find the total area of the shaded region shown to the right given by the curve ( y = 6(sin x)sqrt{1+cos x}…

Explanation:

Step1: Set up the integral

The area (A) under a curve (y = f(x)) from (x = a) to (x = b) is given by (A=\int_{a}^{b}|y|dx). Here, (y = 6\sin x\sqrt{1+\cos x}), (a =-\pi), (b = 0). Since (y\leqslant0) on the interval ([-\pi,0]), (|y|=- 6\sin x\sqrt{1+\cos x}). So, (A=\int_{-\pi}^{0}-6\sin x\sqrt{1 + \cos x}dx).

Step2: Use substitution

Let (u = 1+\cos x), then (du=-\sin xdx). When (x =-\pi), (u=1+\cos(-\pi)=1 - 1=0). When (x = 0), (u=1+\cos(0)=2). The integral becomes (A = 6\int_{0}^{2}\sqrt{u}du).

Step3: Integrate (\sqrt{u})

We know that (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)). For (n=\frac{1}{2}), (\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}+C). So, (6\int_{0}^{2}\sqrt{u}du=6\times\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{0}^{2}).

Step4: Evaluate the definite - integral

(6\times\frac{2}{3}\left[u^{\frac{3}{2}}\right]_{0}^{2}=4\left(2^{\frac{3}{2}}-0^{\frac{3}{2}}\right)). Since (2^{\frac{3}{2}}=\sqrt{2^{3}} = 2\sqrt{2}), then (4\times2\sqrt{2}=8\sqrt{2}).

Answer:

(8\sqrt{2})