Calculate the volume of the solid of revolution generated by revolving the region bounded by the parabolas $y^2=2(x−3)$ and $y^2=x$ about $y=0$.
I was given this problem. The reason why I am feeling stuck is because the area bounded by these parabolas extends both above and below the x-axis. The limits of integration seem to be $0$ and $6$ - zero is the farthest point to the left on the x-axis and 6 is the point of intersection of the parabolas. Does it matter that the function extends below the x-axis as well?
If not, I would set up my problem like this: $$\pi\int^6_0x-2(x-3)\ dx$$
My outer radius is x because $$y^2=x\\y=\sqrt{x}\\r^2\to x$$ I got $2(x-3)$ as the inner radius pretty much the same way.
This would integrate to: $$\pi\left(\frac{x^2}2-x^2+3x\right)|^6_0$$ Doing the algebra gets zero.
Obviously, the volume is not zero. I must be doing something wrong. Any ideas?
I can see how it can be a bit confusing that the given region lies both above and below the $x$-axis, about which we're supposed to revolve it. The question does make sense, though. And if we imagine actually revolving it, we will see that — thanks to the symmetry of the region — the solid we get is the same as if we revolve the upper half (lying above the $x$-axis) only.
Next, if you decide to use this method (as opposed to the method of cylindrical shells, which I feel may be a little easier here), then you did make a mistake in how you set it up in your post. Look at the picture (where I still shaded the entire region, not just the upper half of it):
The second curve $y^2=2(x-3)$ does NOT extend all the way to the left to $x=0$. In other words, you do NOT have an inner radius of $r=\sqrt{x-3}$ between $x=0$ and $x=3$. So your setup with an inner and outer radius is only correct between $x=3$ and $x=6$, but NOT between $x=0$ and $x=3$. In that part, there's only one radius, since there's no hole in the middle.
Thus, to correct your solution, you need to split the integral into two parts: $$\text{Volume}=\int_0^3\text{something}\cdot dx+\int_3^6\text{something}\cdot dx.$$ What you did in your post goes into the second integral, but there's gonna be something different (and more simple) in the first one.