find the volume of the solid of revolution formed by rotating the region bounded by y = x², y = 2 - x, and x…

Explanation:

Step1: Find intersection points

Set $x^{2}=2 - x$. Rearranging gives $x^{2}+x - 2=0$. Factoring, we have $(x + 2)(x - 1)=0$. So the intersection points are $x=-2$ and $x = 1$. Since we also have $x = 0$ as a boundary, we consider the interval $x\in[0,1]$.

Step2: Apply the shell - method formula

The shell - method formula for rotating about the vertical line $x = a$ is $V=2\pi\int_{c}^{d}(a - x)h(x)dx$, where $a = 2$, $c = 0$, $d = 1$, and $h(x)=(2 - x)-x^{2}=2 - x - x^{2}$. So $V = 2\pi\int_{0}^{1}(2 - x)(2 - x - x^{2})dx$.

Step3: Expand the integrand

$(2 - x)(2 - x - x^{2})=4-2x-2x^{2}-2x+x^{2}+x^{3}=4 - 4x - x^{2}+x^{3}$.

Step4: Integrate term - by - term

$\int(4 - 4x - x^{2}+x^{3})dx=4x-4\times\frac{x^{2}}{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}+C=4x - 2x^{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}+C$.

Step5: Evaluate the definite integral

$V = 2\pi\left[4x - 2x^{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}\right]_{0}^{1}$. $V = 2\pi\left(4\times1-2\times1^{2}-\frac{1^{3}}{3}+\frac{1^{4}}{4}\right)$. $V = 2\pi\left(4 - 2-\frac{1}{3}+\frac{1}{4}\right)$. $V = 2\pi\left(2-\frac{4 - 3}{12}\right)$. $V = 2\pi\left(2-\frac{1}{12}\right)$. $V = 2\pi\times\frac{23}{12}=\frac{23\pi}{6}\approx12.043$.

Answer:

$12.043$