# Volume of polar solid of revolution vs cartesian solid of revolution.

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 $$\theta=\pi/4$$

• 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? – Robert Lee Sep 5 '19 at 12:12
• The formula is only good for integral with lower bound at 0. You have two bounds, with the upper one at x=1/$\sqrt{2}$ – Quanto Sep 5 '19 at 12:15
• 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! – Robert Lee Sep 5 '19 at 12:19
• Okay I’ll add to the answer shortly – Quanto Sep 5 '19 at 12:32
• correct. you have to work with n curves to figures bounds – Quanto Sep 5 '19 at 12:55