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
using discs of thickness dx,
v = ∫[0,1] πr^2 dx
where r=y=3x^2
v = ∫[0,1] π(3x^2)^2 dx = 9π/5
using shells of thickness dy,
v = ∫[0,3] 2πrh dy
where r=y and h=1-x = 1-√(y/3)
v = ∫[0,3] 2πy(1-√(y/3)) dy = 9π/5
Vol = π ∫ y^2 dx from 0 to 1
= π ∫ 9x^4 dx from 0 to 1
= π[ (9/5)x^5 ] from 0 to 1
= π [(9/5)(1) - 0]
= 9/5π
To find 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, we can use the method of cylindrical shells.
First, let's sketch the region to get a better understanding of the problem. The graph of y = 3x^2 is a parabola that opens upwards and passes through the origin (0, 0). The line x = 1 is a vertical line passing through x = 1.
The region bounded by these curves looks like a triangle with its base along the x-axis and its height along the line x = 1.
To find the volume using cylindrical shells, we integrate the circumference of each shell multiplied by its height.
The height of each shell is given by the difference in y-values between the curve y = 3x^2 and the x-axis, which is y = 0.
The circumference of each shell is given by the formula 2πr, where r is the x-coordinate.
Integrating the circumference and height over the interval [0, 1] will give us the total volume.
Using the formula for the volume of cylindrical shells, we get:
V = ∫[0, 1] 2πx(3x^2 - 0) dx
Simplifying the equation, we have:
V = ∫[0, 1] 6πx^3 dx
Now, integrate the function 6πx^3 with respect to x:
V = 6π * [x^4/4] evaluated from 0 to 1
V = 6π * (1/4 - 0)
V = 6π/4
Simplifying further, we have:
V = 3π/2
Therefore, the volume of the solid obtained by rotating the region bounded by y = 3x^2, x = 1, and y = 0 about the x-axis is 3π/2 cubic units.