回復 25# arend 的帖子
或是真得很想用積分的話,如下~
令 \(x+1=2\cos\theta\),其中 \(\displaystyle0\leq\theta\leq\frac{\pi}{3}\)
則 \(dx=-2\sin\theta d\theta\),且
\(\displaystyle 6\int_0^1\sqrt{4-(x+1)^2}dx = 6\int_{\pi/3}^0 2\sin\theta\cdot(-2\sin\theta)d\theta\)
\(\displaystyle =-24\int_{\pi/3}^0 \sin^2\theta d\theta\)
\(\displaystyle =-24\int_{\pi/3}^0 \frac{1-\cos2\theta}{2} d\theta\)
\(\displaystyle =-12\int_{\pi/3}^0 (1-\cos2\theta) d\theta\)
\(\displaystyle =-12\left[\theta-\frac{\sin 2\theta}{2}\right]\Bigg|_{\pi/3}^0\)
\(\displaystyle =4\pi-3\sqrt{3}\)