I'm having trouble with this question:
"Find the area bounded by the parabola $y=2x-x^2$, the y-axis and the line y=1 is rotated about the x-axis. Find the volume"
I assume we start off by finding the value of $y^2$, which is $(2x-x^2)^2$, then the bounds (which are found by substituting y=1, and the y-axis which is x=0). The bounds, I think, are x=1 and x=0.
You then find the integral, and multiply it by pi.
Is what I'm doing right?
Thanks
Best Answer
Right you are!
(See the nice graph created by Wolfram Alpha, below.)
So assuming we are seeking the volume of the region between the parabola and the x-axis, rotated about the x-axis, we have the following integral:
$$V = \int_0^1 \pi\Big((2x-x^2)^2 - 0\Big)\,dx = \pi \int_0^1 (4x^2 - 4x^3 + x^4)\,dx$$
On the other hand, if we are interested in the region between the line $y = 1$ and the parabola rotated about the x-axis, then we have the following integral:
$$V = \int_0^1 \pi\Big(1- (2x-x^2)^2\Big)\,dx = \pi \int_0^1 (1-4x^2 + 4x^3 - x^4)\,dx$$