Finding the volume of an object by rotating an area about an axis?

Question

I am given an area bounded by the functions y= 3rdrt(x), y=2, and x=0.

Rotate the area around the line y=2.

I can do this rotating around either the x or y axis, but im not sure how to set this one up. I don't necessarily need the answer, I just am unsure how to get started.

Answer 1

Integrating over the interval [0,8] we have

∫πr²dx = ∫π(2 - x^⅓)²dx = ∫π{4 - 4x^⅓ + x^⅔}dx
= π{4x - 3x^(4/3) + (3/5)x^(5/3)} = π(32 - 3*16 + (3/5)*32)
= π(32 - 48 + 96/5) = π(-16 + 96/5) = π(-80 + 96)/5 = 16π/5

Answer 2

you already know that the quantity of a cylinder is pi * r^2 * h. the mindset of rings tells you to handle the axis, subsequently the x-axis, because of the fact the h, and the y value because of the fact the r. whilst y = x/3 V = pi * crucial from P to Q of (x/3)^2 V = pi * crucial from P to Q of (x^2/9) V = pi * [x^3/27] evaluated at P and Q V = pi * [3^3/27 - a million^3/27] V = pi * 26/27 which will supply you the quantity purely of the forged defined by skill of the curve y = x/3. So do the comparable element for the quantity defined by skill of y = -x + 4 and subtract it, a lot as you're able to for calculating the quantity of a hollowed out cylinder.

Answer 3

well the equation is going to start with the integral of 2*pi* x * the area
in this case it is x because it is opposite of what you are rotating around which happens to be a y line.
the area would be 2 - 3rdrt(x) in the interval [0,8] because that is where y=2 and y=3rdrt(x) intersect
the final equation would be the integral of 2*pi*x*(2-3rdrt(x)) dx over the interval [0,8]

I think the other way you mean is if you did it in terms of y and that would get you
x = y^3
but since you are still rotating around an x line
you start with the integral of 2*pi*x*(y^3) dy over the interval of [0,2]
you want it in terms of y so substitute y^3 in for x
and it is the integral of 2*pi*(y^3)*(y^3) dy over the interval of [0,2]

I think that is right but i would say go with the first way