Mass of solid inside paraboloid $z=x^2+y^2$ and beneath plane $z=1$ I need to find the mass of the solid inside paraboloid $z=x^2+y^2$ and beneath plane $z=1$ if the density is 1.
Here's what I tried:
$$
M=\int\int_R\int^1_{r^2}r^2dzdA=\int\int_Rr^2z|^1_{r^2}dA=\int\int_R(r^2-r^4)dA=\int^{2\pi}_0\int^1_0(r^2-r^4)rdrd\theta=\int^{2\pi}_0\frac{r^4}{4}-\frac{r^6}{6}|^1_0=\int^{2\pi}_0\frac{1}{12}d\theta=\frac{1}{6}\pi
$$
I have a feeling I'm not correct and if so, where did I go wrong?

To find the mass, the quantity we should be integrating is $1$. You appear to be integrating $r^2$! 
\begin{align*} \int_{\theta = 0}^{\theta = 2\pi} \int_{r = 0}^{r = 1} \int_{z = r^2}^{z = 1} 1\times r dz dr d\theta  =  \int_{\theta = 0}^{\theta = 2\pi} \int_{r = 0}^{r = 1} (r - r^3)dr  d\theta  = \int_{\theta = 0}^{\theta = 2\pi}  \frac 1 4 d\theta = \frac \pi 2. \end{align*}
(The volume element, as always, is $dV = rdrd\theta dz$.)
To double check, we could try changing the order of integration:
$$ \int_{\theta = 0}^{\theta = 2\pi} \int_{z = 0}^{z=1} \int_{r = 0}^{r = \sqrt {z}} 1\times rdr dz d\theta =  \int_{\theta = 0}^{\theta = 2\pi} \int_{z = 0}^{z=1}  \frac z 2 dz d\theta  = \int_{\theta = 0}^{\theta = 2\pi}   \frac 1 4 d\theta = \frac \pi 2 .$$
