The solid R is bounded by a paraboloid $$ z = x^2 + y^2 $$ from below and bounded above by the cone $$ z = \sqrt{x^2 + y^2} $$
How do I write the integral in terms of spherical coordinates? So far I've figured that $$ 0 \leq \theta \leq 2\pi$$ but I'm stuck on how to calculate the bounds for $\rho$ and $\phi$.
The solid $R$ is what you get rotating around the $z$-axis (the vertical axis from the picture below) the region (in blue, in the picture below) bounded by the lines $z=x$ and $z=x^2$ ($0\leqslant x\leqslant1$). So, $\phi$ can take values from $\frac\pi4$ to $\frac\pi2$. For each such $\phi$, the line $z=\cot(\phi)x$ intersects the line $z=x^2$ when $x^2=\cot(\phi)x$, which means that $x=0$ or that $x=\cot(\phi)$. So, $\rho$ can take values from $0$ to $\cot(\phi)$.
So, if $f\colon R\longrightarrow\Bbb R$ is a continuous function, then\begin{multline}\iiint_Rf(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=\\=\int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_0^{\cot\phi}f(\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi)\rho^2\sin\phi\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta.\end{multline}