find the volume when the area under the curve x^2y=v3 between x=1 and x=2 is rotated about?

Question

the x axis through 180 degrees.

Answer 1

What is the equation?
Let's assume it is y=x^2

Then

A disc is being generated by the radius which is equal to f(x) from x=1 to x=2.
The radius follows a f(x) so the volume becomes dV=pi(r)^2dx

So we have to integrate Pi (f(x))^2dx from x=1 to x=2.
If y=x^2

Then V=1/3(x^3) x=1 to x=2

V=1/3((3)^3-(1)^3)=1/3(27-1)=26/3 units cubed.

Answer 2

You'll probably want to start by making this explicit in y:
y=v³/x²
This gives you the radius of your solid at x. To find the area of the semicircle, take πr²/2 (since we only rotated it through 180° - normally we would be rotating through 360° and would simply use πr²). This is:
A(x) = πv^6/(2x^4).
Now integrate this function with respect to x to find the formula for the volume:
V = ∫πv^6/(2x^4) dx = πv^6/2∫x^(-4) dx = -πv^6/(6x³)
Then evaluate this from 1 to 2:
-πv^6/6*(1/8-1) = (7/48)πv^6

Note that I assume that by x^2y you mean x²y and not x^(2y). If you meant the latter:
2y ln x=ln v³
y=ln v³/ln x²
Then A(x) = π(ln v³)²/(2(ln x²)²)
Again, the integral of A(x) dx from 1 to 2 gives you your answer. Finding an antiderivative of this function is somewhat tedious, and is thus left as an exercise for the reader (there IS a closed form antiderivative here, though. If you get stuck, try looking at the table of integrals of logarithmic functions on wikipedia).

Answer 3

y = x^2
y = 3
x = 1
x = 2

V = integral(1,sqrt(3), 0.5 * pi * x^4 * dx) + integral(sqrt(3),2,0.5 * pi * 9 * dx)
V = [0.5*pi*(x^5)/5]{1,sqrt(3)} + [0.5*pi*(9x)]{sqrt(3),2}
V = 0.5*pi * (9*sqrt(3) - (1/5) + 18 - 9*sqrt(3))
V = 0.5*pi * (17.8)
V = 27.960174616949159822317526111188 cubic units

I think...