It's not a cone, did you lose the parabola? It's a strange rotated displaced parabola (an overturned curved cup with an indentation at the top), with a cone cut out of it at the bottom.
The curves intersect at $x=0$ and $x=3$, so you maximal radius will be $3$. Your volume lies between the strange outer thingy and the inner cone.
You can solve for volumes of surfaces of revolution in more than one way. If you slice the volume into thin disks and integrate over them (best for revolution around $x$ axis, $V=\int \pi y(x)^2\,dx$ where $y(x)$ is the radius of the current disk). However, the method of cylindrical shells works better for revolution around $y$. Instead of thin disks, it sums over cylindrical shells, as suggested by the name:
$$2\pi\int xf(x)\,dx$$
Which uses $f(x)$ as the height of the cylinder and $2\pi x$ as its circumference. This implies the bottom side of the volume sits at $y=0$ (flat bottom on the floor).
If the surface extends between curves $f(x)$ and $g(x)$, the height of the shell in question is the difference $f(x)-g(x)$.
$$2\pi\int x(f(x)-g(x))\,dx$$
Now just plug it in:
$$2\pi \int_0^3 x((3+2x-x^2)-(3-x))\,{\rm d}x$$
Since we need to rotate about the $Y$-axis, we are forming infinitesimal disks which are aligned on the $y$-axis. (Imagine that the $y$-axis is a rod that goes through each center of a thin disk). These disks go from $y=3$ to $y=4$. The $x=0$ is just the $y$-axis.
Geometrically, these thin disks have a radius perpendicular to the $y$-axis and so to find this radius, we can solve $y=\ln{7x}$ as a function of $x$. By doing this, we can use the distance from the $y$-axis to the curve as the radius. (Go ahead and turn your paper counter-clockwise to show the "heights" from the $y$-axis to the function. These "heights" are the radii of your disks.
We solve for $x$ and get:
$$y=\ln{7x}$$
$$e^y=e^{\ln{7x}}$$
$$e^y=7x$$
$$x=\frac{1}{7}e^y$$
We are summing up the volume of all these REALLY thin disks. So we are adding each infinitesimal disk between $3$ and $4$ on the $y$-axis. This is the same as integrating a volume function from $3$ to $4$ of an area function at each point. Since rotating this gives us a circle, $\pi{r^2}$.
Our function $x=\frac17e^y$ is now a function that outputs a radius. (Quite convenient but not a coincidence!)
So we integrate and get $$V = \int_3^4{\pi(\frac17e^y)^2}dy$$
$$V = \frac{\pi}{49}\int_3^4e^{2y}dy$$
$$V= \frac{\pi}{49}\left[\frac12e^{2y}\right]_3^4$$
$$V= \frac{\pi}{98}\left(e^8-e^6\right)\approx82.628$$
Best Answer
For A, you can revolve either the whole semicircle or the half in the first quadrant. The resulting solid is the same. In a sense, if you revolve the whole semicircle by angle $2\pi$ you double cover the hemisphere in that if you calculate the volume swept out you will count everything twice. But I would do this by the disk method.
For part B you revolve the whole thing around $x=4$ you get a shape that is half of a doughnut with the central hole shrunk to zero. Cylindrical shells seem the way to go here.