# Thread: Rotated parametric curve : calculate the volume

1. ## Rotated parametric curve : calculate the volume

I'm unsure of the way to approach the problem : Let $\displaystyle C:[0,\pi]\to \mathbb{R}^3$ be the curve given by $\displaystyle C(t)=(0,\sin t , t)$. Calculate the volume enclosed by the curve if we rotate $\displaystyle C$ with respect to the $\displaystyle z$ axis.
I think a way would be to determine the solid and then evaluate the volume via a triple/double or even simple integral.
I wonder if there's an easier way to solve the problem.
Otherwise I'm not sure what solid the rotated curve represent. An ellipsoid?

2. Originally Posted by arbolis
I'm unsure of the way to approach the problem : Let $\displaystyle C:[0,\pi]\to \mathbb{R}^3$ be the curve given by $\displaystyle C(t)=(0,\sin t , t)$. Calculate the volume enclosed by the curve if we rotate $\displaystyle C$ with respect to the $\displaystyle z$ axis.
I think a way would be to determine the solid and then evaluate the volume via a triple/double or even simple integral.
I wonder if there's an easier way to solve the problem.
Otherwise I'm not sure what solid the rotated curve represent. An ellipsoid?
Hey, ain't that $\displaystyle \pi r^2 dx$ long-ways? I mean up the z-axis. The parametric representation you gave is just a sine wave along the y-z axis in the form of $\displaystyle y=\sin(z)$. So takin' disks, that's just $\displaystyle \pi \sin^2(z) dz$ from 0 to $\displaystyle \pi$. Think so anyway.

3. Thanks! Ok so if I understood well, $\displaystyle V=\pi \int_0^{\pi} \sin ^2 (z)dz = \frac{\pi ^2}{2}$.