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#