Calc II - Disk vs Shell method different volumes

[MonkeyDLuffy]

So I'm getting ready for an exam on tuesday, and I'm using each method for volumes of revolutions for every problem but I'm not getting the same answers. So, let's use this as an example:

y = 5x; the shaded region is from [1,2]

Using the disk method (about the x-axis) I find:

R(x) = 5x; r(x) = x

V = π ∫ 25x² dx from [1,2] = 175π / 3

Using the shell method (about the x-axis) I find:

r(y) = y/5; h(y) = y

V = 2π ∫ (y² / 5) dy from [5,10] = 350π / 3

I'd like to know why the shell method gave me a volume that is twice that of the one I found using the disk method.

 

[Mark44]

So I'm getting ready for an exam on tuesday, and I'm using each method for volumes of revolutions for every problem but I'm not getting the same answers. So, let's use this as an example:

y = 5x; the shaded region is from [1,2]

Using the disk method (about the x-axis) I find:

R(x) = 5x; r(x) = x

V = π ∫ 25x² dx from [1,2] = 175π / 3

Using the shell method (about the x-axis) I find:

r(y) = y/5; h(y) = y

V = 2π ∫ (y² / 5) dy from [5,10] = 350π / 3

I'd like to know why the shell method gave me a volume that is twice that of the one I found using the disk method.

Using shells, you need to have two integrals, because the shapes of your shells are changing.

For $0 \le y \le 5$ the shell widths extend uniformly from x = 1 to x = 2. For $5 \le y \le 10$, the shells extend from x = y/5 to x = 2. It helps to have a sketch of the solid of revolution.