Ahh mah help.

[RESmonkey]

"Find the volume of the solid generated by revolving the plane region bounded by the equations around y= -2.

y = arccos(x)
y = -(x-1)(x+2)

There are TWO ways to do this problem, and I have to do BOTH.

I already did the DISK method, and confirmed the answer to be 8.81585 ish.

I used x inside my integrand for the disk method, so that leaves me to use Y for the SHELL method.

I can't seem to do the shell method. I have the radius (y+2), but fail at getting the height.

Please help.


cliffs:
Fail at doing SHELL method to find volume.
What would be the height? (think cylinder?)

Thanks

 

[RESmonkey]

bump

 

[OdiN]

I like pi.

 

[RESmonkey]

Just when I thought I would get a somewhat intelligent response...ololll.

 

[SleepWalkerX]

Its been a while and I don't like to think, but don't you just swap the x and y? Then your bounds will be in terms of y.

 

[RESmonkey]

Swap out? You mean taking y = -(x-1)(x+2) and making an X = "" out of it?

 

[hypn0tik]

Draw a picture if you haven't already and visualize what's going on.

 

[DrPizza]

I always tell my students that for the shell method, imagine carving the shape out of a big log (complete with tree rings.) Boil it to separate the rings, then add up the volumes of the rings. In this case, your log will be oriented in the horizontal direction. Each ring, unrolled, is going to be a thin rectangle. The thickness of the rectangular solid is either dx or dy. In your case, it's dy. You probably already knew that, since you have the radius of each ring already: (y+2) The circumference is 2pi(y+2) of each ring.

Look at the graph of these two functions together. (I just hand sketched them; I'm assuming they only intercept at (1,0) in the first quadrant.) Turn your graph 90 degrees counter-clockwise. The separation of these curves is the height of each of the individual tree rings. It'll be the value of x for the top of the ring minus the value of x for the bottom of the ring. In this case, at the top of the ring, x comes from the parabolic function. And at the bottom of the ring, x comes from the inverse trig function. So, it's x minus x. Since it's "dy", simply write each x as a function of y.
y=arccosx becomes x=cosy
y=-(x-1)(x+2) is a minor pain in the ass.

There are several techniques - I'd use either the quadratic formula or complete the square to solve for x.
So, once you get that one as a function of x...
x = some square root function with a y in it
the height of your cylinderical shells = that function minus cosy.

But, be careful! You can only do that from y=0 to y=pi/2. From y=pi/2 to y=2, the top and bottom of your cylindrical shell are both formed from the parabola.

I'm just going on the assumption that the 10 seconds I put into a sketch is correct. Also, I'd consider the cylindrical shell method for this particular problem to be a p.i.t.a.

Do you get to use a graphing calculator for the actual integration? Or do you have to integrate this by hand. If by hand, then it's a huge pita!

 

[RESmonkey]

I just have to write the integral, and then check both integrals with a calc.

Hmmm...I guess I was headed in the right direction I just felt weird trying to solve for x in that messy quadratic.

Thanks.

And BTW, you sound like my teacher, where do you teach?

 

[DrPizza]

western NY. But, don't worry - I'm not your teacher. I didn't assign that problem.
We're not even up to that point yet.

 

[RESmonkey]

Thanks again. I'll solve for x in a little while. I'm trying to figure out how to compare polarities of different molecules and determien which is 'more polar.'

 

[RESmonkey]

I just realized that I misspelled "math" ---> "mah"

Hmmm...