A tricky area

We can parametrize the curve as follows: $\begin{cases} \vec{r}(\theta) = a\cos^5 \theta\hat{i}+a\sin^5 \theta\hat{j}\\ 0 \leq \theta \leq 2\pi \end{cases}$ Now, let's use Green's theorem to find the area enclosed by the curve: $\begin{aligned} A &= \dfrac{1}{2} \oint_C x\mathrm{d}y-y\mathrm{d}x\\ &= \dfrac{1}{2} \int_{0}^{2\pi} \left(\left(a\cos^5 \theta\right)\left(5a\sin^4 \theta \cos \theta \mathrm{d}\theta\right)-\left(a\sin^5 \theta\right)\left(-5a\sin \theta \cos^4 \theta \mathrm{d}\theta\right)\right)\\ &= \dfrac{5a^2}{2} \int_{0}^{2\pi} \left(\sin^4 \theta \cos^6 \theta + \sin^6 \theta \cos^4 \theta\right) \mathrm{d}\theta\\ &= \dfrac{5a^2}{2} \int_{0}^{2\pi} \sin^4 \theta \cos^4 \theta \left(\sin^2 \theta + \cos^2 \theta\right) \mathrm{d}\theta\\ &= \dfrac{5a^2}{2} \int_{0}^{2\pi} \left(\dfrac{\sin 2\theta}{2}\right)^4 \mathrm{d}\theta\\ &= \dfrac{5a^2}{32} \int_{0}^{2\pi} \sin^4 2\theta \mathrm{d}\theta\\ &= \dfrac{5a^2}{32} \int_{0}^{2\pi} \left(\dfrac{1}{8}\cos 8\theta-\dfrac{1}{2}\cos 4\theta+\dfrac{3}{8}\right) \mathrm{d}\theta\\ &= \dfrac{5a^2}{32} \left(\dfrac{1}{64}\sin 8\theta-\dfrac{1}{8}\sin 4\theta+\dfrac{3}{8}\theta\right) \Big|_{0}^{2\pi}\\ &= \dfrac{15}{128} \pi a^2 \end{aligned}$ Finally, $A+B=15+128=143$ , so the answer is $\boxed{8}$ .