Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=0, y=x(3-x) about the axis x = 0
volume = π[integral]y^2 dx from x = 0 to x = 3
y=x(3-x)
y^2 = x^2(3-x)^2
= x^2(9 - 6x + x^2)
= 9x^2 - 6x^3 + x^4
the integral of that is
3x^3 - (3/2)x^4 + (1/5)x^5
so volume
= π(81 - 243/2 + 243/5 - 0)
= (17/2)π
check my arithmetic
To find the volume of the solid obtained by rotating the region bounded by the curves about the specified axis, we can use the method of cylindrical shells.
First, let's plot the region bounded by the curve y = x(3-x) and the x-axis:
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To find the volume, we will integrate the radius of the cylindrical shells multiplied by the height of the shells, over the range of x-values that describes the region. The radius of each cylindrical shell can be determined by the distance between the curve y = x(3-x) and the axis of rotation x = 0, which is simply x. The height of each shell is given by the differential dx.
To set up the integral, we need to determine the limits of integration. The region bounded by y = x(3-x) and the x-axis is defined by the values of x where the curve intersects the x-axis. So we set y = 0:
0 = x(3-x)
This equation can be factored as follows:
0 = x(3-x)
0 = x(3) - x^2
0 = 3x - x^2
Solving for x, we find two solutions: x = 0 and x = 3.
Therefore, our limits of integration are x = 0 and x = 3.
Now, let's set up the integral:
V = ∫[a,b] 2πrh dx
where a = 0, b = 3, r = x, and h = y = x(3-x).
V = ∫[0,3] 2πx * x(3-x) dx
Simplifying the expression, we get:
V = ∫[0,3] 2πx^2(3-x) dx
Now, we can integrate this expression and evaluate the definite integral:
V = 2π * ∫[0,3] (3x^2 - x^3) dx
V = 2π * [x^3/3 - x^4/4] evaluated from 0 to 3
V = 2π * [(3^3/3 - 3^4/4) - (0^3/3 - 0^4/4)]
V = 2π * [(27/3 - 81/4)]
V = 2π * [(9 - 20.25)]
V = 2π * [-11.25]
V ≈ -70.69
Therefore, the volume of the solid obtained by rotating the region bounded by y = x(3-x) and the x-axis about the axis x = 0 is approximately -70.69 cubic units.