let r be the region bounded by the curves x = 2 - y² and x = y⁴. what is the volume of the solid generated…

Explanation:

Step1: Find intersection points

Set $2 - y^{2}=y^{4}$, rewrite as $y^{4}+y^{2}-2 = 0$. Let $u = y^{2}$, then $u^{2}+u - 2=0$. Factoring gives $(u + 2)(u - 1)=0$. Since $u=y^{2}\geq0$, we take $u = 1$, so $y=-1$ and $y = 1$.

Step2: Use disk - washer method

The volume $V$ of the solid of revolution about the $y$-axis using the formula $V=\pi\int_{a}^{b}([R(y)]^{2}-[r(y)]^{2})dy$. Here, $R(y)=2 - y^{2}$ and $r(y)=y^{4}$, $a=-1$, $b = 1$. [ \begin{align*} V&=\pi\int_{-1}^{1}((2 - y^{2})^{2}-(y^{4})^{2})dy\ &=\pi\int_{-1}^{1}(4-4y^{2}+y^{4}-y^{8})dy \end{align*} ]

Step3: Integrate term - by - term

Since the integrand is an even function, we can use $\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx$. [ \begin{align*} V&=2\pi\int_{0}^{1}(4-4y^{2}+y^{4}-y^{8})dy\ &=2\pi\left[4y-\frac{4}{3}y^{3}+\frac{1}{5}y^{5}-\frac{1}{9}y^{9}\right]_{0}^{1}\ &=2\pi\left(4-\frac{4}{3}+\frac{1}{5}-\frac{1}{9}\right)\ &=2\pi\left(\frac{540 - 180+ 54 - 10}{135}\right)\ &=2\pi\times\frac{404}{135}\ &=\frac{808\pi}{135} \end{align*} ]

Answer:

$\frac{808\pi}{135}$