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
use discs, stacked up around the y-axis, from y=0 to 9
V = Int(pi x^2 dy)[0,9]
= pi*Int(y dy)[0,9]
= pi * (1/2 y^2)[0,9]
= 81pi/2
You could also use shells of thickness dx:
V = Int(2pi * r * h dx)[0,3]
= 2pi Int(x * x^2 dx)[0,3]
= 2pi Int(x^3 dx)[0,3]
= 2pi x^4/4 [0,3]
= 2pi * 81/4
= 81pi/2
Actually, the height of the shells is (9-y), rather than y, but as it turns out, the final answer is the same!
To find 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, we can use the method of cylindrical shells.
The equation for the volume of a solid obtained by rotating a curve about the y-axis using cylindrical shells is given by:
V = 2π ∫[a,b] x f(x) dx
where [a,b] is the interval over which the region is bounded, and f(x) represents the vertical distance from the curve to the axis of rotation (in this case, the y-axis).
In this case, the interval [a,b] is [0,3] as given by the boundary curves x = 0 and x = 3.
The equation for the volume of the solid then becomes:
V = 2π ∫[0,3] x (x^2) dx
Using the power rule of integration, we can integrate the function x^3 with respect to x:
V = 2π * [(1/4) * x^4] evaluated from 0 to 3
V = 2π * [(1/4) * 3^4 - (1/4) * 0^4]
V = 2π * [(1/4) * 81]
V = 2π * (81/4)
V = 81π/2
Therefore, the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis is 81π/2 cubic units.
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 visualize the region bounded by the curves y = x^2, y = 0, x = 0, and x = 3. The region is a bounded area between the x-axis and the parabola y = x^2, from x = 0 to x = 3.
To calculate the volume, we need to integrate the area of each cylindrical shell formed when rotating the region about the y-axis.
At any given y-coordinate, the corresponding x-values for the parabola are:
x = sqrt(y) (taking the positive square root since we're only considering the region above the x-axis)
Therefore, the radius of the cylindrical shell at height y will be:
r = sqrt(y)
The height of the cylindrical shell will be the difference in x-values between x = 3 and x = sqrt(y):
h = 3 - sqrt(y)
The area of each cylindrical shell is given by:
dV = 2πrh * dy
To find the total volume, we need to integrate this expression with respect to y over the range of y-values that bounds the region.
∫[0,1] 2π(sqrt(y))(3 - sqrt(y)) dy
Integrating this expression will give us the volume of the solid obtained by rotating the region about the y-axis.
Let's evaluate the integral:
V = ∫[0,1] 2π(sqrt(y))(3 - sqrt(y)) dy
Integrating this expression will give us the volume of the solid obtained by rotating the region about the y-axis. Evaluating the integral will yield the final answer.