Who ate that apple? Dr. Worm ate it.

In order to find the volume of revolution, let us divide the "apple core" into parts by the blue lines $y = \dfrac{\sqrt{3}}{2}$ and $y = \dfrac{-\sqrt{3}}{2}$ .

The volume generated between the blue lines (called $V_{1}$ ) is bounded by the circle graph on the right: $(x - 1.5)^2 + y^2 = 1$ .

Then $x - 1.5 = -\sqrt{1 - y^2}$ (Note that $(0.5 , 0)$ is on the graph, so the sign must be minus.)

Hence, $x^2 = \dfrac{13}{4} - 3\sqrt{1 - y^2} - y^2$ .

Then $V_{1} = \int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \pi(x^2) \ \mathrm{d}y$

= $2 \int_{0}^{\frac{\sqrt{3}}{2}} \pi( \dfrac{13}{4} - 3\sqrt{1 - y^2} - y^2) \ \mathrm{d}y$

= $2\pi[\dfrac{13}{4}y - \dfrac{3}{2}(arcsin(y) + y\sqrt{1 - y^2}) - \frac{y^3}{3}|_{0}^{\frac{\sqrt{3}}{2}}]$

= $2\pi(\dfrac{9}{8}(\sqrt{3}) - \dfrac{\pi}{2})$

Now the volume above the blue line is created from the area bounded between the plus & minus signs of the graph: $(x - 0.5)^2 + y^2 = 1$ .

Thus, the rest of the volume = $2\pi\int_{\frac{\sqrt{3}}{2}}^{1} {(\dfrac{5}{4} + \sqrt{1 - y^2} - y^2) - (\dfrac{5}{4} - 3\sqrt{1 - y^2} - y^2)} \ \mathrm{d}y$

= $2\pi\int_{\frac{\sqrt{3}}{2}}^{1} 2\sqrt{1 - y^2} \ \mathrm{d}y$

= $2\pi[arcsin(y) + y\sqrt{1 - y^2}|_{\frac{\sqrt{3}}{2}}^{1}]$

= $2\pi[\dfrac{\pi}{2} - (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4})]$

Adding the parts together, the total volume = $2\pi[\dfrac{7\sqrt{3}}{8}- \dfrac{\pi}{3} ] \approx 2.943$