I will assume that the x-axis was the other boundary of the region we are rotating
Using discs ...
radius = y = x^3
volume = π∫ y^2 dx from 0 to 2
= π∫ x^6 dx from 0 to 2
= π[ (1/7)x^7 ] from 0 to 2
= π((1/7)(2^7) - 0 ]
= 128Ï€/7
Using discs ...
radius = y = x^3
volume = π∫ y^2 dx from 0 to 2
= π∫ x^6 dx from 0 to 2
= π[ (1/7)x^7 ] from 0 to 2
= π((1/7)(2^7) - 0 ]
= 128Ï€/7
Graph a sketch of it first. The corners are at (0,0) , (2,0) and (2,8)
Lets add up a bunch of thin cylinders with axes along x axis
volume of thin cylinder = pi y^2 dx
so
in from x = 0 to x = 2 of pi (x^3)^2 dx
or pi x^6 dx
or pi (1/7)x^7
or 1/7 (128)
about 18.3
Step 1: Determine the limits of integration.
Since the given curve intersects the line x = 2, the limits of integration will be from 0 to 2.
Step 2: Set up the integral.
The volume of a cylindrical shell can be calculated using the formula:
dV = 2Ï€rh * dx
Where "r" is the radius of the shell, "h" is the height of the shell, and "dx" represents an infinitesimally small width in the x-direction.
To calculate the radius (r) and height (h) of each shell, we need to express them in terms of x.
The radius (r) can be determined as the distance between the x-axis and the curve y = x^3. Therefore, r = x^3.
The height (h) of each shell can be considered as the length of the curve y = x^3 for a given x-coordinate. Hence, h = y = x^3.
Step 3: Set up the integral and evaluate the volume.
The volume (V) can be found by integrating the expression for dV over the interval [0, 2]:
V = ∫[0, 2] 2πrh * dx
Since r = x^3 and h = x^3, the integral expression becomes:
V = ∫[0, 2] 2π(x^3)(x^3) * dx
V = 2π ∫[0, 2] x^6 * dx
Integrate x^6 with respect to x:
V = 2Ï€ * [x^7/7] evaluated from 0 to 2
V = 2Ï€ * [(2^7/7) - (0^7/7)]
V = 2Ï€ * (128/7)
Thus, the volume of the solid obtained by rotating the area bounded by the curve y = x^3 and the line x = 2 around the x-axis is (256Ï€/7) cubic units.