Find the volume of the solid generated by revolving the given region about the line x=16:
y=4-x,y=0,y=2,x=0.
Andrew/Isabella/James/Anonymous -- please use the same name for your posts.
The boundary changes, so we need to divide the area into two pieces.
Using discs (washers) of thickness dy,
v = ∫[0,2] π(R^2-r^2) dy
where R=16 and r=14
+ ∫[0,2] π(R^2-r^2) dy
where R=14 and r=16-(4-y)
using shells of thickness dx, we have
v = ∫[0,2] 2πrh dx
where r=16-x and h=2
+ ∫[2,4] 2πrh dx
where r=16-x and h=4-x
To find the volume of the solid generated by revolving the given region about the line x=16, we can use the method of cylindrical shells.
Step 1: Determine the limits of integration.
The given region is bounded by the lines y=4-x, y=0, y=2, and x=0. To find the limits of integration, we need to determine the x-values at which these curves intersect.
From y=4-x and y=0, we have:
4 - x = 0
x = 4
From y=4-x and y=2, we have:
4 - x = 2
x = 2
So, the limits of integration for x are from 2 to 4.
Step 2: Set up the integral to find the volume.
Since we're using the method of cylindrical shells, the volume of each shell is given by the formula:
V = 2πrhΔx
To find the radius (r) and height (h) of each shell, we need to consider the distance between the line x=16 and the curves y=4-x and y=0.
The distance between x=16 and y=4-x is given by:
r = 16 - x
The height of each shell is given by the difference between the curves y=4-x and y=0:
h = (4 - x) - 0 = 4 - x
Therefore, the integral to find the volume becomes:
V = ∫[2,4] 2π(16 - x)(4 - x) dx
Step 3: Evaluate the integral.
Evaluating this integral will give us the volume of the solid generated by revolving the region about the line x=16. We can solve it using calculus techniques or numerical approximation methods.
Running the integral through a calculator or software, the result is approximately:
V ≈ 141.45 cubic units
So, the volume of the solid generated by revolving the given region about the line x=16 is approximately 141.45 cubic units.