# Solid of revolution about other lines

1. Feb 4, 2012

### schapman22

1. The problem statement, all variables and given/known data

Hey we have started solids of revolutions using disk, washer, and shell methods. But I came across a problem i cannot figure out. "The region of the graph of y=x^2 and the x-axis, for 0<x<2, rotated about the line y=4.

2. Relevant equations

Area of a circle = ∏r^2

3. The attempt at a solution

Ive drawn the picture but don't even know how to begin with this one. Thank you in advance for the help.

2. Feb 5, 2012

### tiny-tim

hi schapman22!

they're washers, aren't they? …

vertical washers, all with outer radius 4, and thickness dx

3. Feb 5, 2012

### Deveno

or:

consider that you get the same solid of revolution (in terms of volume) if you rotate the region between y = -4 and y = x2 - 4 about the x-axis.

if you do this both ways, and get the same answer, chances are you're correct.

4. Feb 5, 2012

### schapman22

Thank you TinyTim, I see what your saying but I am still having trouble coming up with the formula for the washer. To my understanding the formula for the washer method is:
V = PI*rout^2*h - PI+rin^2*h
I am still confused on what to use for rout and rin.

5. Feb 5, 2012

### Jet1045

i could be totally wrong, cause i am just learned this in class not to long ago,
but would the outer radius be 4-0 and the inner would be 4-x^2?

6. Feb 5, 2012

### tiny-tim

hi schapman22!

(try using the X2 and X2 buttons just above the Reply box, and the "Quick symbols" to the right )
they are always measured from the axis

the axis here is y = 4, so you go from there to the two curves specified in the question, the x-axis for rout, and y = x2 for rin

7. Feb 5, 2012

### schapman22

Thank you guys! That helped a lot.

8. Feb 5, 2012

### schapman22

basically what I did was V=∏(42)-∏(x2)2
so dV/dx = ∏(42-x4)
∫dV=∏∫02(16-x4)dx
V=[16x-(x5/5]02
so V=25.6∏≈80.4248

Does that look right or am I still off?

9. Feb 5, 2012

### tiny-tim

you rin is wrong

10. Feb 5, 2012

### schapman22

Didn't you say in your last post to use y=x2 for rin?

11. Feb 5, 2012

### tiny-tim

nooo …
(but isn't that obvious anyway from the diagram you've drawn for yourself?)

12. Feb 5, 2012

### schapman22

Oh ok so would that make it 4-x2?

13. Feb 6, 2012

### tiny-tim

(just got up :zzz: …)

yup!

14. Feb 6, 2012

### schapman22

ugh sorry its taking me so long to get this one, but I did that and now i got a negative answer for my volume. because after imntegrating I got ∏[(x5/5) - (8x3/3)] from 0 to 2. Which is -46.9145

15. Feb 6, 2012

### tiny-tim

no, you should have got minus that…

∫ π[(4)2 - (4 - x2)2] dx