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
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.
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