Consider the solid obtained by rotating the region bounded by y=2x, y=2sqrtx about the line y = 2, how do you find the volume?

1 Answer
Jun 11, 2015

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