I have given myself two parameters; $\theta$, and $\alpha$. (see figure, it is a side view):
The idea is to find an expression for the radius of the circles as $x$ varies on that line (figure), then sum up infinitely many cylinders of infinitesimal thickness.
The radii all have an angle $\theta$ with the $x$-axis, and meet with the line making an angle $\alpha$ with the same axis at $x=b$, thus:
$$r(x)=\sqrt{(b-x)^2+(b\tan\alpha)^2}$$
To find $b$, I need to solve $(\tan\alpha)x-\tan\theta(x-b)=0$, which gives me:
$$b=\frac{\tan\theta-\tan\alpha}{\tan\theta}x$$
If the line upon which I put my $x$-axis has a length $a$, then the volume of the cone is:
$$\pi\int_0^ar^2(x)dx$$
Is the reasoning correct?
I could've also remembered triangles' similarities and notice that $\frac{R}{a}=\frac{r(x)}{x}$, but this gives a result as if $a=h$, where $h$ is the height perpendicular to the biggest circle, and that seems incorrect.
Before commenting on your attempted solution I will give mine for your to go through.
Consider the Fig.1. We are going to calculate the volume of the portion of the oblique circular cone, which lies between the two read lines. The perpendicular distance between these two lines is $\Delta y$. For a small value of $\Delta y$, we can assume that this portion is a circular cylinder having radius $r+\frac{1}{2}\Delta r$ and height $\Delta y$, both of which are functions of $x$. The volume of this thin cylinder is given by $$\Delta V =\pi \left(r+\frac{1}{2}\Delta r\right)^2\Delta y.$$
If we leave out the higher order terms, because they are very small, we are left with, $$\Delta V =\pi r^2\Delta y. \tag{1}$$
Before integrating $\Delta V$ to obtain the volume $V$ of oblique cone, we need to express $r$ in terms of $x$ and $\Delta y$ in terms of $\Delta x$. First, let us apply the $sine\space rule$ to the triangle $ABC$. $$\frac{r\left(x\right)}{\sin\left(\alpha\right)}=\frac{x}{\sin\left(\theta - \alpha\right)} \tag{2}$$
The relationship between $\Delta x$ and $\Delta y$ is given by, $$\Delta y = \Delta x \sin\left(\theta\right) \tag{3}$$
When we substitute the values from equations (2) and (3) in equation (1), we get, $$\Delta V = \pi\frac{\sin^2\left(\alpha\right)\sin\left(\theta\right)}{\sin^2\left(\theta - \alpha\right)}x^2\Delta x.$$
Now the volume of the cone is given by, $$V=\pi\frac{\sin^2\left(\alpha\right)\sin\left(\theta\right)}{\sin^2\left(\theta - \alpha\right)}\int_0^H x^2 dx=\frac{\pi}{3}\frac{\sin^2\left(\alpha\right)\sin\left(\theta\right)}{\sin^2\left(\theta - \alpha\right)} H^3. \tag{4a}$$
If you want to bring the radius $R$ of the base in to the picture, we can use the equation (2) to do that. $$V=\frac{\pi}{3}\sin\left(\theta\right) R^2 H \tag{4b}$$
We can check the formula (4b) by using it to calculate the volume of the right cone shown in Fig. 2. For that, we need to substitute $\theta = 90^0$, i.e. $$V_\mathrm{\space right\space cone}=\frac{\pi}{3}\sin\left(90^0\right) R^2 H=\frac{\pi}{3}R^2H.$$
Your idea of summing up of infinitely many cylinders of infinitesimal thickness is correct. However, I could not understand why you brought a variable or constant named $b$ in to your calculation. From that point onward, I am sorry to say that I cannot follow your reasoning at all. Your equation $$\pi \int_0^a r\left(x\right)^2 dx$$ is also wrong, because $r\left(x\right)$ is not perpendicular to $x$. The relationship between $x$, $a$, and $r$ obtained by using similar triangles is correct though.
$$\underline{\mathrm{The\space following\space text\space was\space added\space as\space a\space reply\space to\space a\space question\space posed\space by\space the\space OP\space in\space his\space comment}}$$ Although we assumed that the thin portion of the cone is cylindrical, the truth is that it is not. If we let the radius of this cylinder to be $r$ or $r+\Delta r$, the calculated value of the volume is either smaller or larger than the actual value. Now, if we use the average of these radii, i.e. $r+\frac{1}{2}\Delta r$, instead, we get a value that lies in between those mentioned above and, hence, a better approximation. This is how things are usually done when you use the basic laws, rules, and principles to work out a problem. However, in the case of calculating the volume of thin portion of a cone, the formula for the volume reduces to that of a cylinder with the radius $r$, once we discard the higher order terms.