Washer or Disk Method ? - # 2

[Jason76]

\(\displaystyle y = 5 e^{-x}, y = 5, x = 6\) about \(\displaystyle y = 10\) find the volume (since it's \(\displaystyle y = 10\) we know to revolve around the x axis)

Find limits of integration:

\(\displaystyle 5e^{-x} = 5\)

\(\displaystyle e^{-x} = 1\)

\(\displaystyle \ln(e^{-x}) = \ln(1)\)

\(\displaystyle -x = 0\)

\(\displaystyle x = 0\)

Limits of integration are \(\displaystyle x = 0\) lower bound and \(\displaystyle x = 6\) upper bound

How do we know whether to use the disk or washer method? Can we tell by the \(\displaystyle 0\) being an intersection. But in this case, we aren't revolving around \(\displaystyle y = 0\), but rather, \(\displaystyle y = 10\)

 

[HallsofIvy]

\(\displaystyle y = 5 e^{-x}, y = 5, x = 6\) about \(\displaystyle y = 10\) find the volume (since it's \(\displaystyle y = 10\) we know to revolve around the x axis)

That's a very strange remark. You are told to revolve around the line y= 10, NOT the x-axis, y= 0. Did you mean "revolve around an axis parallel to the x- axis"? So that, using the washer or disk method, you will be integrating with respect to x?

Find limits of integration:

\(\displaystyle 5e^{-x} = 5\)

\(\displaystyle e^{-x} = 1\)

\(\displaystyle \ln(e^{-x}) = \ln(1)\)

\(\displaystyle -x = 0\)

\(\displaystyle x = 0\)

You should have been able to see that x= 0 is a boundary without all that!

Limits of integration are \(\displaystyle x = 0\) lower bound and \(\displaystyle x = 6\) upper bound

Yes, that is correct.

How do we know whether to use the disk or washer method? Can we tell by the \(\displaystyle 0\) being an intersection. But in this case, we aren't revolving around \(\displaystyle y = 0\), but rather, \(\displaystyle y = 10\)

Have you given this any thought? And "Can we tell by the \(\displaystyle 0\) being an intersection" is meaningless. What \(\displaystyle 0\) are you talking about?
You can tell by the fact that the upper bound on the plane figure is y= 5 while we are rotating about the axis y= 10. There will be part of the graph from \(\displaystyle y= 5e^{-x}\) up to y= 5 but then a hole from y= 5 to y= 10. What do we call a disk with a hole in the middle?

(We could do this as two disk calculations. First do the entire disk given by take y from \(\displaystyle 5e^{-x}\) up to 10, then subtract the disk from 5 to 10. The latter part is easy- since the lower bound is the horizontal line y= 5, the volume of rotation is just a cylinder with radius r= 10- 5= 5 and height h= 6- 0= 6 so volume \(\displaystyle \pi r^2 h= 150\pi\).)

 

[Jason76]

Again it's all about observation (Does it touch the axis of rotation?), and not calculation, when deciding whether to use the "washer" or "disk" method.