Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=3x^2, x=1, y=0 about the x-axis
volume = π∫(3x^2)^2 dx from x - 0 to x=1
= π∫9x^4 dx from 0 to 1
= π [ (9/5)x^5 ] from 0 to 1
= π(9/5 - 0)
= 9π/5
To find the volume of the solid obtained by rotating the region bounded by the curves about the x-axis, you can use the method of cylindrical shells.
First, let's sketch the region bounded by the curves y=3x^2, x=1, and y=0. This region is a parabolic shape that is symmetric about the y-axis and opens upward.
Next, we need to determine the limits of integration. Since we are rotating about the x-axis, the x-values will be the limits of integration. From the given curves, we know that x ranges from 0 to 1.
The volume of each cylindrical shell can be calculated as follows:
- The height of each shell will be the difference in y-values between the upper and lower curves at a given x-value. In this case, the upper curve is y=3x^2 and the lower curve is y=0. Therefore, the height of each shell is given by h=3x^2 - 0 = 3x^2.
- The radius of each shell is equal to the x-value at which it is formed. Therefore, the radius of each shell is r=x.
The volume of each shell is given by V=2πrh, where r is the radius and h is the height. Therefore, the volume of each cylindrical shell is V=2πx(3x^2).
To find the total volume, we need to integrate the volume of each shell from x=0 to x=1:
V_total = ∫[0,1] 2πx(3x^2) dx.
Integrating the function, we get:
V_total = 2π ∫[0,1] 3x^3 dx.
Evaluating the integral, we have:
V_total = 2π [3/4 * x^4] evaluated from 0 to 1.
Substituting the limits of integration, we get:
V_total = 2π [3/4 * 1^4 - 3/4 * 0^4].
Simplifying, we have:
V_total = 2π * 3/4.
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y=3x^2, x=1, and y=0 about the x-axis is V_total = (3/2)π cubic units.