Use the substitution $x=1+\sin u$ to find $\int\sqrt{2x-x^{2}}\,dx$ . The finite region $R$, which includes the point (0,1), is bounded by the curves $y=|x|^{3}+|x|$ and $y=|x|+2-\sqrt{2}|x|-x^{2}$. Find the exact area of $R$. the numerical value of the volume of revolution formed when $R$ is rotated completely about the x-axis, giving your answer to 3 decimal places.

1. Find the differential $dx$ in terms of $du$.

Given the substitution $x = 1 + \sin u$, differentiating with respect to $u$ gives:
$\frac{dx}{du} = \cos u \implies dx = \cos u \, du$

2. Substitute $x$ and $dx$ into the integral and simplify the expression under the square root.

$\int\sqrt{2x-x^{2}}\,dx = \int\sqrt{2(1+\sin u)-(1+\sin u)^{2}}\,(\cos u \, du)$
Expanding and simplifying the terms under the square root:
$\sqrt{2+2\sin u - (1 + 2\sin u + \sin^2 u)} = \sqrt{1 - \sin^2 u}$
Using the Pythagorean identity $1 - \sin^2 u = \cos^2 u$:
$\int\sqrt{\cos^2 u}\,(\cos u)\,du = \int \cos^2 u \, du$

3. Evaluate the trigonometric integral using the double-angle formula.

Using the double-angle identity $\cos^2 u = \frac{1}{2}(1+\cos 2u)$:
\begin{aligned}
\int\cos^{2}u,du &= \int\frac{1}{2}(1+\cos 2u),du \
&= \frac{1}{2}u + \frac{1}{4}\sin 2u + C \
&= \frac{1}{2}u + \frac{1}{2}\sin u \cos u + C
\end{aligned}

4. Substitute $u$ back in terms of $x$ to obtain the final answer.

From $x = 1 + \sin u$, we have $\sin u = x - 1$, which gives $u = \sin^{-1}(x-1)$.
Also, using the identity $\cos u = \sqrt{1-\sin^2 u}$:
$\cos u = \sqrt{1-(x-1)^2} = \sqrt{2x-x^2}$
Substitute these expressions back:
\begin{aligned}
\frac{1}{2}u + \frac{1}{2}\sin u \cos u + C &= \frac{1}{2}\sin^{-1}(x-1) + \frac{1}{2}(x-1)\sqrt{2x-x^2} + C
\end{aligned}

1. Use symmetry and find the intersection points / identify upper and lower curves

Both curves depend only on $|x|$ and $x^2$, so they are symmetric about the $y$-axis. For $x\ge 0$, $|x|=x$, so the curves are
$y = x^3 + x \quad\text{(lower)}, \qquad y = x + 2 - \sqrt{2x - x^2}\quad\text{(upper)}.$
Setting them equal:
$x^3 + x = x + 2 - \sqrt{2x-x^2} \;\Rightarrow\; x^3 = 2 - \sqrt{2x-x^2}.$
At $x=1$: $1 = 2-\sqrt{2-1}=2-1=1$ ✓. So the curves meet at $x=\pm 1$. At $x=0$ the upper curve gives $y=2$ and the lower gives $y=0$, consistent with $(0,1)$ lying inside $R$.

2. Set up the area integral exploiting symmetry

By symmetry, the area is twice the area for $x\in[0,1]$:
$A = 2\int_0^1\Big[(x+2-\sqrt{2x-x^2}) - (x^3+x)\Big]\,dx = 2\int_0^1\big(2 - x^3 - \sqrt{2x-x^2}\big)\,dx.$

3. Use part (i) to evaluate the surd integral exactly

From part (i), with $x=1+\sin u$:
$\int\sqrt{2x-x^2}\,dx = \tfrac12\arcsin(x-1) + \tfrac12(x-1)\sqrt{2x-x^2} + C.$
Evaluating from $0$ to $1$:
$\Big[\tfrac12\arcsin(x-1)+\tfrac12(x-1)\sqrt{2x-x^2}\Big]_0^1 = (0+0) - \Big(\tfrac12\!\cdot\!(-\tfrac{\pi}{2}) + 0\Big) = \frac{\pi}{4}.$

4. Combine the integrals for the exact area

$\int_0^1 2\,dx = 2,\qquad \int_0^1 x^3\,dx = \tfrac14,\qquad \int_0^1\sqrt{2x-x^2}\,dx = \tfrac{\pi}{4}.$
$A = 2\left(2 - \tfrac14 - \tfrac{\pi}{4}\right) = 2\left(\tfrac{7}{4} - \tfrac{\pi}{4}\right) = \frac{7-\pi}{2}.$

1. Set up the integral for the volume of revolution.

Using the intersection points $x = -1$ and $x = 1$ from part (a), the volume of revolution about the $x$-axis is given by:
$V = \pi \int_{-1}^{1} \left( y_{\text{upper}}^2 - y_{\text{lower}}^2 \right) \, dx$
Substituting the equations of the curves $y = |x| + 2 - \sqrt{2|x| - x^2}$ (correcting the apparent typographical error in the question stem using context from part (i)) and $y = |x|^3 + |x|$, and noting the symmetry of both even functions about the $y$-axis:
$V = 2\pi \int_{0}^{1} \left[ \left( x + 2 - \sqrt{2x - x^2} \right)^2 - \left( x^3 + x \right)^2 \right] \, dx$

2. Use a graphing calculator to evaluate the numerical value of the volume.

Using a graphing calculator to numerically evaluate the definite integral:
$\int_{0}^{1} \left[ \left( x + 2 - \sqrt{2x - x^2} \right)^2 - \left( x^3 + x \right)^2 \right] \, dx \approx 2.078087$

$\implies V \approx 2\pi (2.078087) \approx 13.057007$
$V = 13.057 \text{ (to 3 d.p.)}$