Find the volume if the region enclosing y=2x, y=0, x=3 is rotated a)about the y-axis b)about the line x=3 c)about the line x=4.
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To find the volume of the region enclosed by the curves and rotated around different axes, we can use the method of cylindrical shells.
a) Rotating about the y-axis:
In this case, we need to express the curves in terms of x. Since the region is bounded by y = 2x, y = 0, and x = 3, we can express the curves as x = y/2, x = 0, and x = 3.
To find the volume, we integrate the formula for the surface area of a cylindrical shell:
V = ∫[a, b] 2πx * h(x) * dx
where a and b are the x-values of the boundaries of the region, h(x) represents the height of the shell, and dx is an infinitesimally small width of the shell.
The height, h(x), can be determined by subtracting the lower function from the upper function. In this case, h(x) = (2x - 0) = 2x.
The boundaries of the region are x = 0 and x = 3.
Therefore, the volume (V) is given by:
V = ∫[0, 3] 2πx * 2x * dx
= 4π ∫[0, 3] x^2 dx.
Using integration, we have:
V = 4π * [x^3/3] evaluated from 0 to 3
= 4π * [(3^3/3) - (0^3/3)]
= 4π * 9.
So, the volume of the region when rotated about the y-axis is 36π units^3.
b) Rotating about the line x = 3:
In this case, we express the curves in terms of y since the rotation is happening around a vertical line.
To find the volume, we integrate the formula for the surface area of a cylindrical shell again:
V = ∫[c, d] 2πy * h(y) * dy
where c and d are the y-values of the boundaries of the region, h(y) represents the height of the shell, and dy is an infinitesimally small height of the shell.
The height, h(y), can be determined by subtracting the right-side function (x = 3) from the left-side function (y = 2x). In this case, h(y) = (2(3) - y/2) = 6 - y/2.
The boundaries of the region are y = 0 and y = 6.
Therefore, the volume (V) is given by:
V = ∫[0, 6] 2πy * (6 - y/2) * dy
= 12π ∫[0, 6] (y - y^2/2) dy.
Using integration, we have:
V = 12π * [(y^2/2) - (y^3/6)] evaluated from 0 to 6
= 12π * [(6^2/2) - (6^3/6) - (0^2/2) + (0^3/6)]
= 12π * (18 - 72/6)
= 12π * (18 - 12)
= 12π * 6.
So, the volume of the region when rotated about the line x = 3 is 72π units^3.
c) Rotating about the line x = 4:
In this case, we again express the curves in terms of y since the rotation is happening around a vertical line.
To find the volume, we use the same formula for the surface area of a cylindrical shell:
V = ∫[e, f] 2πy * h(y) * dy
where e and f are the y-values of the boundaries of the region, h(y) represents the height of the shell, and dy is an infinitesimally small height of the shell.
The height, h(y), can be determined by subtracting the left-side function (y = 2x) from the right-side function (x = 4). In this case, h(y) = (4 - y/2) = 4 - y/2.
The boundaries of the region are y = 0 and y = 6.
Therefore, the volume (V) is given by:
V = ∫[0, 6] 2πy * (4 - y/2) dy
= 4π ∫[0, 6] (2y - y^2/2) dy.
Using integration, we have:
V = 4π * [(y^2 - y^3/6) evaluated from 0 to 6
= 4π * [(6^2 - 6^3/6) - (0^2 - 0^3/6)]
= 4π * (36 - 72/6)
= 4π * (36 - 12)
= 4π * 24.
So, the volume of the region when rotated about the line x = 4 is 96π units^3.
To find the volume of the region enclosed by the curves, we can use the method of cylindrical shells. Here's how you can find the volume for each case:
a) About the y-axis:
In this case, the region is rotated around the y-axis and the axis of rotation is vertical. To calculate the volume, we integrate the circumference of the shell multiplied by the height along the y-axis.
The height of the cylindrical shell is given by the difference between the two functions, y = 2x and y = 0, so the height is 2x.
Since the curves intersect at (3, 6), we need to find the limits of integration. To do this, we set the two functions equal to each other: 2x = 0. Solving for x, we get x = 0.
So the limits of integration for x are 0 to 3.
The circumference of the shell is given by 2πy, so in this case, it becomes 2πx.
The integral for the volume is then:
V = ∫(2πx)(2x) dx from x = 0 to x = 3.
Simplifying the integral, we get:
V = 4π∫x^2 dx from 0 to 3.
Evaluating the integral, we get:
V = 4π [(1/3)x^3] from 0 to 3.
Substituting the limits of integration, we have:
V = 4π [(1/3)(3)^3 - (1/3)(0)^3]
Simplifying further, we get:
V = 36π
Therefore, the volume of the region when rotated about the y-axis is 36π.
b) About the line x=3:
In this case, the region is rotated about the vertical line x = 3.
Since the axis of rotation is vertical, we can use the method of cylindrical shells to find the volume.
The height of the cylindrical shell is given by the difference between the two functions, y = 2x and y = 0. So the height is 2x.
The limits of integration for x remain the same as in part a, which are 0 to 3.
The distance between the axis of rotation and the curve is given by 3 - x, since the axis is at x = 3.
So the circumference of the shell becomes 2π(3 - x).
The integral for the volume is then:
V = ∫(2π(3 - x))(2x) dx from x = 0 to x = 3.
Simplifying the integral, we get:
V = 4π ∫(3x - x^2) dx from 0 to 3.
Evaluating the integral, we get:
V = 4π [(3/2)x^2 - (1/3)x^3] from 0 to 3.
Substituting the limits of integration, we have:
V = 4π [(3/2)(3)^2 - (1/3)(3)^3]
Simplifying further, we get:
V = 63π/2
Therefore, the volume of the region when rotated about the line x=3 is 63π/2.
c) About the line x=4:
In this case, the region is rotated about the vertical line x = 4.
Similar to part b, we use the method of cylindrical shells to find the volume.
The height of the cylindrical shell is still given by the difference between the two functions, y = 2x and y = 0. So the height is 2x.
The limits of integration for x remain the same, which are 0 to 3.
The distance between the axis of rotation and the curve is given by 4 - x, since the axis is at x = 4.
So the circumference of the shell becomes 2π(4 - x).
The integral for the volume is then:
V = ∫(2π(4 - x))(2x) dx from x = 0 to x = 3.
Simplifying the integral, we get:
V = 4π ∫(8x - 2x^2) dx from 0 to 3.
Evaluating the integral, we get:
V = 4π [4x^2 - (2/3)x^3] from 0 to 3.
Substituting the limits of integration, we have:
V = 4π [(4)(3)^2 - (2/3)(3)^3]
Simplifying further, we get:
V = 108π - 54π/3
Therefore, the volume of the region when rotated about the line x = 4 is 54π/3.
Note: Make sure to simplify the final answers or use an appropriate calculator to get the exact numerical value.