Find the volume of the solid generated by revolving region bound by y=x^2, y=0, and x=2 about the x axis.
Volume = ∫ y^2 dx from 0 to 2
= π∫ x^4 dx from 0 to 2
easy to integrate, and evaluate.
to check enter "π∫ x^4 dx from 0 to 2" into
https://www.wolframalpha.com/
using discs of thickness dx,
v = ∫[0,2] πr^2 dx = ∫[0,2] πx^4 dx = 32π/5
using shells of thickness dy,
v = ∫[0,4] 2πrh dy = ∫[0,4] 2πy(2-√y) dy = 32π/5
Ah, the thrilling world of geometry! Let's get ready to revolve, my friend. To find the volume of the solid generated by revolving the region bound by y = x^2, y = 0, and x = 2 about the x-axis, we'll have to use the method of cylindrical shells. Are you ready for some mathematical excitement? Hold onto your pi's!
First, let's visualize the situation. Imagine taking that region between the curve x^2 and the x-axis from x = 0 to x = 2 and spinning it around the x-axis. We're creating a solid that's going to make heads spin... or something like that.
To find the volume, we'll integrate. Now, each cylindrical shell will have a radius equal to the x-coordinate (because we're revolving around the x-axis), a height equal to the y-coordinate (since we're considering the curve y = x^2), and a thickness of dx. Follow along, math-magician!
The formula for the volume of a cylinder is V = 2πrh, where r is the radius and h is the height. In our case, r = x (since we're revolving around the x-axis) and h = x^2 (because our region is defined by y = x^2). So, V = 2πx * x^2 = 2πx^3.
Now we'll integrate this expression from x = 0 to x = 2. Ready? Wait for it... integration time!
∫(0 to 2) of 2πx^3 dx
Integrating term by term, we get:
2π * [(x^4)/4] evaluated from 0 to 2
Plugging in the limits, we have:
2π * [(2^4)/4] - [(0^4)/4]
= 2π * (16/4) - 0
= 2π * 4
= 8π cubic units
Voilà! The volume of the solid generated by revolving the region bound by y = x^2, y = 0, and x = 2 about the x-axis is 8π cubic units. I hope you enjoyed this mathematical circus act!
To find the volume of the solid generated by revolving the region bound by the curves y = x^2, y = 0, and x = 2 about the x-axis, we can use the method of cylindrical shells.
Step 1: Determine the limits of integration
Since the region is bounded by y = 0 and y = x^2, we need to find the x-values where these curves intersect.
Setting y = 0, we find:
0 = x^2
x = 0
Setting y = x^2, we find:
x^2 = x^2
x = 0
So, the limits of integration are from x = 0 to x = 2.
Step 2: Set up the integral for the volume using cylindrical shells
The volume of a thin cylindrical shell is given by the formula V = 2πrhΔx, where h is the height of the shell and Δx is the width of the shell.
In this case, the radius r will be the distance from the axis of rotation (x-axis) to the function y = x^2, which is given by r = x^2. The height h will be the difference between the upper curve y = x^2 and the lower curve y = 0, which is h = x^2 - 0 = x^2.
The integral for the volume is then given by:
V = ∫[0 to 2] 2π(x^2)(x^2) dx
Step 3: Evaluate the integral
V = 2π ∫[0 to 2] x^4 dx
V = 2π [ (1/5)x^5 ] [0 to 2]
V = 2π ( (1/5)(2^5) - (1/5)(0^5) )
V = 2π ( (1/5)(32) - (1/5)(0) )
V = 2π ( 32/5 )
V = (64π/5) cubic units
Therefore, the volume of the solid generated by revolving the region about the x-axis is (64π/5) cubic units.
To find the volume of the solid generated by revolving the region bound by the curves y = x^2, y = 0, and x = 2 about the x-axis, we can use the method of cylindrical shells.
First, let's graph the region to get a better understanding of the problem. The region is the area between the curve y = x^2, the x-axis, and the vertical line x = 2.
Now, to find the volume, we'll divide the region into thin vertical strips and rotate each strip about the x-axis to create cylindrical shells. The height of each cylindrical shell will be the difference between the y-values of the curve y = x^2 and the x-axis, while the radius will be the x-coordinate of each strip.
To determine the limits of integration, we need to find the x-values where the curves intersect. The curve y = x^2 intersects the x-axis when y = 0, so we set x^2 = 0, which gives us x = 0. The curve also intersects the line x = 2, so our limits of integration for x will be from 0 to 2.
Now, let's set up the integral to find the volume. The volume of each cylindrical shell is given by:
V = ∫(2πrh)dx
where r is the x-coordinate of each strip, h is the height of each strip, and the integral is taken with respect to x.
In this case, the radius r is equal to the x-coordinate of each strip, which is just x. The height h is the difference between the y-values of the curve y = x^2 and the x-axis, which is x^2.
Therefore, we have:
V = ∫(2πx * x^2)dx
= 2π∫(x^3)dx, with limits from 0 to 2
Evaluating the integral:
V = 2π(1/4 * x^4) [0,2]
= 2π(1/4 * 2^4 - 1/4 * 0^4)
= π * (8 - 0)
= 8π
So the volume of the solid generated by revolving the region about the x-axis is 8π cubic units.