Find the volume of the solid obtained by rotating the region bounded by y= x^2 and x=y^2 about the x-axis.
The bounded region is x =0 to x = 1, so those are your limits of integration.
The integral is of
pi*(x^2 - sqrt x)^2 dx.
Visualize it as a stack of washers with holes in the middle.
I think that
the volume=pi*Int(from 0 to 1) (sqrt(x))^2dx-pi*Int(0 to 1)(x^2)^2dx=
pi(1/2-1/5)
Mgraph is correct. I should have written
Integral of pi[(sqrtx)^2 - (x^2)^2] dx
0 to 1
which is pi*(1/2 - 1/5)
Thank you so much that makes alot of since.
To find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the x-axis, we can use the method of cylindrical shells.
First, let's find the points of intersection of the two curves by setting them equal to each other:
y = x^2 (equation 1)
x = y^2 (equation 2)
Setting equation 1 equal to equation 2, we have:
x^2 = x
Rearranging, we get:
x^2 - x = 0
Factoring out an x, we get:
x(x - 1) = 0
So x = 0 or x = 1. These are the x-coordinates of the intersection points.
Next, we need to determine the limits of integration for our cylindrical shells. We can do this by finding the y-coordinate of the intersection points. Plugging the x-values we obtained earlier into either equation 1 or 2, we get:
For x = 0: y = (0)^2 = 0
For x = 1: y = (1)^2 = 1
Therefore, the limits of integration for our cylindrical shells will be from y = 0 to y = 1.
Now, let's set up the integral to calculate the volume using the formula for the volume of a cylindrical shell:
V = ∫ 2πxy dx,
where x represents the radius of the shell and y represents the height of the shell.
Since we are rotating the region about the x-axis, the radius of the shell is x, and the height is the difference in y-coordinates of the two curves at each x-value.
Therefore, the integral becomes:
V = ∫[0 to 1] 2πxy dx.
Integrating the function 2πxy with respect to x, we get:
V = 2π ∫[0 to 1] xy dx.
To simplify the integral, we can pull out the constant π:
V = 2π ∫[0 to 1] x(dx ∫[0 to 1] y dy).
The inner integral, with respect to y, evaluated from y = 0 to y = 1, can be replaced by x^2 - y^2:
V = 2π ∫[0 to 1] x(x^2 - y^2) dx.
Now, we can expand the integral and solve for the volume.