这个式子是怎么算出来的？

$$\int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{R} \left( \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} \right) r^3 \mathrm{d}r$$
$$=\int_{0}^{2\pi} \mathrm{d}\theta \left( \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} \right)\int_{0}^{R} r^3 \mathrm{d}r$$
$$=\int_{0}^{2\pi} \mathrm{d}\theta \left( \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} \right)\frac{R^4}{4}$$
$$=\frac{R^4}{4}\int_{0}^{2\pi} \left( \frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2} \right)\mathrm{d}\theta$$
$$=\frac{R^4}{4}\int_{0}^{2\pi} \left( \frac{\cos^2 \theta}{a^2} + \frac{1-\cos^2 \theta}{b^2} \right)\mathrm{d}\theta$$
$$=\frac{R^4}{4}\int_{0}^{2\pi} \left(\cos^2 \theta( \frac{1}{a^2} -\frac{1}{b^2} )+\frac{1}{b^2} \right)\mathrm{d}\theta$$
$$=\frac{R^4}{4}\int_{0}^{2\pi} \left(\frac{1+\cos(2\theta)}{2} ( \frac{1}{a^2} -\frac{1}{b^2} )+\frac{1}{b^2} \right)\mathrm{d}\theta$$
$$=\frac{R^4}{4}\left(\int_{0}^{2\pi} \left(\frac{\cos(2\theta)}{2} ( \frac{1}{a^2} -\frac{1}{b^2} ) \right)\mathrm{d}\theta+\int_{0}^{2\pi} \left(\frac{1}{2} ( \frac{1}{a^2} +\frac{1}{b^2} )
\right)\mathrm{d}\theta\right)$$
$$=\frac{R^4}{4}\left(0+\pi ( \frac{1}{a^2} +\frac{1}{b^2} )
\right)$$
$$=\frac{\pi R^4}{4}\left( \frac{1}{a^2} +\frac{1}{b^2}\right)$$

很基本的计算，应该自己完成。