * √(4ax) * Δx.
To find the total volume, we integrate this expression over the interval [0, a/4]:
V = ∫[0, a/4] 2πx * √(4ax) dx.
To evaluate this integral, we can use the substitution u = 4ax, du = 4a dx:
V = ∫[0, a] π/2 * √u du.
Using the power rule, we can evaluate this integral:
V = π/2 * (2/3) * u^(3/2) |[0, a]
= π/3 * (a)^(3/2).
Therefore, the volume of the solid formed by revolving part of the parabola y^2 = 4ax cut off by the latus rectum about the tangent at the vertex is given by V = π/3 * (a)^(3/2).
So the answer is not 1/5.
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