I was analyzing the polar function $r = \tan(\theta)$, trying to find the volume that is generated by rotating about the $x$-axis with $\theta \in \left[0, \frac{\pi}{4}\right]$.
I transformed it into cartesian coordinates and got the function
$$y = \frac{x^2}{\sqrt{1-x^2}}$$
which is correctly defined to be equal to the polar curve on the domain I'm interested in. By further checking, I determined that the same segment in cartesian coordinates is $x \in \left[0, \frac{1}{\sqrt{2}}\right] $.
Using the volume formulas for a polar solid of revolution around $x$-axis and the disc method for Cartesian, I have following two expressions for the volume:
$$\frac{2\pi}{3} \int_{\theta_0}^{\theta_1} r^3 \sin(\theta) \ d\theta = V = \pi \int_{x_0}^{x_1} \left[f(x)\right]^2 dx$$
Inputting the bounds I previously got, I ended up with:
$$ \frac{2\pi}{3}\int_0^\frac{\pi}{4}[\tan(\theta)]^3 \sin(\theta) \ d\theta = V = \pi \int_{0}^{\frac{1}{\sqrt{2}}} \left[\frac{x^2}{\sqrt{1-x^2}}\right]^2 dx$$
The problem is that after I evaluated them on WolframAlpha I got different results:
$$\frac{2\pi}{3}\int_0^\frac{\pi}{4}[\tan(\theta)]^3 \sin(\theta) \ d\theta \approx 0.1930 $$ $$ \pi \int_{0}^{\frac{1}{\sqrt{2}}} \left[\frac{x^2}{\sqrt{1-x^2}}\right]^2 dx \approx 0.1772$$
I'm not sure, in what part of my analysis. I made a mistake. I hope someone can help me find it.
There is a problem in your polar expression of the volume. Two bounds for $r$ need to considered. The correct integral should be
$$ \frac{2\pi}{3}\int_0^\frac{\pi}{4}\left[ \left( \frac{1}{\sqrt{2}\cos\theta}\right)^3 -\tan^3 \theta \right] \sin \theta \ d\theta $$
A general polar volume formula to use is
$$ \frac{2\pi}{3}\int_0^\frac{\pi}{4}\left[ r_2^3(\theta) - r_1^3(\theta)\right] \sin \theta \ d\theta $$
where the volume is bounded between the two curves $r_1(\theta)$ and $r_2(\theta)$. In your case, the volume is sandwiched between
$$r_1= \tan\theta$$ $$r_2 \cos \theta = \frac{1}{\sqrt{2}}$$
In contrast, the volume you calculated with
$$ \frac{2\pi}{3}\int_0^\frac{\pi}{4}r_1^3(\theta) \sin \theta \ d\theta $$
is sandwiched between $r=\tan\theta$ and $\theta=\pi/4$