find the volume of the solid figure generated by rotating the area of the region bounded by the graphs (y =…

Explanation:

Step1: Find intersection points

Set $2x^{2}+2 = 3 - 2x$. Rearrange to $2x^{2}+2x - 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b = 2$, $c=-1$, we have $x=\frac{-2\pm\sqrt{4 + 8}}{4}=\frac{-2\pm2\sqrt{3}}{4}=\frac{-1\pm\sqrt{3}}{2}$.

Step2: Use disk - washer method

The volume $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer - radius and $r(x)$ is the inner - radius. Here, $R(x)=3 - 2x$ and $r(x)=2x^{2}+2$, and $a=\frac{-1-\sqrt{3}}{2}$, $b=\frac{-1 + \sqrt{3}}{2}$. $V=\pi\int_{\frac{-1-\sqrt{3}}{2}}^{\frac{-1+\sqrt{3}}{2}}((3 - 2x)^{2}-(2x^{2}+2)^{2})dx$. Expand the integrand: $(3 - 2x)^{2}=9-12x + 4x^{2}$ and $(2x^{2}+2)^{2}=4x^{4}+8x^{2}+4$. So, $(3 - 2x)^{2}-(2x^{2}+2)^{2}=9-12x + 4x^{2}-(4x^{4}+8x^{2}+4)=-4x^{4}-4x^{2}-12x + 5$.

Step3: Integrate

$\int(-4x^{4}-4x^{2}-12x + 5)dx=-\frac{4}{5}x^{5}-\frac{4}{3}x^{3}-6x^{2}+5x+C$. Evaluate the definite integral: $V=\pi\left[-\frac{4}{5}x^{5}-\frac{4}{3}x^{3}-6x^{2}+5x\right]_{\frac{-1-\sqrt{3}}{2}}^{\frac{-1+\sqrt{3}}{2}}$. After substituting the upper and lower limits and simplifying: [ \begin{align*} V&=\pi\left[\left(-\frac{4}{5}\left(\frac{-1 + \sqrt{3}}{2}\right)^{5}-\frac{4}{3}\left(\frac{-1+\sqrt{3}}{2}\right)^{3}-6\left(\frac{-1+\sqrt{3}}{2}\right)^{2}+5\left(\frac{-1+\sqrt{3}}{2}\right)\right)-\left(-\frac{4}{5}\left(\frac{-1-\sqrt{3}}{2}\right)^{5}-\frac{4}{3}\left(\frac{-1-\sqrt{3}}{2}\right)^{3}-6\left(\frac{-1-\sqrt{3}}{2}\right)^{2}+5\left(\frac{-1-\sqrt{3}}{2}\right)\right)\right]\ &=\frac{8\sqrt{3}\pi}{5} \end{align*} ]

Answer:

$\frac{8\sqrt{3}\pi}{5}$