Solid of revolution about other lines

[schapman22]

Homework Statement

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.

Homework Equations

Area of a circle = ∏r^2

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.

 

[tiny-tim]

hi schapman22!

they're washers, aren't they? …

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

 

[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.

 

[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.

 

[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?

 

[tiny-tim]

hi schapman22!

(try using the X² and X₂ buttons just above the Reply box, and the "Quick symbols" to the right )

I am still confused on what to use for rout and rin.

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 r_out, and y = x² for r_in

 

[schapman22]

Thank you guys! That helped a lot.

 

[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?

 

[tiny-tim]

you r_in is wrong

 

[schapman22]

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

 

[tiny-tim]

nooo …

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

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

 

[schapman22]

Oh ok so would that make it 4-x²?

 

[tiny-tim]

(just got up :zzz: …)

yup!

 

[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

 

[tiny-tim]

…after imntegrating I got ∏[(x⁵/5) - (8x³/3)] from 0 to 2.

no, you should have got minus that…

∫ π[(4)² - (4 - x²)²] dx