discs and washers are the same. Washers are just discs with holes in them. Since x is difficult to express as a function of y, we want to integrate using dx.
Since we're revolving around a horizontal line, using dx means working with discs, since the thickness of each disc will be dx. Each disc will have a hole in it.
v = ∫[0,3] π(R^2-r^2) dx
where R = (5-x)+1 and r = (x^2-4x+5)+1
So,
v = π∫[0,3] ((5-x)+1)^2-((x^2-4x+5)+1)^2) dx
= π∫[0,3] -x^4 + 8x^3 - 27x^2 + 36x
= 162/5 π
If you like pain, you could use shells, integrating on dy. This is awkward, since the parabola has two branches. Its vertex is at (2,1). So, we have to break the integrand up into two parts, one where the height of the shells is the distance between the two sides of the parabola, and the rest where the height is the distance between the curve and the line.
y=5-x means x = 5-y
y=x^2- 4x+ 5 means x = (4±√(16-4(5-y)))/2 = 2±√(y-1)
v = ∫ 2πrh dy in each part, making it
v = 2π∫[1,2] (y+1)((2+√(y-1))-(2-√(y-1))) dy
+ 2π∫[2,5] (y+1)((5-y)-(2-√(y-1))) dy
= 2π(52/15 + 191/15)
= 162/5 π
whew
So I'm trying to solve this problem but one thing I don't get is, how do I know which method to use just by looking at the graph (disk, washer, or shell)?
Find the volume of the solid generated by revolving the region bounded by the graphs y=x^2- 4x+ 5 and y= 5-x about the line y=-1.
I found the points of intersection between the parabola and the y=5-x, and I get x=3 and x=0. So my a will be 0, and my b will be 3. And on this interval 5-x has greater y values than those of the parabola, so I do: (5-x)-(x^2-4x+5) and I get 3x-x^2. How do I proceed though? I think I get all the steps UP to this point, which is choosing which method to use. Thanks!
4 answers
Wow, thank you so much! But I'm wondering where the + 1 for the R and r, comes into play?
rotated around y = -1, not y=0. So the radius of each disc/shell is y+1, not just y.
Oh, so that value is to be added. Okay, thank you!
1 answer