let $r$ be the region bounded above by the graph of $y = 1 - x^{2}$ and below by the graph of $y = x^{2}-1$…

let $r$ be the region bounded above by the graph of $y = 1 - x^{2}$ and below by the graph of $y = x^{2}-1$, for $-1leq xleq1$, as shaded in the figure above. what is the volume of the solid generated when region $r$ is revolved about the horizontal line $y = 3$?

let $r$ be the region bounded above by the graph of $y = 1 - x^{2}$ and below by the graph of $y = x^{2}-1$, for $-1leq xleq1$, as shaded in the figure above. what is the volume of the solid generated when region $r$ is revolved about the horizontal line $y = 3$?

Answer

Explanation:

Step1: Use the washer - method formula

The formula for the volume $V$ of a solid of revolution about a horizontal line $y = k$ using the washer - method is $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer radius and $r(x)$ is the inner radius. Here, $a=-1$, $b = 1$, $k = 3$, $R(x)=3-(x^{2}-1)$ and $r(x)=3-(1 - x^{2})$.

Step2: Simplify the radii

$R(x)=3-(x^{2}-1)=4 - x^{2}$ and $r(x)=3-(1 - x^{2})=2 + x^{2}$.

Step3: Expand the integrand

$[R(x)]^{2}-[r(x)]^{2}=(4 - x^{2})^{2}-(2 + x^{2})^{2}$. Using the formula $(A - B)^{2}=A^{2}-2AB + B^{2}$ and $(A + B)^{2}=A^{2}+2AB + B^{2}$, we have: [ \begin{align*} (4 - x^{2})^{2}-(2 + x^{2})^{2}&=(16-8x^{2}+x^{4})-(4 + 4x^{2}+x^{4})\ &=16-8x^{2}+x^{4}-4 - 4x^{2}-x^{4}\ &=12-12x^{2} \end{align*} ]

Step4: Calculate the integral

$V=\pi\int_{-1}^{1}(12 - 12x^{2})dx$. Since $\int_{-1}^{1}(12 - 12x^{2})dx=\int_{-1}^{1}12dx-\int_{-1}^{1}12x^{2}dx$. We know that $\int_{-1}^{1}12dx=12x\big|{-1}^{1}=12(1-( - 1)) = 24$ and $\int{-1}^{1}12x^{2}dx=12\times\frac{x^{3}}{3}\big|{-1}^{1}=4(x^{3})\big|{-1}^{1}=4(1-( - 1)) = 8$. So $\int_{-1}^{1}(12 - 12x^{2})dx=24 - 8=16$. Then $V = 16\pi$.

Answer:

$16\pi$