Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=x^2, y = 0, x = 0, x = 3, about the y-axis
To find the volume of the solid obtained by rotating the region bounded by the curves about the y-axis, we can use the method of cylindrical shells.
First, let's set up the integral. The volume can be calculated by integrating the circumference of each cylindrical shell multiplied by its height, over the interval of y-values that the region spans.
The region bounded by the curves is defined by y = x^2, y = 0, x = 0, and x = 3.
To rotate this region about the y-axis, we imagine vertical lines being rotated to form the shells.
The height of each shell will be the difference between the y-values of the upper and lower curves, which is y = x^2 - 0 = x^2.
The radius of each shell will be the x-value of the curve that is being rotated, which is x.
The circumference of each shell is given by 2πr, where r is the radius.
Therefore, the equation for the volume will be:
V = ∫2πr * h dy
where r = x and h = x^2.
Since the region spans from y = 0 to y = 9 (obtained by substituting y = x^2 into the equation y = 9), the integral will be evaluated within this range:
V = ∫[0,9] 2πx * x^2 dy
Now, let's perform the integration:
V = 2π ∫[0,9] x^3 dy
V = 2π ∫[0,9] x^3 dy
V = 2π ∫[0,9] x^3 dy
Using the power rule for integration, we can integrate x^3 as (1/4)x^4:
V = 2π * (1/4) ∫[0,9] x^4 dy
V = π/2 ∫[0,9] x^4 dy
Evaluating the integral, we get:
V = π/2 * [(1/5)x^5] evaluated from 0 to 9
V = π/2 * [(1/5)(9^5 - 0^5)]
V = π/2 * (59049/5)
V = 59049π/10
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x^2, y = 0, x = 0, and x = 3, about the y-axis is 59049π/10 cubic units.