Find the volume of the solid generated by revolving the region bounded by y=x+(x/4), the x-axis, and the lines x=1 and x=3 about the y-axis.
I've drawn the graph and I understand which part is being rotated, but I'm having trouble setting up the equation. Because there is a hole between the y-axis and the end of the region of the graph being rotated, I know I have to use ((R^2)-(r^2)), but I don't know which equations to plug in for R and r. Also, because it's being revolved around the y-axis, am I supposed to use 2π or just π? Also, I have the integral from 5/4 to 15/4 but I'm also not sure that this is right.
y=x+(x/4) ? Why not just 5/4 x?
Once you get your curves figured out,
r = the distance to the inside curve
R = the distance to the outside curve
You need to solve for x, so you know r.
Clearly, R=3, since that's the far boundary.
The limits of integration (along dy) are the intersections of the curve with y(1) and y(3)
The volume of a disc πr^2 dy.
If using shells, then you have 2πrh dx
To find the volume of the solid generated by revolving the region around the y-axis, we can use the shell method, which involves integrating the volume of cylindrical shells.
To set up the equation, we need to determine the outer radius (R) and the inner radius (r) of each cylindrical shell.
First, let's find the outer radius (R). In this case, the outer radius will be the distance from the y-axis to the curve. To find the equation for the curve, we need to solve y = x + (x/4) for x.
y = x + (x/4)
4y = 4x + x
4y = 5x
x = (4y)/5
So, the equation for the curve is x = (4y)/5.
Now, let's find the inner radius (r). The inner radius will be the distance from the y-axis to the x-axis. Since we are revolving the region bounded by y = x + (x/4) and the x-axis, the inner radius will be zero.
In the shell method, the volume of each cylindrical shell is given by the equation V = 2πrhΔx, where r is the radius, h is the height of the shell, and Δx is the width of each shell.
Since we're revolving around the y-axis, we use 2π as the coefficient of the volume equation.
To determine the limits of integration, we need to find the y-values at which the region is bounded. From the given information, the region is bounded by the lines x = 1 and x = 3. Let's substitute these limits into the equation for the curve (x = (4y)/5) to find the corresponding y-values:
x = (4y)/5
When x = 1:
1 = (4y)/5
5 = 4y
y = 5/4
When x = 3:
3 = (4y)/5
15 = 4y
y = 15/4
Therefore, the limits of integration for y are from 5/4 to 15/4.
Now, we can set up the integral to find the volume:
V = ∫ (from 5/4 to 15/4) 2πrh dy
Since the height (h) is the difference between the upper and lower boundaries of y, we have:
V = ∫ (from 5/4 to 15/4) 2π[(4y)/5 - 0] dy
Simplifying:
V = ∫ (from 5/4 to 15/4) (8πy)/5 dy
After integrating, we get:
V = (8π/5) * [y^2/2] (from 5/4 to 15/4)
Finally, substituting the limits and simplifying:
V = (8π/5) * [(15/4)^2/2 - (5/4)^2/2]
= (8π/5) * [(225/16)/2 - (25/16)/2]
= (8π/5) * [(225 - 25)/32]
= (8π/5) * (200/32)
= (8π/5) * (25/4)
= 40π
Therefore, the volume of the solid generated by revolving the region bounded by y = x + (x/4), the x-axis, and the lines x = 1 and x = 3 about the y-axis is 40π cubic units.