I'm not sure about your setup, but in any case you should be integrating in a direction perpendicular to the $r$ in your cross section.
I'll use a setup that's hopefully what you had in mind: the tip of the cone is at the origin, and the cone is lying on its side. Suppose the radius of the base is $r$ (this is the $y$ value at the end of the triangle you're rotating) and the height is $h$ (this is the $x$ value at the end of the triangle).
Then, consider a circular cross section at some point $x$. The radius is the corresponding $y$ value, which by similar triangles is $\frac{rx}{h}$. So, the area is
$$
A(x) = \frac{\pi r^2}{h^2} x^2,
$$
and the volume is
$$
V = \int_0^h \frac{\pi r^2}{h^2} x^2\ dx = \frac{\pi r^2 h^3}{3h^2} = \frac{1}{3} \pi r^2h.
$$
Note the cone lies on its side, so the $x$ values we integrate over range from $0$ to the "height" of the cone, $h$.
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first
$$
\begin{aligned}
J_1
&=
\int_0^\rho
dt
\int_0^\pi
\frac
{t^2\sin\phi\cdot t\cos\phi}
{\left[t^2\sin^2\phi+\left(t\cos\phi-D\right)^2\right]^{3/2}}\; d\phi
\\
&=
\int_0^\rho
dt
\int_0^\pi
\frac
{t^2(-\cos\phi)'\cdot t\cos\phi}
{\left[t^2-2Dt\cos\phi+D^2\right]^{3/2}}\; d\phi
\\
&\qquad\text{ Substitution: }u=\cos \phi\ ,
\\
&=
\int_0^\rho
dt
\int_{-1}^1
\frac
{t^3\; u}
{\left[t^2-2Dt\;u+D^2\right]^{3/2}}\; du
\\
&\qquad\text{ Substitution (for $u$, fixed $t$) of the radical }v=\sqrt{t^2-2Dt\;u+D^2}\ ,
\\
&\qquad u=\frac 1{2Dt}(t^2+D^2-v^2)\ ,\ du=-\frac v{Dt}\; dv\\ ,
\\
&=
-
\int_0^\rho
dt
\int_{\sqrt{t^2+2Dt+D^2}}^{\sqrt{t^2-2Dt+D^2}}
\frac
{t^3\; \frac 1{2Dt}(t^2+D^2-v^2)}
{v^3}\; \frac v{Dt}\; dv
\\
&=
\int_0^\rho
t\;dt
\int_{D-t}^{D+t}
\frac 1{2D^2}
\cdot
\frac {t^2+D^2-v^2}
{v^2}\; dv
\\
&=
\int_0^\rho
t\;dt
\;\frac 1{2D^2}
\left[
-(t^2+D^2)\frac 1v
-1
\right]_{v=D-t}^{v=D+t}
\\
&=
\int_0^\rho
dt
\;\frac t{2D^2}
\left[
(t^2+D^2)\left(\frac 1{D-t}-\frac 1{D+t}\right)
-
2t
\right]
\\
&=
\int_0^\rho
dt
\left[
\frac D{D+t}
+\frac D{D-t}
-2\frac{D^2+t^2}{D^2}
\right]
\\
&=
D\ln\frac {D+t}{D-t}
-
2\rho\left(1+\frac {\rho^2}{3D^2}\right)
\ .
\end{aligned}
$$
Computer check for $D=2$, $\rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$
\begin{aligned}
J_2
&=
-D
\int_0^\pi
d\phi
\int_0^\rho
\frac
{t^2}
{\left[(t-D\cos\phi)^2+D\sin^2\phi\right]^{3/2}}
\; dt
\\
&\qquad\text{ and we consider separately (without the factor $-D$)}
\\
J_2(\phi)
&=
\int_0^\rho
\frac
{t^2}
{\left[(t-D\cos\phi)^2+D\sin^2\phi\right]^{3/2}}
\; dt
\\
&=
\int_{0-D\cos\phi}^{\rho-D\cos\phi}
\frac
{(u+D\cos\phi)^2}
{(u^2+a^2)^{3/2}}
\; du\ ,\qquad a:= D\sin\phi
\ .
\\
&\qquad
\text{ Now the integrals can be computed}
\\
\int \frac{u^2}
{(u^2+a^2)^{3/2}}
\; dt
&=
-\frac t{(u^2+a^2)^{1/2}}+\operatorname{arcsinh} \frac ta+C\ ,
\\
\int \frac{u}
{(u^2+a^2)^{3/2}}
\; dt
&=
-\frac 1{(u^2+a^2)^{1/2}}+C\ ,
\\
\int \frac{1}
{(u^2+a^2)^{3/2}}
\; dt
&=
-\frac {a^2\;u}{(u^2+a^2)^{1/2}}+C\ ,
\end{aligned}
$$
and the computation goes on.
If my calculus is ok, then
$$
\begin{aligned}
J_2(\phi)
&=
\int_0^\pi
d\phi\;
\Bigg[
\operatorname{arcsinh} \frac{t-D\cos \phi}{D\sin\phi}
\\&\qquad\qquad\qquad+
\frac{t-D\cos\phi}{(t^2-2Dt\cos\phi+D^2)^{1/2}\sin^2\phi}
\\&\qquad\qquad\qquad\qquad\qquad\qquad
+\frac2{(t^2-2Dt\cos\phi+D^2)^{1/2}}
\Bigg]_0^\rho\ .
\end{aligned}
$$
I have to submit, hope this is helpful to check with the own computations.
I'll be back, but typing kills a lot of time.
Best Answer
Sure there is. I think the possible confusion might come from thinking that the functions are defined in terms of what we usually use as the dependant variable.
Since we are calculating the area it doesn't matter if we do any rotation to them. How about considering the area defined by:
$x \in [-1,1]$ and bounded by the curves $f_1(x) = x^2 -2$ and $f_2(x) = e^x$ Then, since $f_2 > f_1$ in the given interval, the area would be:
$$\int_{-1}^{1} e^x dx - \int_{-1}^{1}( x^2 -2 )dx$$
EDIT: Corrected a couple of typos.