Volume of a surface of revolution with curve in parametric form 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?
 A: 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.
A: 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
$$ dV = \pi a^2 dz\,,V_{positive} = \pi \int_a^b a^2 dz \quad V_{negative}= \pi \int_b^a a^2 dz $$
A: 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
$$V=2\pi RA=2\pi\int\!\!\!\int x~dx~dy=\pi\int x^2~dy=\pi\int x^2\frac{dy}{dt}~dt$$
A: 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
$$V= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi x^2\frac{dy}{dt}dt=\pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\cos^3t)dt= \pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos t \,(1-\sin^2t)dt$$
$$
=\pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos t \, dt - \pi r^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 t \cos t \, dt=\pi r^3 \left [ \sin t -\frac{\sin^3 t}{3} \right ]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
$$
$$
=\pi r^3\left ( 1-\frac{1}{3}-(-1)-\frac{1}{3} \right )=\frac{4}{3} \pi r^3
$$
