5

On Wikipedia, I recently stumbled upon a method of obtaining the volume of a solid of revolution generated by a curve in parametric form, which was useful in my case because I had a curve I had trouble representing as an equation of 2 variables. However, when I got strange results (integrating an odd function from $-a$ to $a$, volume should have been nonzero), I tried testing the formula on something simpler: a sphere.

Wikipedia gives a formula for the volume of a solid of revolution generated by taking a curve with $x$ and $y$ given as functions of $t$ and rotating it around the $y$-axis as

$$V=\int_a^b\pi x^2\frac{dy}{dt}dt$$

which I attempted to use on the sphere generated by rotating

$$x^2+y^2=r^2,x>0$$

around the y-axis. To change this to parametric form, I applied a substitution

$$x=r\sin t,y=r\cos t,\frac{dy}{dt}=-r\sin t$$ $$\int_0^\pi\pi x^2\frac{dy}{dt}dt=\pi r^3\int_0^\pi-\sin^3tdt=\pi r^3\int_0^\pi-\sin t(1-\cos^2t)dt$$ $$u=\cos t,du=-\sin tdt$$ $$\pi r^3\int_1^{-1}1-u^2du=\pi r^3(u-\frac{u^3}3]^{-1}_1)=$$ $$\pi r^3(-1+\frac13-1+\frac13)=-\frac43\pi r^3$$

Which is right except for the minus sign. Why did it come out negative? Was it a mistake on my end or a problem with Wikipedia's formula?

Mike
  • 13,318
  • 4
    The volume is the absolute value of this integral. Otherwise you can always change sign by taking the symmetric of your curve, and you get a kind of "algebraic" volume, that is, signed volume. – Jean-Claude Arbaut Dec 18 '13 at 16:16
  • @arbautjc So if I take the absolute value of the formula from Wikipedia, it should be correct for all such curves? – Mike Dec 18 '13 at 16:23
  • 1
    Yes (by the way, you forgot a $\pi$ in your first formula). See also Guldin theorems – Jean-Claude Arbaut Dec 18 '13 at 16:27
  • @arbautjc Oops! Yes, you're correct. I'll fix it. – Mike Dec 18 '13 at 16:28
  • The problem here is that $dy/dt<0$ because of your parametrization. To go from $y=-r$ to $y=r$, you really need $t$ to go from $\pi$ to $0$, or to set $y=-r\cos t$. – Ted Shifrin Dec 18 '13 at 16:32
  • @TedShifrin So $\frac{dy}{dt}$ should remain positive to yield a positive result? – Mike Dec 18 '13 at 16:45
  • 2
    Yes, Mike, to give the right result. :) This formula on Wikipedia is coming from chopping the region up by planes perpendicular to the $y$-axis and integrating cross-sectional areas, i.e., $\int_{y=y_0}^{y_1} (\pi x^2),dy$, and then making the parametric substitution, with $dy = (dy/dt)dt$ and $t$ going from $a$ to $b$. – Ted Shifrin Dec 18 '13 at 16:47
  • @TedShifrin All right, thanks. I think you and arbautjc have given me all I need to know. Of course, this means I probably have some errors to track down with my other curve... – Mike Dec 18 '13 at 17:09
  • 1
    $\frac{dy}{dt}$ don't need to be positive all the time (e.g. you will encounter this when you calculate the volume of a torus ). In general, to use the formula $\int \pi x^2 \frac{dy}{dt} dt$, the direction of the parametrization $t$ need to be chosen such that $\frac{dy}{dt} > 0 / < 0$ when the region bounded by the curve is on the left-hand side/right-hand side of the curve. If you look at the parametrization you use for the sphere, you will find your have chosen a parametrization in the wrong direction. This explains why you get an overall minus side in your result. – achille hui Dec 18 '13 at 17:12
  • U may split the integrals 0-π/2. &π/2-π. Then take absolute values. – Aditya Kumar Dec 29 '15 at 16:10

4 Answers4

1

Yes, last answer is right, since $\cos x$ decreases as your $x$ increases from $0$ to $\pi$, then the curve will start at the top and go down, whereas you generally go outwards.

Eg, Integral from $2$ to $1$ of $x^2$, is the negative of the integral from $1$ to $2$ of $x^2$. I think for the volume of revolution you should go outwards, so as to get a positive volume.

Yiyuan Lee
  • 14,435
colin
  • 11
  • 1
0

It is about sign convention of rotation. If anti-clockwise rotational convention for $t$ is followed the interchanged integral sign limits gets you correct positive sign for volume.

EDIT1:

Independent variable of physical dimension or the parameter that determines it, decides sign. Volume of cylinder $V$ from the simplest consideration

$$ dV = \pi a^2 dz\,,V_{positive} = \pi \int_a^b a^2 dz \quad V_{negative}= \pi \int_b^a a^2 dz $$

Narasimham
  • 40,495
0

There is nothing surprising here. The result follows directly from Pappus's $2^{nd}$ Centroid theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$. The centroid of an area is given by

$$\mathbf{R}=\frac{\int_A \mathbf{r}dA}{\int_A dA}=\frac{1}{A} \int_A \mathbf{r}dA$$

So, in your case we have

$$V=2\pi RA=2\pi\int\!\!\!\int x~dx~dy=\pi\int x^2~dy=\pi\int x^2\frac{dy}{dt}~dt$$

Cye Waldman
  • 7,524
0

Actually the right parametrization of the half circle is $x(t)= r \cos t, y(t)= r \sin t$ with $t \in [-\pi/2, \pi/2]$,then

$$V= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi x^2\frac{dy}{dt}dt=\pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos^3t)dt= \pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos t \,(1-\sin^2t)dt$$ $$ =\pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos t \, dt - \pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 t \cos t \, dt=\pi r^3 \left [ \sin t -\frac{\sin^3 t}{3} \right ]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} $$ $$ =\pi r^3\left ( 1-\frac{1}{3}-(-1)-\frac{1}{3} \right )=\frac{4}{3} \pi r^3 $$

Matheman
  • 551