find the volume of the solid generated by revolving the area by the given curves about the indicated axis of revolution y^2=4ax,x=a;about the y-axis
To find the volume of the solid generated by revolving the area bounded by the curves y^2 = 4ax and x = a about the y-axis, we can use the method of cylindrical shells.
The equation y^2 = 4ax represents a parabola opening to the right. We know that the vertex of the parabola is at the origin (0, 0) and its axis of symmetry is the y-axis.
Let's break down the steps to find the volume of the solid:
Step 1: Determine the limits of integration.
Since the parabola y^2 = 4ax intersects the line x = a at the point (a, 2a), the limits of integration for the variable y will be from -2a to 2a.
Step 2: Set up the integral.
The volume of the solid can be calculated using the following integral:
V = ∫[a,-a] 2πxy dy
Since x = a, we can simplify the integral:
V = 2πa ∫[-2a, 2a] y dy
Step 3: Evaluate the integral.
Integrating 2πa with respect to y leaves only y:
V = 2πa ∫[-2a, 2a] y dy
V = 2πa [(1/2)y^2] from -2a to 2a
V = 2πa [(1/2)(4a^2) - (1/2)(4a^2)]
V = 2πa (2a^2 - 2a^2)
V = 2πa(0)
V = 0
Step 4: Interpret the result.
The volume of the solid generated by revolving the area bounded by the curves y^2 = 4ax and x = a about the y-axis is zero. This suggests that the resulting solid is either empty or degenerate, having no volume.
Therefore, the volume of the solid is 0.
To find the volume of the solid generated by revolving the area between the curves y^2 = 4ax and x = a about the y-axis, we can use the method of cylindrical shells.
First, let's visualize the curves. The equation y^2 = 4ax represents a parabola with the vertex at the origin. The line x = a is a vertical line passing through the point (a,0).
To find the volume, we integrate the area of the cylindrical shells formed when revolving the curves around the y-axis.
The radius of each shell can be determined by calculating the distance between the y-axis and the curve y^2 = 4ax at a given y-coordinate. Since the curve is symmetrical, we only need to consider the positive portion of the curve.
For a given value of y, we have y^2 = 4ax. Solving for x, we get x = y^2 / (4a).
Since we are revolving around the y-axis, the height of each shell is the differential element dy, representing an infinitesimally small change in y.
The volume of each cylindrical shell is given by the product of the height, circumference, and thickness.
The circumference of each shell is given by 2π(radius), which simplifies to 2π(y^2 / (4a)).
Therefore, the volume dV of an infinitesimally small cylindrical shell located at y can be calculated as dV = 2π(y^2 / (4a)) dy.
To find the total volume V, we integrate dV from the lower limit of y to the upper limit of y, which are the values where the parabola and the line intersect.
The intersection points can be found by setting the two equations equal to each other: y^2 = 4ax and x = a.
Substituting x = a into the equation y^2 = 4ax, we get y^2 = 4a^2.
Taking the square root of both sides, we have y = 2a (since y cannot be negative for this particular curve).
So, the volume V can be calculated by integrating dV from y = 0 to y = 2a:
V = ∫(0 to 2a) [2π(y^2 / (4a))] dy
Simplifying the integral:
V = π/a ∫(0 to 2a) y^2 dy
Evaluating the integral:
V = π/a * [y^3 / 3] (0 to 2a)
V = π/a * [(2a)^3 / 3 - 0]
V = π/a * (8a^3 / 3)
V = (8πa^2)/3
Therefore, the volume of the solid generated by revolving the area between the curves y^2 = 4ax and x = a about the y-axis is (8πa^2)/3.
using shells,
v = ∫[0,a] 2πrh dx
where r = x and h=2y=4√(ax)
v = ∫[0,a] 2πx(4√(ax)) dx = 16/5 πa^3
using discs,
v = ∫[0,2a] π(R^2-r^2) dy
where R=a and r=x=y^2/(4a)
v = ∫[0,a] 2π(a^2-(y^2/(4a))^2) dy = 16/5 πa^3