Could someone help me out with calculating this integral.
$\int_{-a}^a \sqrt{a^2-t^2}dt$
Where $a>0$.
Could someone help me out with calculating this integral.
$\int_{-a}^a \sqrt{a^2-t^2}dt$
Where $a>0$.
$ \int_{-a}^a dt \sqrt{a^2-t^2} = -a \int_{\pi}^0 d\theta \ \sin \theta \sqrt{a^2-\left(a \cos \theta\right)^2} = a^2 \int_{0}^{\pi} d\theta \ \sin^2 \theta $ You can finish it. $\sin^2 \theta = \left[1 - \cos\left(2\theta\right)\right]/2$ should help.