Calculus Integral Questins

The volume of the solid defined by the region below the Paraboloid $z = 9 ?x^2 ?y^2$ and above the unit circle on the xy plane is given by
$$V=\int\int_RzdA$$
where R is the region on the xy-plane governed by the unit circle which in polar coordinates can be expressed as
$$R=\{(r,\theta) | 0\le r\le 1,0\le\theta\le2\pi\}$$ where $r^2=x^2+y^2$ and $\theta=\frac yx$
We know that the Jacobian of the area element in integration is given by
$$dA=dxdy=rdrd\theta$$
Thus, our integral reduces to
$$\displaystyle V=\int\int_RzdA=\int\limits_0^{2\pi}\int\limits_0^1(9-r^2)rdrd\theta$$
$$\displaystyle V=\int\limits_0^{2\pi}\int\limits_0^1(9r-r^3)drd\theta$$
Since the limits are constants we can split the integral as
$$\displaystyle V=\color{red}{\int\limits_0^{2\pi}d\theta }\color{blue}{\int\limits_0^1(9r-r^3)dr}$$
$$\displaystyle V=\color{red}{[\theta]_0^{2\pi} }\color{blue}{\left[ \frac92 r^2-\frac{r^4}{4} \right]_0^1}$$
$$\displaystyle V=\color{red}{[2\pi-0] }\cdot \color{blue}{\left[ \frac92 -\frac{1}{4} \right]}$$
$$\displaystyle V=\color{red}{[2\pi] }\cdot \color{blue}{\left[ \frac{17}{4} \right]}$$
$$\displaystyle V=\boxed{\frac{17\pi}{2}}$$