If you imagine this, it's like a cone that has been somewhat hollowed out, with a round lip.
Strange as it is, since the opening of the cone faces the left, I want to actually flip it horizontally. Also, I want to shift it down so that it is rotated around the x-axis (#y = 0#). Now, we are graphing according to the equation #y = -2sqrt(1-x) + 2# instead of #y = 2sqrtx#.
It was flipped vertically (#2sqrtx -> -2sqrtx#), horizontally (#x -> -x#), shifted right 1 (#x -> x - 1 -> 1-x#), and up 2 (#-2sqrt(1-x) -> -2sqrt(1-x) + 2#).
#y = 2x# stays the same. The shape did not change, however---only in direction. It went from this to this.
Alright, now that it looks nicer, we can integrate from #0# to #1#. You could have done it as it was, but I think better when the first circle starts out at a radius of #0#. Since we are layering circles, we would use a variant on volume:
#V = A(x)*h = pi(r(x))^2*x = int_0^1pi[r(x)]^2dx#
where the horizontal height #x# is represented by the accumulation of "circular" layers from #x = 0# to #x = 1#. #r(x)# is a special function that alters how the revolved shape is made.
We are using the idea that #pir^2# implies a constant radius revolved around an axis going through the center of a soon-to-be-formed circle. Instead of a constant radius, it will be varying according to the two equations, #y = 2x# and #y = -2sqrt(1-x) + 2#. We will be acquiring the area enclosed below #y = 2x# and above #y = -2sqrt(1-x) + 2#, and then revolving that around the x-axis. Therefore, we need to subtract the two:
#y = 2x - (-2sqrt(1-x) + 2) = 2x + 2sqrt(1-x) - 2#
Using the above equation in the volume formula:
#int_0^1 pi[r(x)]^2dx#
#= piint_0^1(2x + 2sqrt(1-x) - 2)^2dx#
Expansion:
#= piint_0^1 4x^2 - 12x + 8 + 8xsqrt(1-x)-8sqrt(1-x) dx#
Splitting into separate integrals:
#= pi[int_0^1 4x^2dx - int_0^1 12xdx + int_0^1 8 dx + int_0^1 8xsqrt(1-x)dx-int_0^1 8sqrt(1-x)dx]#
Evaluating some of them, and leaving some harder ones for next step:
#= pi[4/3x^3 - 6x^2 + 8x]|_(0)^(1)#
#+ pi [int_0^1 8xsqrt(1-x)dx-int_0^1 8sqrt(1-x)dx]#
These last two need some extra thinking.
Let's try integration by parts on the first one.
Let:
#u = 8x#
#dv = sqrt(1-x)dx#
#du = 8dx#
#v = -2/3(1-x)^(3/2)#
#= uv - intvdu#
#= -(16x)/3(1-x)^(3/2) + 16/3int(1-x)^(3/2)dx#
Evaluating the integral here:
#= pi{-(16x)/3(1-x)^(3/2) + 16/3[-2/5(1-x)^(5/2)]}|_(0)^(1)#
Now the second one is not so bad. Just a variant of the integral right above:
#8int_0^1 sqrt(1-x)dx = [-16/3(1-x)^(3/2)]|_(0)^(1)#
Overall:
# = pi[4/3x^3 - 6x^2 + 8x + {-(16x)/3(1-x)^(3/2) + [-32/15(1-x)^(5/2)]} - {-(16)/3(1-x)^(3/2)}]|_(0)^(1)#
Distributing the negative signs (parentheses are important!):
# = pi[4/3x^3 - 6x^2 + 8x - (16x)/3(1-x)^(3/2) - 32/15(1-x)^(5/2) + (16)/3(1-x)^(3/2)]|_(0)^(1)#
Plugging in #1# and #0# and doing #F(1) - F(0)#, with #pi# still factored out:
# = pi[(4/3 - 6 + 8cancel( - (16)/3(1-1)^(3/2) - 32/15(1-1)^(5/2) + (16)/3(1-1)^(3/2))) - (cancel(4/3*0 - 6*0 + 8*0 - (16*0)/3(1-0)^(3/2)) - 32/15(1-0)^(5/2) + (16)/3(1-0)^(3/2))]#
Home free from here:
# = pi[4/3 - 6 + 8 + 32/15 - 16/3]#
# = pi[20/15 - 90/15 + 120/15 + 32/15 - 80/15]#
# =(2pi)/15 = 0.1bar33pi ~~ 0.4189 "u"^3#
