Find the volume of the solid bounded by the curves y = 4 - x^2 and y = x revolved about the x axis.
intersect them first
x = 4-x^2
x^2 + x - 4 = 0
x = (-1 ± √32)/2
I am skeptical about your typing of the problem
The region you are rotating has part above and part below the x-axis.
This is not your usual kind of question of this type.
Just try to visualize the kind of solid you could get, I can't.
Yeah that's why I'm confused about this one, I saw that the area went above and below. So I'm guessing the equation was supposed to be y = (4-x)^2.?
The region R is bounded by the curves x=y2+2, y=x-4, and y=0
Write a single integral that gives the volume of the solid generated when R is revolved about the x-axis, as well as an intergral revolved around the y-axis. Do not evaluate the intergrals.
To find the volume of the solid formed by revolving the region between two curves about the x-axis, we can use the method of cylindrical shells.
First, let's graph the two given curves, y = 4 - x^2 and y = x:
To find the points of intersection between the two curves:
Set the two equations equal to each other:
4 - x^2 = x.
Rearrange the equation to form a quadratic equation:
x^2 + x - 4 = 0.
Now, solve the quadratic equation to find the x-values of the points of intersection.
Next, we need to determine which equation is above the other in order to correctly set up our integral.
To do this, we compare the y-values of the two equations at the points of intersection.
Substitute the x-values we found into both equations and compare the resulting y-values.
Finally, we can set up the integral to calculate the volume.
The volume can be found using the formula:
V = integral [from a to b] 2πx * (y1(x) - y2(x)) dx,
where a and b are the x-values of the points of intersection, and y1(x) and y2(x) are the functions representing the top and bottom boundaries of the solid, respectively.
Evaluate this integral to find the volume of the solid formed by revolving the given region about the x-axis.