Paraboloid Volume

The volume is $V \;=\; \iint_{x^2 + 4y^2 \le x + 4} \Big[1 + \tfrac14x - \big(\tfrac14x^2 + y^2\big)\Big]\,dxdy \; = \; \iint_{(x-\frac12)^2 + 4y^2 \le \frac{17}{4}} \Big[\tfrac{17}{16} - \tfrac14(x - \tfrac12)^2 - y^2\Big]\,dxdy$ Parametrizing the region of integration by $x = \tfrac12 + r\cos\theta \qquad y \; = \; \tfrac12r\sin\theta \qquad \qquad 0 \le r \le \frac{\sqrt{17}}{2}\,,\,0 \le \theta \le 2\pi$ we see that $V \; = \; \tfrac12 \int_0^{2\pi} d\theta \int_0^{\frac{\sqrt{17}}{2}} \left(\tfrac{17}{16} - \tfrac14r^2\right)\,r\,dr \; = \; \frac{289}{256}\pi$ making the answer $17+16 = \boxed{33}$ .