note [MATH]-2 \le y \le 0[/MATH] for all [MATH]0 \le \theta \le 2\pi[/MATH]
"Hence", may be easier integration using the Cartesian equation ...
[MATH]x\left(\dfrac{\pi}{2}\right) = 0[/MATH][MATH]y\left(\dfrac{\pi}{2}\right) = -2[/MATH]
[MATH]x\left(\dfrac{3\pi}{2}\right) = 0[/MATH][MATH]y\left(\dfrac{3\pi}{2}\right) = 0[/MATH]
[MATH]V = -\pi \int_{-2}^0 y^4+2y^3 \, dy[/MATH]