let r be the first quadrant region bounded by the graphs of $f(x) = \\cos\\left(\\frac{\\pi x}{2}\\right)$…

Explanation:

Step1: Find intersection points

Set $f(x)=g(x)$: $$\cos\left(\frac{\pi x}{2}\right)=(x-1)^2$$ Test $x=0$: $\cos(0)=1$, $(0-1)^2=1$; $x=1$: $\cos\left(\frac{\pi}{2}\right)=0$, $(1-1)^2=0$; $x=2$: $\cos(\pi)=-1$, $(2-1)^2=1$. So bounds are $x=0$ to $x=1$.

Step2: Side length of square

Side length = $f(x)-g(x)$: $$s(x)=\cos\left(\frac{\pi x}{2}\right)-(x-1)^2$$

Step3: Area of cross section

Area of square $A(x)=[s(x)]^2$: $$A(x)=\left[\cos\left(\frac{\pi x}{2}\right)-(x-1)^2\right]^2$$ Expand: $$A(x)=\cos^2\left(\frac{\pi x}{2}\right)-2(x-1)^2\cos\left(\frac{\pi x}{2}\right)+(x-1)^4$$

Step4: Set up volume integral

Volume $V=\int_{0}^{1}A(x)dx$: $$V=\int_{0}^{1}\cos^2\left(\frac{\pi x}{2}\right)dx - 2\int_{0}^{1}(x-1)^2\cos\left(\frac{\pi x}{2}\right)dx + \int_{0}^{1}(x-1)^4dx$$

Step5: Evaluate first integral

Use $\cos^2\theta=\frac{1+\cos2\theta}{2}$, $\theta=\frac{\pi x}{2}$: $$\int_{0}^{1}\frac{1+\cos(\pi x)}{2}dx=\frac{1}{2}\int_{0}^{1}1dx+\frac{1}{2}\int_{0}^{1}\cos(\pi x)dx$$ $$=\frac{1}{2}[x]{0}^{1}+\frac{1}{2}\left[\frac{\sin(\pi x)}{\pi}\right]{0}^{1}=\frac{1}{2}+0=\frac{1}{2}$$

Step6: Evaluate second integral

Let $u=(x-1)^2$, $dv=\cos\left(\frac{\pi x}{2}\right)dx$ $du=2(x-1)dx$, $v=\frac{2}{\pi}\sin\left(\frac{\pi x}{2}\right)$ First integration by parts: $$uv|{0}^{1}-\int{0}^{1}vdu=\frac{2}{\pi}(x-1)^2\sin\left(\frac{\pi x}{2}\right)\bigg|{0}^{1}-\frac{4}{\pi}\int{0}^{1}(x-1)\sin\left(\frac{\pi x}{2}\right)dx$$ First term: $0 - \frac{2}{\pi}(1)\sin(0)=0$ Now let $u=x-1$, $dv=\sin\left(\frac{\pi x}{2}\right)dx$ $du=dx$, $v=-\frac{2}{\pi}\cos\left(\frac{\pi x}{2}\right)$ Second integration by parts: $$-\frac{4}{\pi}\left[-\frac{2}{\pi}(x-1)\cos\left(\frac{\pi x}{2}\right)\bigg|{0}^{1}+\frac{2}{\pi}\int{0}^{1}\cos\left(\frac{\pi x}{2}\right)dx\right]$$ $$=-\frac{4}{\pi}\left[0+\frac{2}{\pi}(1)\cos(0)+\frac{2}{\pi}\cdot\frac{2}{\pi}\sin\left(\frac{\pi x}{2}\right)\bigg|_{0}^{1}\right]$$ $$=-\frac{4}{\pi}\left[\frac{2}{\pi}+\frac{4}{\pi^2}(1-0)\right]=-\frac{8}{\pi^2}-\frac{16}{\pi^3}$$ Multiply by -2: $$-2\left(-\frac{8}{\pi^2}-\frac{16}{\pi^3}\right)=\frac{16}{\pi^2}+\frac{32}{\pi^3}$$

Step7: Evaluate third integral

Let $u=x-1$, $du=dx$: $$\int_{-1}^{0}u^4du=\left[\frac{u^5}{5}\right]_{-1}^{0}=0-\frac{(-1)^5}{5}=\frac{1}{5}$$

Step8: Sum all parts

$$V=\frac{1}{2}+\frac{16}{\pi^2}+\frac{32}{\pi^3}+\frac{1}{5}=\frac{7}{10}+\frac{16}{\pi^2}+\frac{32}{\pi^3}$$

Answer:

$\frac{7}{10}+\frac{16}{\pi^2}+\frac{32}{\pi^3}$