volume by the disc method

[Jaina]

Here's the problem (direct wording from the book):

Find the volume of the solid generated by revolving the region about the given line.
The region in the first quadrant bounded above by the line y=2, below by the curve y=2sinx, 0≤x≤(п/2), and on the left by the y-axis, about the line y=2.

I drew a picture and this is a pretty trippy looking graph--kind of like an apple core in the cartoons. The only thing I don't know is how the fact that the axis of rotation is y=2 changes the equation. I know that normally, I would use this:

(integral from 0 to п/2) (2sinx)² dx

and go from there. I also know that the fact that it is y=2 changes it somehow, I just don't know how.

I have a couple of similar problems that use the washer method later, and I'm stuck on those too. What do I do with that weird axis?

Oh, and this thing: п is pi, but it doesn't really look like it...

 

[soroban]

Hello, Jaina!

Find the volume of the solid generated by revolving about the line \(\displaystyle y\,=\,2\)
the region in the first quadrant bounded above by the line \(\displaystyle y\,=\,2\),
below by the curve \(\displaystyle y\,=\,2\cdot\sin x,\;0\,\leq\,x\,\leq\,\frac{\pi}{2}\)

          |
        2 +-------* 
          |   *                The solid would be half
          | *                   of that "apple core".
          |*      
          |       
      - - * - - - + -
          0      π/2

We will use "washers": . \(\displaystyle \L V\;=\;\pi\int^{\;\;\;b}_a(\text{radius})^2\,dx\)

For this problem, the radius is: .\(\displaystyle 2\,-\,2\cdot\sin x\)

 

[Jaina]

Okay, how did you get the 2? Is it just whatever the axis is minus the original equation?

edit: And should the answer be π²?

 

[wjm11]

Okay, how did you get the 2? Is it just whatever the axis is minus the original equation?

edit: And should the answer be π²

In *this* case it is the axis minus the function value, because that's what the r of your "washer" is.

No, the answer is not pi^2. It's

pi[(-2)((sinx - 4)*cosx - 3x)] from 0 to pi/2
= 3pi - 8.

(If I entered it correctly into my "magic" box! Please check the solution.)