Find the volume under the paraboloid $z=x^2+y^2$ within the cylinder $x^2+y^2\le1$, $z\ge0$ .
Could someone tell me if I got it right or wrong? Here is my attempt in solving that problem.
Initial conversions:
$$z=x^2+y^2$$
$$z=r^2$$
$$x^2+y^2\le1$$
$$r^2\le1$$
$$r\le1$$
The double integration part:
$$V=\int\int z dA$$
$$V=\int^{2\pi}_{0}\int^{1}_{0}r^2rdrd\theta$$
$$V=\int^{2\pi}_{0}\int^{1}_{0}r^3drd\theta$$
$$V=\int^{2\pi}_{0}(\frac{r^4}{4})\bigg|^{1}_{0} d\theta$$
$$V=\int^{2\pi}_{0}\frac{1}{4}d\theta$$
$$V=\frac{\theta}{4}\bigg|^{2\pi}_{0}$$
$$V=\frac{\pi}{2}$$
There's an obvious dimensional inconsistency in your calculations. Suppose we change the problem to $x^2+y^2\le a^2$ for some length $a>0$, so you'd be saying $V=\int_0^{2\pi}\int_0^ar^3drd\theta$, which would have the dimension of $a^4$, not $a^3$. Note that the infinitesimally thick ring of radius $r$ has area $2\pi rdr$, with a volume $2\pi rz_\max dr=2\pi r^2dr$ above it. So $V=\int_0^12\pi r^2dr=\frac{2\pi}{3}$. Again, if we introduce a length scale $a$, we can check whether this works.