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.

Plot of the function in polar/cartesian forms on Desmos


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 $\theta=\pi/4$

  • $\begingroup$ I'm still confused about where the $\left(\frac{1}{\sqrt{2} \cos \theta}\right)^3$ appears from. Is the formula as a wrote it above wrong then? $\endgroup$ – Robert Lee Sep 5 '19 at 12:12
  • $\begingroup$ The formula is only good for integral with lower bound at 0. You have two bounds, with the upper one at x=1/$\sqrt{2}$ $\endgroup$ – Quanto Sep 5 '19 at 12:15
  • $\begingroup$ But the lower bound from the polar form is $0$, right? So the formula should work? Can you tell me what the corrected formula is in general form or point me to a place where I can see the derivation? Again, thank you for your answer! $\endgroup$ – Robert Lee Sep 5 '19 at 12:19
  • 1
    $\begingroup$ Okay I’ll add to the answer shortly $\endgroup$ – Quanto Sep 5 '19 at 12:32
  • 1
    $\begingroup$ correct. you have to work with n curves to figures bounds $\endgroup$ – Quanto Sep 5 '19 at 12:55

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